Major exam 1 revision test by Abul Lais

KFUPM PYP001 Old Exams Revision for Major Exam 1, 172

Instructions • These are the solutions for Physical Science KFUPM old exams • It covers Chapter 2.1 – 4.3, almost all of major exam 1 (except 14.1) • For each question, first try the question yourself, then click next to see the answer, after that the solutions follow • Try to read and understand the solutions even if you get the answers correct, as there is a lot of exciting physics concepts and theories explained clearly within the solutions. • After this is the contents page, just click to go to the chapter you want • Good luck!!!

Table of Contents 2.1 2.2 2.3 3.1 3.2 3.3 4.1 4.2 4.3

Describing Motion Velocity and Momentum Acceleration Forces Newtonâ&#x20AC;&#x2122;s Laws of Motion Using Newtonâ&#x20AC;&#x2122;s Laws Work and Machines Describing Energy Conservation of Energy

2.1 Describing Motion

Graph questions Travelling-type questions Displacement questions

General

Find positions

Find displacement

3, 4

Find instantaneous speed 5, 6 Find average speed

7-9

Find average speed

10 – 14

Find total time

15 – 16

Definitions

17 – 18

Calculating

2.1 Describing Motion 1. Which of the following graphs represents a motion with constant (nonzero) speed? A) I B) III

C) II D) IV (152 Major 1, Q1)

2.1 Describing Motion 1. Which of the following graphs represents a motion with constant (nonzero) speed? A) I B) III

C) II D) IV (152 Major 1, Q1)

Solutions • Look carefully at the axes • (I) and (II) are position-time graphs • but (III) and (IV) are speed-time graphs • Slope of position-time graphs = speed or velocity • Slope of speed or velocity-time graph = acceleration • So constant (nonzero) speed means: (i) position-time graphs with constant, nonzero slope or (ii) speed or velocity-time graph with zero slope

2.1 Describing Motion 2. The position-time graph shown in the figure below represents the motion of a ball being rolled back and forth by two children. At what positions are the two children sitting? A) Both are sitting at 0.0 m. B) At 0.0 m and 6.0 m. C) At –6.0 m and 6.0 m. D) At – 6.0 m and 0.0 m.

E) Both are sitting at –6.0 m.

(171 Final, Q5)

E) Both are sitting at –6.0 m.

(171 Final, Q5)

Solutions • The position of the ball goes from –6.0 m (0 s) to 6.0 m (4 s) and back to –6.0 m (20 s) • Hence the two children must be sitting at –6.0 m & 6.0 m

2.1 Describing Motion 3. The figure below describes the position of an object as a function of time. How much is the displacement of the object in 8.0 seconds? A) 3.0 m.

B) 8.0 m. C) 2.0 m.

D) 5.8 m. (171 Major 1, Q1)

2.1 Describing Motion 3. The figure below describes the position of an object as a function of time. How much is the displacement of the object in 8.0 seconds? A) 3.0 m.

B) 8.0 m. C) 2.0 m.

D) 5.8 m. (171 Major 1, Q1)

Solutions

• Displacement = net change in position from start to end • Displacement is not determined by total distance or actual path travelled • Object goes from 0 m (start) to 5 m (middle) and back to 3 m (end)

• Displacement = end position – start position = 3 m – 0 m = 3 m

2.1 Describing Motion 4. The figure below represents the position of an abject as a function of time, what is the displacement of the object after 5 seconds? A) 6.0 m east. B) 5.0 m west. C) 1.0 m west. D) Zero (162 Major 1, Q2)

Solutions • Displacement = net change in position from start to end

• Displacement is not determined by total distance or actual path travelled • Object goes from 0 m (start) to 5 m (middle) and back to 3 m (end) • Displacement = end position – start position = 3 m – 0 m = 3 m

2.1 Describing Motion 5. The motion of a swimmer during 30.0 minutes workout is represented by the graph below. What is the speed of the swimmer at 2:15 PM? A) Zero.

B) 40.0 m/min. C) 120 m/min.

D) 279 m/min. (161 Major 1, Q1)

2.1 Describing Motion 5. The motion of a swimmer during 30.0 minutes workout is represented by the graph below. What is the speed of the swimmer at 2:15 PM? A) Zero.

B) 40.0 m/min. C) 120 m/min.

D) 279 m/min. (161 Major 1, Q1)

Solutions â&#x20AC;˘ speed at a certain time (2:25 PM in this question) actually means instantaneous speed = Slope of position-time graphs â&#x2C6;&#x2020;đ???đ??˘đ??Źđ??đ??&#x161;đ??§đ??&#x153;đ??&#x17E; final distance â&#x2C6;&#x2019; start distance đ??Źđ??Ľđ??¨đ??Šđ??&#x17E; = = = â&#x2C6;&#x2020;đ??đ??˘đ??Śđ??&#x17E; final time â&#x2C6;&#x2019; start time (600 m â&#x20AC;&#x201C; 600 m)/(2:20 PM â&#x20AC;&#x201C; 2:10 PM) = 0 m /10 min = 0 m/min

2.1 Describing Motion 6. Figure (1) represents the motion of a swimmer during a 30-minute workout. What is the speed of the swimmer at 2:25 PM A) 80 m/min. B) 40 m/min. C) 47 m/min. D) 27 m/min. E) Zero.

(142 Major 1, Q1)

(1400 m â&#x20AC;&#x201C; 600 m)/(2:30 PM â&#x20AC;&#x201C; 2:20 PM) = 800 m /10 min = 80 m/min

2.1 Describing Motion 7. The figure below describes the position of an object as a function of time. What is the average speed of the object in 5 seconds? A) 0.8 m/s

B) Zero C) 0.6 m/s

D) 0.4 m/s (162 Major 1, Q1)

2.1 Describing Motion 7. The figure below describes the position of an object as a function of time. What is the average speed of the object in 5 seconds? A) 0.8 m/s

B) Zero C) 0.6 m/s

D) 0.4 m/s (162 Major 1, Q1)

Solutions • Average speed = Total distance/total time • Total distance does not depend on direction of travel (whether north, south, east, west), just add all distances for each part of travel from vertical (distance) axis:

|1 – 3| m + |1 – 1| m + |3 – 1| m = 4 m • Total time: read from horizontal (time) axis:

5s–0s=5s • Average speed = 4 m/5 s = 0.8 m/s

2.1 Describing Motion 8. The figure below represents the position-time graph for the motion of a vehicle during a 5.0-minute trip. What is the average speed of the vehicle? A) 72 km/h.

B) 1.2 km/h C) 24 km/h

D) 120 km/h (152 Final, Q1)

2.1 Describing Motion 8. The figure below represents the position-time graph for the motion of a vehicle during a 5.0-minute trip. What is the average speed of the vehicle? A) 72 km/h.

B) 1.2 km/h C) 24 km/h

D) 120 km/h (152 Final, Q1)

|4 – 0| km + |2 – 4| km = 6 km • Total time: read from horizontal (time) axis:

5 min – 0 min = 5 min = (5/60) hr • Average speed = 6 km/(5/60) hr = 72 km/h

2.1 Describing Motion 9. The position-time graph of an objectâ&#x20AC;&#x2122;s motion is represented as in the figure below. What is the average speed of this object during the 100-second motion? A) 0.40 m/s B) Zero

C) 0.20 m/s D) 0.50 m/s E) 1.0 m/s (151 Major 1, Q2)

|20 – 0| m + |20 – 20| m + |10 – 20| m + |0 – 10| m = 40 m • Total time: read from horizontal (time) axis:

100 s – 0 s = 100 s • Average speed = 40 m/100 s

2.1 Describing Motion 10. A bus travels 280 km south along a straight path with an average speed of 80.0 km/h. The bus stops for 30 min, then it travels 210 km north with an average speed of 75.0 km/h. What is the average speed for the total trip? A) 72.0 km/h. B) 10.3 km/h. C) 77.5 km/h. D) 5.0 km/h, South. E) 77.7 km/h

(171 Major 1, Q2)

Solutions • Average speed = Total distance/total time • Total distance does not depend on direction of travel (whether north, south, east, west), just add all distances • Total time includes travelling time and resting time

• Total distance = 280 km + 210 km = 490 km • Total time = (280/80) hr + (30/60) hr + (210/75) hr = 6.8 hrs • Average speed = 490 km/6.8 hrs = 72.0 km/h

2.1 Describing Motion 11. A man drives to a town, 120 km away, at an average speed of 60 km/h. He returns at an average speed of 40 km/h. What is his average speed for the entire trip? A) 35 km/h

B) 50 km/h C) 65 km/h

D) 48 km/h (161 Major 1, Q4)

2.1 Describing Motion 11. A man drives to a town, 120 km away, at an average speed of 60 km/h. He returns at an average speed of 40 km/h. What is his average speed for the entire trip? A) 35 km/h

B) 50 km/h C) 65 km/h

D) 48 km/h (161 Major 1, Q4)

Solutions • Average speed = Total distance/total time • Total distance does not depend on direction of travel (whether north, south, east, west), just add all distances • Total distance = 120 km + 120 km = 240 km

• Total time = (120/60) hr + (120/40) hr = 5 hrs • Average speed = 240 km/5 hrs = 48 km/h

2.1 Describing Motion 12. A man jogs for 2.0 km at 8.0 km/h and then walks for 3.0 km at 6.0 km/h. What is his average speed for the entire trip? A) 14 km/h

B) 6.0 km/h C) 6.7 km/h

D) 3.8 km/h (152 Major 1, Q4)

2.1 Describing Motion 12. A man jogs for 2.0 km at 8.0 km/h and then walks for 3.0 km at 6.0 km/h. What is his average speed for the entire trip? A) 14 km/h

B) 6.0 km/h C) 6.7 km/h

D) 3.8 km/h (152 Major 1, Q4)

• Total time = (2/8) hr + (3/6) hr = 0.75 hrs • Average speed = 5 km/0.75 hrs = 6.7 km/h

2.1 Describing Motion 13. A car traveled 50 km east in 30 minutes, stopped at a gas station for 15 minutes, then traveled again 100 km north in 45 minutes. What is the average speed of the car? A) 120 km/h. B) 100 km/h.

C) 75 km/h. D) 150 km/h.

E) 50 km/h. (142 Major 1, Q4)

C) 75 km/h. D) 150 km/h.

E) 50 km/h. (142 Major 1, Q4)

• Total time = (30/60) hr + (15/60) hr + (45/60) hr = 1.5 hrs • Average speed = 150 km/1.5 hrs = 100 km/h

2.2 Velocity and Momentum 14. A man drives to a town, 120 km away, at an average speed of 60 km/h. He returns at an average speed of 40 km/h. What is his average speed for the entire trip? A) 35 km/h

B) 50 km/h C) 65 km/h

D) 48 km/h (161 Major 1, Q4)

2.2 Velocity and Momentum 14. A man drives to a town, 120 km away, at an average speed of 60 km/h. He returns at an average speed of 40 km/h. What is his average speed for the entire trip? A) 35 km/h

B) 50 km/h C) 65 km/h

D) 48 km/h (161 Major 1, Q4)

• Total time = (120/60) hr + (120/40) hr = 5 hrs • Average speed = 240 km/5 hrs = 48 km/h

2.1 Describing Motion 15. A cyclist must travel 800 km. How many days will the trip take if the cyclist travel 5.00 h/day at an average speed of 16.0 km/h? A) 5.00 days B) 10.0 days C) 50.0 days D) 160 days (162 Final, Q26)

Solutions • Average speed = 16 km/h x 5 h/day = 80 km/day • total time = Total distance/Average speed = (800/80) days = 10 days

2.1 Describing Motion 16. A cyclist must travel 800 km. How many days will the trip take if the cyclist travels 10 h/day at an average speed of 16 km/h? A) 5.0 days. B) 50 days.

C) 80 days. D) 8.0 days.

E) 0.5 days. (142 Final, Q22)

2.1 Describing Motion 16. A cyclist must travel 800 km. How many days will the trip take if the cyclist travels 10 h/day at an average speed of 16 km/h? A) 5.0 days. B) 50 days.

C) 80 days. D) 8.0 days.

E) 0.5 days. (142 Final, Q22)

Solutions • Average speed = 16 km/h x 10 h/day = 160 km/day • total time = Total distance/Average speed = (800/160) days = 5 days

2.1 Describing Motion 17. Which one of the following is always true about the magnitude of a displacement?

A) It is greater than the distance traveled. B) It is equal to the distance traveled.

C) It is less than the distance traveled. D) It is less than or equal to the distance traveled. E) None of these statements are always true. (171 Final, Q3)

2.1 Describing Motion 17. Which one of the following is always true about the magnitude of a displacement?

A) It is greater than the distance traveled. B) It is equal to the distance traveled.

C) It is less than the distance traveled. D) It is less than or equal to the distance traveled. E) None of these statements are always true. (171 Final, Q3)

Solutions • Total displacement is always less than or equal to total distance travelled

• i.e. in Mathematical form we write: displacement ≤ distance • If object move in straight line, displacement = distance • If object moves in a closed loop, and starts and finishes at the same point, displacement = 0 • So these are the 2 extreme cases: closed loop (displacement = 0) and straight line (displacement = distance) • for other cases, in general, displacement < distance

2.1 Describing Motion 18. Which one of the following is always true about the magnitude of a displacement? A) It is less than or equal to the distance traveled. B) It is greater than the distance traveled. C) It is equal to the distance traveled. D) It is less than the distance traveled. (161 Major 1, Q2)

Solutions • Total displacement is always less than or equal to total distance travelled

2.1 Describing Motion 19. A cyclist moves 5 km east, 3 km south, then 5 km west. What is the displacement of the cyclist? A) 3 km, north. B) 3 km, south.

C) 13 km, south. D) 13 km, north.

E) Zero. (142 Major 1, Q3)

2.1 Describing Motion 19. A cyclist moves 5 km east, 3 km south, then 5 km west. What is the displacement of the cyclist? A) 3 km, north. B) 3 km, south.

C) 13 km, south. D) 13 km, north.

E) Zero. (142 Major 1, Q3)

Solutions 5 km east

3 km south

Displacement = 3 km south

5 km west

• Just draw the 3 paths “head to tail” like above • then find the line joining the start and end point • That is the answer for displacement •

2.2 Velocity and Momentum

2.2 Velocity and Momentum 1. A hurricane is approaching a house at 90 km/h due west as shown in the figure below. The car is moving at 80 km/h away from the house due west. How long does it take the hurricane to strike the car if the distance between them is 70 km and their velocities remain constant? A) 0.77 hour. B) 7.0 hours. C) 0.41 hour.

D) It will never strike the car. (171 Major 1, Q6)

Solutions • Velocity of hurricane relative to car = velocity of hurricane – velocity of car = 90 km/h (left) – 80 km/h (left) = 10 km/h (left) • Speed = distance/time,

• so time = distance/speed = (70/10) hrs = 7 hrs • Note: Here both car and hurricane are travelling in the same direction (both arrows point left), so we let left be positive, and write both velocities as positive numbers

2.2 Velocity and Momentum 2. Two cars are driving toward each other on a straight and level road. One car is traveling at 100 km/h north and the other car is traveling at 42.0 km/h south, both velocities measured relative to the road. At a certain instant, the distance between the cars is 10.0 km. How long will take for the two cars to meet? A) 6.00 minutes. B) 10.3 minutes. C) 4.23 minutes.

D) 0.07 minutes. (162 Major 1, Q6)

Solutions • Velocity of car A (going north) relative to car B (going south) = velocity of car A (going north) – velocity of car B (going south) = 100 km/h (north) – (+42 km/h) (south) = 100 km/h (north) – (– 42 km/h) (north) = 142 km/h (north) • Speed = distance/time, • so time = distance/speed = (10/142) hrs = 0.0704 hrs = 4.23 mins

Solutions • Be careful! Whenever you change east to west, or north to south, or up to down, or right to left, basically any direction change to its opposite direction, change the sign also i.e. + become – and – become +. • So for this question, (+42 km/h) (south) becomes (– 42 km/h) (north)

2.2 Velocity and Momentum 3. A bus is moving with 80 km/h in front of a car that is moving with 100 km/h. If the distance between the bus and the car is 5 km, how long it takes for the car to reach the bus if their velocities remain unchanged? A) 4 hours B) 15 minutes C) 3 minutes

D) 30 minutes (161 Major 1, Q5)

Solutions • Velocity of car relative to bus = velocity of car – velocity of bus = (+100 km/h) – (+80 km/h) = 20 km/h • Speed = distance/time, • so time = distance/speed = (5/20) hrs = 15 mins • Note: Here both car and bus are travelling in the same direction (both forward), so we let forward be positive, and write both velocities as positive numbers

2.2 Velocity and Momentum 4. A boat can travel with a maximum speed of 12 m/s in still water is moving at maximum speed against the current of a stream that flows with a velocity of 5.0 m/s relative to the bank of the stream. What is the velocity of the boat relative to the bank of the stream? A) 7.0 m/s with the flow of the stream.

B) 17 m/s with the flow of the stream. C) 7.0 m/s against the flow of the stream. D 17 m/s against the flow of the stream. (152 Major 1, Q3)

Solutions • Suppose the boat moving north, then current stream must be moving south as it says boat is moving against the current stream • Velocity of boat relative to current stream = vboat,cs = 12 m/s (north) • Velocity of current stream relative to bank = vcs,bank 5 m/s (south) • Velocity of boat relative to bank = vboat,bank • vboat,bank = vboat,cs + vcs,bank

= 12 m/s (north) + 5 m/s (south) = 12 m/s (north) – 5 m/s (north) = 7 m/s (north) Now, current stream is moving south, which means north is against the flow of the stream

2.2 Velocity and Momentum 5. A passenger rides on a train that is moving at 80.0 km/h eastward. If the passenger walks toward the back of the train at a speed of 1.2 km/h relative to the train, what would be his velocity relative to the ground? A) 81.2km/h east.

B) 78.8 km/h west. C) 78.8 km/h east. D) 81.2 km/h west. E) 80.0 km/h east.

(151 Major 1, Q4)

B) 78.8 km/h west. C) 78.8 km/h east. D) 81.2 km/h west. E) 80.0 km/h east.

(151 Major 1, Q4)

Solutions • Velocity of passenger relative to train = vp,t= 1.2 km/h (west) • Velocity of train relative to ground = vt,g 80 km/h (east) • Velocity of passenger relative to ground = vp,g • vp,t + vt,g = vp,g • vp,g = + 1.2 km/h (west) + 80 km/h (east)

= – 1.2 km/h (east) + 80 km/h (east) = 78.8 km/h (east)

Solutions • Be careful! Whenever you change east to west, or north to south, or up to down, or right to left, basically any direction when you change to its opposite direction, you have to change the sign also i.e. (+ become – and – become +) • So for this question, + 1.2 km/h (west) becomes – 1.2 km/h (east)

2.2 Velocity and Momentum 6. A hurricane moves toward a house with 30 km/h west. A car moves away from the house with 100 km/h west. What is the velocity of the hurricane relative to the car? A) 30 km/h, west. B) 70 km/h, west. C) 130 km/h, west. D) 130 km/h, east. E) 70 km/h, east.

(142 Major 1, Q6)

Solutions • Velocity of hurricane relative to car = velocity of hurricane – velocity of car = 30 km/h (west) – 100 km/h (west) = – 70 km/h (west) = + 70 km/h (east) • Be careful! Whenever you change east to west, or north to south, or up to down, or right to left, any direction when change to its opposite, you have to change the sign also (+ become – and – become +)

2.2 Velocity and Momentum The mass of a truck is three times greater than the mass of a car. Given that the speed of the car is three times greater than the speed of the truck, how do their momentums compare? 7.

A) The momentum of the truck is nine times greater than the momentum of the car B) The momentum of the car is three times greater than the momentum of the truck. C) The momentum of the truck is three times greater than the momentum of the car. D) Their momentums are equal.

(171 Major 1, Q8)

Solutions â&#x20AC;˘ For momentum questions, just use p = mv â&#x20AC;˘ Remember that direction of momentum (p) = direction of velocity (v) â&#x20AC;˘ Pcar = mcarvcar â&#x20AC;˘ Ptruck = mtruckvtruck Pcar mcarvcar mcar3vtruck â&#x20AC;˘ = = =đ?&#x;? Ptruck mtruckvtruck 3mcarvtruck

â&#x20AC;˘ So Pcar = Ptruck

2.2 Velocity and Momentum 8. An 80.0-kg person is driving a 1000-kg car at a constant velocity of 20.0 m/s due north. What is the momentum of the person? A) Zero. B) 1,600 kg.m/s North. C) 21,600 kg.m/s North.

D) 20,000 kg.m/s North. (162 Major 1, Q8)

Solutions • For momentum questions, just use p = mv • For momentum of person, just use mass of person alone, not car • P = mv = (80 kg)(20 m/s) = 1600 m/s • Remember that direction of momentum (p) = direction of velocity (v) = north

2.2 Velocity and Momentum 9. How fast and in which direction a 50-kg object must travel to have a momentum of 2,000 kg.m/s east? A) 40.0 m/s east. B) 40.0 m/s west.

C) 400 m/s east. D) 4.00 m/s west.

(152 Final, Q2)

2.2 Velocity and Momentum 9. How fast and in which direction a 50-kg object must travel to have a momentum of 2,000 kg.m/s east? A) 40.0 m/s east. B) 40.0 m/s west.

C) 400 m/s east. D) 4.00 m/s west.

(152 Final, Q2)

Solutions • For momentum questions, just use p = mv • P = mv = (50 kg)(v) = 2000 kg.m/s east • So v = P/m = 2000/50 = 40 m/s • Remember that direction of velocity (v) = direction of momentum (p) = east

2.2 Velocity and Momentum 10. When a car moves with a speed of 25 m/s, its momentum is 50,000 kg.m/s. What is the mass of the car?

A) 2500 kg. B) 1000 kg.

C) 1500 kg. D) 2000 kg. E) None of these. (151 Major 1, Q7)

2.2 Velocity and Momentum 10. When a car moves with a speed of 25 m/s, its momentum is 50,000 kg.m/s. What is the mass of the car?

A) 2500 kg. B) 1000 kg.

C) 1500 kg. D) 2000 kg. E) None of these. (151 Major 1, Q7)

Solutions • For momentum questions, just use p = mv • P = mv = (25 kg)(v) = 50000 kg.m/s • So m = P/v = 50000/25 = 2000 kg

2.3 Acceleration

2.3 Acceleration 1. In which of the following examples is the motion of the car NOT accelerated? A) A car climbs a steep hill on a straight road at the constant speed of 40 km/h. B) A car turns a corner at the constant speed of 20 km/h .

C) A car climbs a steep hill on a straight road with its speed dropping from 60 km/h at the bottom to 15 km/h at the top . D) A car climbs a steep hill and goes over the crest and down on the other side, all at the same speed of 40 km/h (152 Major 1, Q2)

Solutions Acceleration â&#x20AC;&#x201C; either magnitude or direction of velocity changes, or both A) Direction never changes (climbs a steep hill on a straight road straight road) and magnitude also constant at 40 km/h . B) Direction changes (turns a corner) C) magnitude changes because speed dropping means (negative) acceleration, also known as deceleration D) The direction keeps changing â&#x20AC;&#x201C; from upwards, to flat, to downwards â&#x20AC;&#x201C; so there is acceleration

2.3 Acceleration 2. The figure below describes the speed of an object as a function of time during 8.0 s. What can you conclude about the acceleration of the object? A) Only the magnitude of the acceleration changes after 4.0 s. B) Only the direction of the acceleration changes after 4.0 s. C) The acceleration changes in magnitude and direction after 4.0 s.

D) The acceleration remains constant in magnitude and direction. (171 Major 1, Q3)

Solutions • Acceleration is the slope of the speed-time graph • The slope remains fixed for the entire journey

• Likewise acceleration also remains constant • Actually this is not a very good question • All we can say is acceleration remains constant in magnitude • but we cannot say anything about the direction of acceleration • unless the question also states that the object is moving in a straight line ( but that's not written anywhere), So we cannot really conclude anything about direction of acceleration from the graph itself • But anyway options A – C are clearly wrong, so the best answer is D

2.3 Acceleration 3. The figure below describes the speed of an object as a function of time. What is the acceleration of the object as it is slowing down? A) -1.6 m/s2

B) -2.0 m/s2 C) -1.3 m/s2

D) -4.0 m/s2 (162 Major 1, Q3)

2.3 Acceleration 3. The figure below describes the speed of an object as a function of time. What is the acceleration of the object as it is slowing down? A) -1.6 m/s2

B) -2.0 m/s2 C) -1.3 m/s2

D) -4.0 m/s2 (162 Major 1, Q3)

2.3 Acceleration 4. The figure below represents the speedtime graph for the motion of a vehicle during a 7.0-minute trip. Which of the following statements is correct about the motion of this vehicle?

A) The acceleration of the vehicle has changed several times. B) The acceleration of the vehicle has never changed. C) The vehicle has never accelerated. D) The vehicle has never stopped. (161 Major 1, Q8)

2.3 Acceleration 5. The figure below represents the speed-time graph for the motion of a vehicle during a 7.0minute trip. Which of the following statements is correct about the motion of this vehicle? A) The acceleration of the vehicle has changed several times. B) The acceleration of the vehicle has never changed. C) The vehicle has never accelerated. D) The vehicle has never stopped.

E) The vehicle has never moved. (151 Major 1, Q1)

2.3 Acceleration 6. The figure below represents the speed-time graph for the motion of a vehicle during a 7.0-minute trip. How many times the vehicle had decelerated? A) 3

B) 2 C) 4

D) 5 (151 Final, Q1)

2.3 Acceleration 6. The figure below represents the speed-time graph for the motion of a vehicle during a 7.0-minute trip. How many times the vehicle had decelerated? A) 3

B) 2 C) 4

D) 5 (151 Final, Q1)

2.3 Acceleration 7. A car whose acceleration is constant reaches a speed of 80.0 km/h in 20.0 s starting from rest. How much more time is required for it to reach a speed of 130 km/h? A) 20.0 s

B) 32.5 s C) 12.5 s

D) 10.0 s (171 Major 1, Q4)

2.3 Acceleration 7. A car whose acceleration is constant reaches a speed of 80.0 km/h in 20.0 s starting from rest. How much more time is required for it to reach a speed of 130 km/h? A) 20.0 s

B) 32.5 s C) 12.5 s

D) 10.0 s (171 Major 1, Q4)

2.3 Acceleration 8. A car can go from 0 to 100 km/h in 12 s. A second car is capable of twice the acceleration of the first car. Assuming that it could maintain the same acceleration at higher speeds, how much time will this second car take to go from 0 to 200 km/h? A) 24 s. B) 48 s. C) 6.0 s. D) 3.0 s.

E) 12 s. (171 Final, Q4)

2.3 Acceleration 9. A car is traveling due west at 20.0 m/s. What is the velocity of the car after 37.0 seconds if its acceleration is 1.0 m/s2 due east? A) 17.0 m/s west.

B) 17.0 m/s east. C) 57.0 m/s east.

D) 57.0 m/s west. (162 Major 1, Q4)

2.3 Acceleration 9. A car is traveling due west at 20.0 m/s. What is the velocity of the car after 37.0 seconds if its acceleration is 1.0 m/s2 due east? A) 17.0 m/s west.

B) 17.0 m/s east. C) 57.0 m/s east.

D) 57.0 m/s west. (162 Major 1, Q4)

2.3 Acceleration 10. A motorcyclist traveling at 23 m/s (north) applies the brakes, producing an acceleration of 7.2 m/s2 (south). What is the motorcyclist's velocity after 2.5 s? A) 41 m/s, north

B) 5 m/s, south. C) 41 m/s, south.

D) 5 m/s, north (161 Major 1, Q3)

2.3 Acceleration 10. A motorcyclist traveling at 23 m/s (north) applies the brakes, producing an acceleration of 7.2 m/s2 (south). What is the motorcyclist's velocity after 2.5 s? A) 41 m/s, north

B) 5 m/s, south. C) 41 m/s, south.

D) 5 m/s, north (161 Major 1, Q3)

2.3 Acceleration 11. A car whose brakes can produce an acceleration of -6.0 m/s2 is travelling at 30 m/s when its brakes are applied. What is the carâ&#x20AC;&#x2122;s speed 2.3 s later? A) 44 m/s. B) 16 m/s. C) 14 m/s.

D) 30 m/s. (152 Major 1, Q5)

2.3 Acceleration 12. A cyclist starts at rest and accelerates at 0.5 m/s2 north for 40 s. What is the cyclistâ&#x20AC;&#x2122;s final velocity? A) 20 m/s south. B) 20 m/s north.

C) 80 m/s north. D) 80 m/s south.

(152 Final, Q3)

2.3 Acceleration 12. A cyclist starts at rest and accelerates at 0.5 m/s2 north for 40 s. What is the cyclistâ&#x20AC;&#x2122;s final velocity? A) 20 m/s south. B) 20 m/s north.

C) 80 m/s north. D) 80 m/s south.

(152 Final, Q3)

Solutions â&#x20AC;˘ đ?&#x2019;&#x2014;đ?&#x2019;&#x2021; = đ?&#x2019;&#x2014;đ?&#x2019;&#x160; + đ?&#x2019;&#x201A;đ?&#x2019;&#x2022; â&#x20AC;˘ đ?&#x2019;&#x2014;đ?&#x2019;&#x2021; = đ?&#x;&#x17D; + đ?&#x;&#x17D;. đ?&#x;&#x201C;(40) m/s (north) = đ?&#x;?đ?&#x;&#x17D; m/s (north)

2.3 Acceleration 13. A car accelerates from rest by 5 km/h.s northward for 20 seconds, then accelerates by 4 km/h.s southward for 10 seconds. What is the final velocity of the car? A) 40 km/h southward. B) 140 km/h northward C) 60 km/h northward. D) 100 km/h northward. E) 140 km/h southward. (142 Major 1, Q5)

Solutions â&#x20AC;˘ đ?&#x2019;&#x2014;đ?&#x2019;&#x2021;đ?&#x;? = đ?&#x2019;&#x2014;đ?&#x2019;&#x160; + đ?&#x2019;&#x201A;đ?&#x2019;&#x2022; â&#x20AC;˘ đ?&#x2019;&#x2014;đ?&#x2019;&#x2021;đ?&#x;? = đ?&#x;&#x17D; + đ?&#x;&#x201C;(20) m/s (north) = đ?&#x;?đ?&#x;&#x17D;đ?&#x;&#x17D; m/s (north) â&#x20AC;˘ đ?&#x2019;&#x2014;đ?&#x2019;&#x2021;đ?&#x;? = đ?&#x2019;&#x2014;đ?&#x2019;&#x2021;đ?&#x;? + đ?&#x2019;&#x201A;đ?&#x2019;&#x2022; â&#x20AC;˘ đ?&#x2019;&#x2014;đ?&#x2019;&#x2021; = đ?&#x;?đ?&#x;&#x17D;đ?&#x;&#x17D; m/s (north) +đ?&#x;&#x2019;(10) m/s (south) = đ?&#x;?đ?&#x;&#x17D;đ?&#x;&#x17D; m/s (north) + đ?&#x;&#x2019;đ?&#x;&#x17D; m/s (south)

= 60 m/s (north)

2.3 Acceleration 14. A car goes from rest to a velocity of 108 km/h in 10.0 s. What is the car acceleration? A) 10.8m/s2 B) 3.00 m/s2

C) 30.0 m/s2 D) Zero

E) None of these. (151 Major 1, Q5)

2.3 Acceleration 14. A car goes from rest to a velocity of 108 km/h in 10.0 s. What is the car acceleration? A) 10.8m/s2 B) 3.00 m/s2

C) 30.0 m/s2 D) Zero

E) None of these. (151 Major 1, Q5)

Solutions • 𝒗𝒇 = 𝒗𝒊 + 𝒂𝒕 • 𝟑𝟎 𝐦/𝐬 = 𝟎 + 𝒂(10) • a = 3.00 m/s2

2.3 Acceleration 15. Which of the following statements about centripetal acceleration is correct? A) Centripetal acceleration occurs whenever the speed changes. B) The centripetal acceleration of an object is always perpendicular to its velocity. C) Centripetal acceleration will equal to zero whenever the object moves in a circular path. D) Centripetal acceleration occurs when the object moves at constant velocity. (171 Major 1, Q5)

Solutions For centripetal acceleration questions • just remember that any object moving in a circle at a constant speed, has: • centripetal acceleration of value ac = v2/r (magnitude) • pointing towards centre of circle, and perpendicular to velocity (direction)

2.3 Acceleration 16. A car is moving with constant speed in a circular track as shown in the figure below. In which direction will the acceleration be at point A? A) South B) East C) North D) The acceleration of the car= 0, because the speed of the car is constant. (162 Major 1, Q7)

Solutions â&#x20AC;˘ centripetal acceleration points towards centre of circle, and perpendicular to velocity (direction) â&#x20AC;˘ At A, velocity is east and acceleration is north

2.3 Acceleration 17. Which best describes why projectiles move in a curved path? A) They have constant horizontal velocity and vertical acceleration. B) They have horizontal acceleration and constant vertical velocity.

C) They have horizontal and vertical accelerations. D) They have constant horizontal velocity and constant vertical velocity.

(151 Final, Q2)

C) They have horizontal and vertical accelerations. D) They have constant horizontal velocity and constant vertical velocity.

(151 Final, Q2)

Solutions • For projectile motion questions, normally like throwing ball in the air, just remember that the only force acting on object is gravitational force (= weight). This causes gravitational acceleration, g = 9.8 m/s2 for objects near the earth surface. • There is no other horizontal or vertical force acting on the object. • if no air resistance, and the object has some initial horizontal velocity, then this horizontal velocity will remain constant for the entire motion of object through the air • hence there is constant horizontal velocity and constant vertical acceleration for projectile motion

Solutions • If the object has some initial vertical velocity = vi, then use: vf = vi + gt to find the vertical velocity vf during the flight at some later time t • vf increases by 9.8 m/s downwards every second • Remember! a = g = 9.8 m/s2 downwards for any object in free-fall

2.3 Acceleration 18.

Which best describes why projectiles move in a curved path?

A) They have constant horizontal velocity and vertical acceleration.

B) They have horizontal acceleration and constant vertical velocity. C) They have constant horizontal velocity and constant vertical velocity. D) They have horizontal and vertical accelerations. E) None of these. (142 Major 1, Q2)

2.3 Acceleration 18.

Which best describes why projectiles move in a curved path?

A) They have constant horizontal velocity and vertical acceleration.

Solutions Same question!!!

2.3 Acceleration 19. If you drop a baseball from a height of 1.5 m, it will hit the ground in 0.55 s. If you throw the baseball horizontally with a speed of 30 m/s from the same height, how long will it take the ball to hit the ground? A) 0.75 s. B) 0.35 s C) 0.55 s D) It depends on how hard the ball is thrown horizontally.

E) 1.5 s (151 Major 1, Q6)

Why time of flight doesn’t depend on horizontal speed? • To understand why time of flight does not depend on whether the ball is dropped or thrown horizontally, look at this picture  • The ball falls by same height in equal times for both: • The extra initial horizontal velocity will just:

(a) Make the ball travel a further horizontal distance (b) Increase the final velocity of ball hitting the ground • But the time of flight of both balls in the air remains the same

Why time of flight doesn’t depend on horizontal speed? • Also look at this picture  • Assuming there was no gravity, the ball will continue flying in the air forever at a constant height

• Whereas time of flight is the time for ball from starting to ending point • This means the horizontal motion or speed can never ever cause the ball to stop moving (i.e. affect time of flight)

Why time of flight doesnâ&#x20AC;&#x2122;t depend on horizontal speed? â&#x20AC;˘ This is the main reason why the time of flight is not dependent on initial horizontal velocity, of course assuming the ball is launched with the same (a) initial vertical velocity (b) from the same height â&#x20AC;˘ Try to understand all these points well

Solutions So, it doesn't matter whether a ball is thrown horizontally (0.55 s) or dropped (0.55 s), it will take the same time to reach the ground. Answer = 0.55 s

2.3 Acceleration 20. If you drop a baseball from a height of 1.5 m, it will hit the ground in 0.55 s. If you throw the baseball horizontally with a speed of 30 m/s from the same height, how long will it take the ball to hit the ground? A) 0.75 s. B) 0.35 s C) 0.55 s D) It depends on how hard the ball is thrown horizontally.

E) 1.5 s (151 Major 1, Q6)

2.3 Acceleration 21. Two balls are at the same height and released at the same time. One ball is dropped and hits the ground 3.0 s later. The other initially moves horizontally at 10 m/s as shown the figure below. How far does the second ball travel horizontally before it hits the ground? A) 53.3 m B) 30.0 m. C) 44.1 m

D) None of these. (171 Major 1, Q7)

Solutions Again, it doesn't matter whether a ball is thrown horizontally (3.0 s) or dropped (3.0 s), it will take the same time to reach the ground. Horizontal distance travelled

= Horizontal speed x time of flight = 3.0 s x 10 m/s = 30 m

2.3 Acceleration 22. A rock is dropped at the same instant that a ball at the same height is thrown horizontally. Which of the following statement is correct? (Neglecting air resistance) A) The rock and the ball hit the ground with equal speed. B) The rock hits the ground with higher speed. C) The rock hits the ground first. D) The ball hits the ground with higher speed. (161 Major 1, Q11)

2.3 Acceleration 23. Two balls are at the same height and released at the same time. One ball is dropped and hits the ground 5 s later. The other initially moves horizontally with 13 m/s as in the figure below. When does the second ball hit the ground? (Assume no air resistance) A) 3 s. B) 5 s.

C) 7 s. D) 9 s. E) Cannot be found.

(142 Final, Q23)

C) 7 s. D) 9 s. E) Cannot be found.

(142 Final, Q23)

Solutions â&#x20AC;¢ Same idea!!!

2.3 Acceleration 24. A rock is dropped at the same instant that a ball at the same height is thrown horizontally. Which of the following statement is correct? (Neglecting air resistance) A) The rock and the ball hit the ground with equal speed.

B) The rock hits the ground with higher speed. C) The rock hits the ground first.

D) The ball hits the ground with higher speed. (152 Major 1, Q6)

B) The rock hits the ground with higher speed. C) The rock hits the ground first.

D) The ball hits the ground with higher speed. (152 Major 1, Q6)

Solutions â&#x20AC;˘ Now, if the ball is thrown horizontally, when it reaches the ground, it will have both horizontal and vertical component of velocity, but if dropped, will have only the vertical component. â&#x20AC;˘ Hence, if thrown horizontally, the final speed (horizontal + vertical components) is larger than the final speed if dropped (vertical component only) â&#x20AC;˘ So the ball (thrown horizontally) hits the ground with higher speed than rock (dropped)

2.3 Acceleration 25. Three balls of different masses are thrown horizontally from the same height at the same time with different initial speeds as shown in the figure below (where the speeds are represented by the length of the arrows). Which ball will reach the ground first? (Assume no air resistance)

A) Ball (1)

B) Ball (2) C) Ball (3)

D) All balls will reach the ground at the same time. (162 Major 1, Q5)

A) Ball (1)

B) Ball (2) C) Ball (3)

D) All balls will reach the ground at the same time. (162 Major 1, Q5)

Solutions â&#x20AC;˘ As mentioned many times, the Time of flight does not depend on the initial horizontal velocity â&#x20AC;˘ So, all balls will reach the ground at the same time.

2.3 Acceleration 26. A toy car, as in the figure below, runs off the edge of the table that is 1.225-m high. The car lands 0.400 m from the base of the table, how long does it take the car to fall? A) 0.500 s B) 0.250 s C) 4.900 s D) 1.225 s (161 final, Q30)

Solutions • Use this kinematics equation to find time of flight (t):

s = ut + 0.5at2 • s = vertical distance travelled = 1.225 m

• a = vertical acceleration = g = 9.8 m/s2 • u = initial vertical velocity = 0 m/s (dropped from rest)

• s = (0)(t) + 0.5(9.8)t2 • 1.225 = 0.5(9.8)t2

• t = 0.5 s

2.3 Acceleration 27. Three arrows are shot horizontally. They have left the bow and are traveling parallel to the ground as shown in the figure below. Air resistance is negligible. Rank in order, from largest to smallest, the magnitude of the horizontal forces F(I), F(II), and F(III) acting on the arrows. A) F(III), F(I), F(II) B) F(I), F(II), F(III) C) F(II), F(III), F(I) D) F(I), F(III), F(II)

E) None of these. (171 Final, Q7)

Solutions • Just remember, the only possible horizontal force during projectile motion is air resistance. • since question say air resistance = 0 • so horizontal force = 0

• Hence F(I) = F(II) = F(III) • And none of the above is correct

2.3 Acceleration 28. A ball thrown upward with an initial velocity of 8.30 m/s. Air resistance is negligible. How long will it take to reach the maximum height? A) 1.69 s

B) 1.18 s C) 2.36 s

D) 0.847 s (161 Major 1, Q20)

2.3 Acceleration 28. A ball thrown upward with an initial velocity of 8.30 m/s. Air resistance is negligible. How long will it take to reach the maximum height? A) 1.69 s

B) 1.18 s C) 2.36 s

D) 0.847 s (161 Major 1, Q20)

Solutions â&#x20AC;˘ At maximum height, the vertical velocity = 0 â&#x20AC;˘ Just use a =

đ?&#x2019;&#x2014;đ?&#x2019;&#x2021; â&#x2C6;&#x2019;đ?&#x2019;&#x2014;đ?&#x2019;&#x160;

đ?&#x2019;&#x2022;

â&#x20AC;˘ With a = g = 9.8 m/s2, vf = 0 m/s, vi = 8.30 m/s,

â&#x20AC;˘ We get t = 0.847 s

2.3 Acceleration 29. While moving at constant velocity on a level, snowy surface, a snowmobiler throws a ball vertically upward. Where will the ball land? Assume that the air resistance is negligible. A) It will fall in front of him.

B) It will fall behind him. C) It will return back to his hand.

D) It depends on the initial speed of the ball. (171 Major 1, Q10)

Solutions â&#x20AC;˘ This is the main story: since they have same horizontal velocity, and the ball when flying through the air its horizontal velocity does not change, so in equal time period, both the ball and the motorcycle travel the exact same horizontal distance, and the person catch back his ball

B) It will fall behind him. C) It will return back to his hand.

D) It depends on the initial speed of the ball. (171 Major 1, Q10)

2.3 Acceleration 30. An object falls from the ceiling of a bus that is moving with constant velocity. If the object falls from point P, as shown in the figure below, at what point will it hit the bus floor below? (Assume all windows and doors are closed) A) At point X.

B) At point Y. C) At point Z.

D) It will not hit the bus floor. (162 Major 1, Q13)

B) At point Y. C) At point Z.

D) It will not hit the bus floor. (162 Major 1, Q13)

Solutions â&#x20AC;˘ Exact same story as previous question: in this case both the bus and falling object have the same horizontal velocity, so they travel equal horizontal period in equal time period, hence the ball appears to fall in a straight vertical line when seen by a man inside the bus. To a person sitting outside the bus, this is how he will see the ball and bus moving.

3.1 Forces

3.1 Forces 1. Three balanced forces are applied to a box as in the figure below. If F1 = 2F2, and F3 = 9.0 N, what is the value of F1? A) 6.0 N B) 3.0 N

C) 4.5 N D) 9.0 N

(162 Major 1, Q9)

3.1 Forces 1. Three balanced forces are applied to a box as in the figure below. If F1 = 2F2, and F3 = 9.0 N, what is the value of F1? A) 6.0 N B) 3.0 N

C) 4.5 N D) 9.0 N

(162 Major 1, Q9)

Solutions • The key word is 3 balanced forces • So they must cancel • Use F1 + F2 = F3 = 9 N and F1 = 2F2 • To get F1 = 6.0 N

3.1 Forces 2. A 5.0-kg block rests on top of a 10-kg block, which rests on a horizontal table. What is the force (magnitude and direction) exerted by the 10-kg block on the 5.0-kg block? A) 98 N upward B) 49 N upward C) 49 N downward D) 98 N downward

(152 Major 1, Q14)

Solutions

Normal Reaction

5 kg 10 kg Weight = mg

Solutions • As Seen in the previous, the 5 kg block is at rest • So, the forces acting on it must all balance • hence the downward gravitational force (weight) must be equally balanced by the upward normal reaction/contact force

• Hence normal reaction/contact force = weight = 49 N

3.1 Forces 3. A 10.0-kg block rests on top of a 20.0-kg block, which rests on a horizontal table. What is the force (magnitude and direction) exerted by the 20.0-kg block on the 10.0-kg block? A) 196 N downward.

B) 98.0 N downward. C) 98.0 N upward.

D) 196 N upward. (152 Final, Q7)

B) 98.0 N downward. C) 98.0 N upward.

D) 196 N upward. (152 Final, Q7)

Solutions Exact same idea as previous questionâ&#x20AC;¦

3.1 Forces Which of the following sets of forces are unbalanced forces when act on an object? 4.

A) (5.0 N east, 9.8 N north, 5.6 N west, 9.8 N south) B) (20 N east, 10 N north, 20 N west, 10 N south) C) (1.7 N east, 3.5 N north, 1.7 N west, 3.5 N south) D) (6.4 N east, 10 N north, 6.4 N west, 10 N south) E) (8.5 N east, 15 N north, 8.5 N west, 15 N south) (151 Major 1, Q8)

3.1 Forces Which of the following sets of forces are unbalanced forces when act on an object? 4.

Solutions • Just remember that equal forces in opposite direction will cancel each other • so North cancel south & East cancel West & up cancel down & right cancels left and so on…

• In (a) 9.8 N north cancels 9.8 N south, but 5.0 N east does NOT cancel 5.6 N west, so the forces are unbalanced

3.1 Forces Which of the following sets of forces are balanced forces when act on an object? 5.

A) (5 N east, 10 N north, 5 N west, 10 N south) B) (20 N east, 10 N north, 10 N west, 20 N south) C) (10 N east, 10 N north, 20 N west, 20 N south) D) (30 N east, 10 N north, 20 N west, 20 N south) E) (20 N east, 15 N north, 10 N west, 5 N south) (142 Major 1, Q8)

3.1 Forces Which of the following sets of forces are balanced forces when act on an object? 5.

Solutions Exact same idea as previous questionâ&#x20AC;¦

3.1 Forces 6. A worker is trying to slide a wooden box as in the figure below. He pushes the box by a force of 100 N due east, but the box doesnâ&#x20AC;&#x2122;t slide. There is friction between the box and the floor. What is the magnitude of the force of friction? A) 100 N.

B) Greater than 100 N. C) Less than 100 N.

D) Zero. (162 Major 1, Q10)

B) Greater than 100 N. C) Less than 100 N.

D) Zero. (162 Major 1, Q10)

Solutions • The keyword is box not sliding • So acceleration = 0, & net force on box = 0 N (by Newton’s 2nd law) • Which says F = ma • so the forward pushing force of 100 N is completely opposed by the backward friction force of 100 N.

Fpush = 100 N Friction = 100 N

Solutions • But, this question is not very good… • Suppose the friction force was more than 100 N, e.g. 120 N, even then the box will not be sliding, because forward pushing force (100 N) has been completely opposed by the backward friction (120 N)

• so the more accurate answer should be: more than or equal to 100 N

Fpush = 100 N Friction = 120 N

Solutions â&#x20AC;˘ Unless the question says, that when the man push by 100 N, the box was just/almost about to slide, then of course this will mean that the friction is exactly 100 N, since even 100.1 N (i.e. slightly more than 100 N) will cause it to at least move very slowly.

Fpush = 100 N Friction = 120 N

3.1 Forces 7. If you use a horizontal force of 30 N to slide a sofa across a floor at a constant velocity, what is the sliding friction between the sofa and the floor? A) Less than 30 N.

B) More than 30 N. C) 30 N.

D) Zero. (161 Major 1, Q9)

3.1 Forces 7. If you use a horizontal force of 30 N to slide a sofa across a floor at a constant velocity, what is the sliding friction between the sofa and the floor? A) Less than 30 N.

B) More than 30 N. C) 30 N.

D) Zero. (161 Major 1, Q9)

Solutions • Again, the keyword is box sliding at constant velocity • So acceleration = 0, & net force on box = 0 N (by Newton’s 2nd law) • Which says F = ma • so the forward pushing force of 30 N must balance the backward friction force of 30 N

Fpush = 30 N Friction = 30 N

3.1 Forces 8. If you use a horizontal force of 30 N to slide a sofa across a floor at a constant velocity, what is the sliding friction between the sofa and the floor? A) 30 N.

B) Less than 30 N. C) More than 30 N.

D) Zero. (152 Major 1, Q7)

3.1 Forces 8. If you use a horizontal force of 30 N to slide a sofa across a floor at a constant velocity, what is the sliding friction between the sofa and the floor? A) 30 N.

B) Less than 30 N. C) More than 30 N.

D) Zero. (152 Major 1, Q7)

Fpush = 30 N Friction = 30 N

3.1 Forces 9. A boy pushes on a desk and the desk slides horizontally across the floor with a constant speed. If the sliding friction between the desk and the floor is 10.0 N, how much force must the boy apply to maintain the desk constant speed? A) 20.0 N.

B) 10.1 N. C) 5.00 N.

D) 10.0 N. E) 9.90 N.

(151 Major 1, Q10)

B) 10.1 N. C) 5.00 N.

D) 10.0 N. E) 9.90 N.

(151 Major 1, Q10)

Solutions • Again, the keyword is box sliding at constant velocity • So acceleration = 0, & net force on box = 0 N (by Newton’s 2nd law) • Which says F = ma • so the forward pushing force of 10 N must balance the backward friction force of 10 N

Fpush = 10 N Friction = 10 N

3.1 Forces 10. On Earth, a scale shows that you weigh 585.0 N. What would be the scale read of your weight on another planet whose mass and size are less than their values for Earth by one-half? A) 585.0 N.

B) 73.13 N. C) 292.5 N.

D) 1,170 N. (171 Major 1, Q12)

3.1 Forces 11. On Earth, a scale shows that you weigh 585.0 N. What would be the scale read of your weight on another planet whose mass and size are less than their values for Earth by one-half? A) 585.0 N.

B) 73.13 N. C) 292.5 N.

D) 1,170 N. (171 Major 1, Q12)

Solutions â&#x20AC;˘ Just use đ?? =

đ?&#x2018;´ đ?&#x2018;Ž đ?&#x;? đ?&#x2018;š

â&#x20AC;˘ g = Gravitational acceleration = Gravity field strength (same meaning) â&#x20AC;˘ NEAR EARTH surface, g = 9.8 m/s2 = 9.8 N/kg,

â&#x20AC;˘ but doesn't mean g always 9.8 N/kg. Near moon surface, g = 1.67 N/kg â&#x20AC;˘ G = Universal gravitational constant = 6.67 x 10-11 N m2 kg -2

G is ALWAYS fixed for earth, moon, ANY PLANETS, ANY STARS etc. â&#x20AC;˘ M = mass of planet, R = radius of planet (be careful of units!!)

Solutions â&#x20AC;˘ Just use đ?? =

đ?&#x2018;´ đ?&#x2018;Ž đ?&#x;? đ?&#x2018;š

â&#x20AC;˘ đ?&#x2019;&#x2C6;đ?&#x2019;&#x2020; =

đ?&#x2018;´đ?&#x2019;&#x2020; đ?&#x2018;Ž đ?&#x;? đ?&#x2018;šđ?&#x2019;&#x2020;

for earth

â&#x20AC;˘ đ?&#x2019;&#x2C6;đ?&#x2018;ż =

đ?&#x2018;´đ?&#x2018;ż đ?&#x2018;Ž đ?&#x;? đ?&#x2018;šđ?&#x2018;ż

for other planet X

â&#x20AC;˘

đ?&#x2019;&#x2C6;đ?&#x2018;ż đ?&#x2019;&#x2C6;đ?&#x2019;&#x2020;

đ?&#x2018;´ đ?&#x2018;Ž đ?&#x;?đ?&#x2018;ż đ?&#x2018;šđ?&#x2018;ż đ?&#x2018;´ đ?&#x2018;Ž đ?&#x;?đ?&#x2019;&#x2020; đ?&#x2018;šđ?&#x2019;&#x2020;

đ?&#x2018;´đ?&#x2018;ż đ?&#x2018;šđ?&#x;?đ?&#x2019;&#x2020; đ?&#x2018;´đ?&#x2019;&#x2020; đ?&#x2018;šđ?&#x;?đ?&#x2018;ż

This is the MAIN EQUATION!!

â&#x20AC;˘ It tells you the value of g for some unknown planet by comparing it's mass & radius with the mass & radius for Earth

Solutions đ?&#x2019;&#x2C6;đ?&#x2018;ż â&#x20AC;˘ đ?&#x2019;&#x2C6;đ?&#x2019;&#x2020;

đ?&#x2018;´đ?&#x2018;ż đ?&#x2018;šđ?&#x;?đ?&#x2019;&#x2020; đ?&#x2018;´đ?&#x2019;&#x2020; đ?&#x2018;šđ?&#x;?đ?&#x2018;ż

VERY IMPORTANT EQUATION!!

â&#x20AC;˘ Question says: Me = 2Mx and Re = 2Rx â&#x20AC;˘ Substitute these information into above equation â&#x20AC;˘ We get

đ?&#x2019;&#x2C6;đ?&#x2018;ż đ?&#x2019;&#x2C6;đ?&#x2019;&#x2020;

đ?&#x2018;´đ?&#x2018;ż đ?&#x2018;šđ?&#x;?đ?&#x2019;&#x2020; đ?&#x2018;´đ?&#x2019;&#x2020; đ?&#x2018;šđ?&#x;?đ?&#x2018;ż

đ?&#x2018;´đ?&#x2018;ż đ?&#x;?đ?&#x2018;šđ?&#x2018;ż đ?&#x;? đ?&#x;?đ?&#x2018;´đ?&#x2018;ż đ?&#x2018;šđ?&#x;?đ?&#x2018;ż

đ?&#x;?đ?&#x;? đ?&#x;?

=đ?&#x;?

â&#x20AC;˘ So đ?&#x2019;&#x2C6;đ?&#x2018;ż = đ?&#x;?đ?&#x2019;&#x2C6;đ?&#x2019;&#x2020; = đ?&#x;?đ?&#x2019;&#x2C6; NOTE: Basically, on earth we call ge as simply g (so đ?&#x2019;&#x2C6;đ?&#x2019;&#x2020; = đ?&#x2019;&#x2C6;)

Solutions â&#x20AC;˘ Since đ?&#x2019;&#x2C6;đ?&#x2018;ż = đ?&#x;?đ?&#x2019;&#x2C6; and weight, W = mg â&#x20AC;˘ Note: W = mg applies for ALL PLANETS, STARS etc. â&#x20AC;˘ m = mass is ALWAYS FIXED, but g changes when go to another planet â&#x20AC;˘ So to find new weight on new planet, just find new g â&#x20AC;˘ So W = mgx = m(2g) = 2mg = 2(585) = 1170 N.

3.1 Forces 12. Rank in order, from largest to smallest, the gravitational field strengths on the surfaces of the following planets.

A) (IV), (II), (III), (I). B) (II), (I), (IV), (III).

C) (IV), (III), (II), (I). D) (I), (II), (IV), (III). E) (II), (IV), (III), (I). (171 Final, Q1)

3.1 Forces 12. Rank in order, from largest to smallest, the gravitational field strengths on the surfaces of the following planets.

A) (IV), (II), (III), (I). B) (II), (I), (IV), (III).

C) (IV), (III), (II), (I). D) (I), (II), (IV), (III). E) (II), (IV), (III), (I). (171 Final, Q1)

Solutions Similar to previous questionsâ&#x20AC;Ś đ?&#x2018;´ đ?&#x2018;Ž đ?&#x;? đ?&#x2018;š

Just use đ?? = and find the values of g for all 4 planets and rank them in orderâ&#x20AC;Ś

3.1 Forces 13. How does the gravitational force change as two objects move farther apart? A) It increases. B) It remains constant.

C) It is zero. D) It decreases.

(162 Major 1, Q11)

3.1 Forces 13. How does the gravitational force change as two objects move farther apart? A) It increases. B) It remains constant.

C) It is zero. D) It decreases.

(162 Major 1, Q11)

Solutions â&#x20AC;˘ This time question mentions gravitational force (F)

â&#x20AC;˘ So, use đ??&#x2026; =

đ?&#x2018;´đ?&#x2019;&#x17D; đ?&#x2018;Ž đ?&#x;? đ?&#x2018;š

â&#x20AC;˘ F = gravitational force (F) between 2 objects separated by distance R

â&#x20AC;˘ G = Universal gravitational constant = 6.67 x 10-11 N m2 kg -2 â&#x20AC;˘ M = mass of object 1, m = mass of object 2

â&#x20AC;˘ R = distance between 2 objects â&#x20AC;˘ So, obviously F decrease when R increases due to inverse relationship

3.1 Forces 14. Assume your mass is m, the mass of Earth is M, and the radius of Earth is R. Which of the following equations represent your weight at Earthâ&#x20AC;&#x2122;s surface? (G is the universal gravitational constant) A) đ?&#x2019;&#x17D;đ?&#x2018;Ž đ?&#x2018;´ B) đ?&#x2018;Ž đ?&#x;? đ?&#x2018;š đ?&#x2019;&#x17D; C) đ?&#x2018;Ž đ?&#x;? đ?&#x2018;š đ?&#x2019;&#x17D;đ?&#x2018;´ D) đ?&#x2018;Ž đ?&#x;? đ?&#x2018;š

(161 Major 1, Q17)

Solutions Similar to previous questionsâ&#x20AC;Ś Just use đ?&#x2019;&#x2C6; =

đ?&#x2018;´ đ?&#x2018;Ž đ?&#x;? đ?&#x2018;š

So W = mg =

đ?&#x2018;´đ?&#x2019;&#x17D; đ?&#x2018;Ž đ?&#x;? đ?&#x2018;š

and W = mg

â&#x20AC;˘ Notice that this weight formula is exactly the same as the gravitational force equation between two bodies M & m â&#x20AC;˘ This proves weight is indeed a force!

â&#x20AC;˘ So your weight on the earth surface is basically the gravitational force by the Earth on you. Make sure this point is very clearâ&#x20AC;Ś

3.1 Forces Assume your mass is m, the mass of Earth is M, and the radius of Earth is R. Which of the following equations represent your weight at Earthâ&#x20AC;&#x2122;s surface? (G is the universal gravitational constant) 15.

A) đ?&#x2018;Ž

đ?&#x2019;&#x17D;đ?&#x2018;´ đ?&#x2018;šđ?&#x;?

B) đ?&#x2019;&#x17D;đ?&#x2018;Ž đ?&#x2018;´ C) đ?&#x2018;Ž đ?&#x;? đ?&#x2018;š đ?&#x2019;&#x17D; D) đ?&#x2018;Ž đ?&#x;? đ?&#x2018;š

(152 Major 1, Q8)

A) đ?&#x2018;Ž

đ?&#x2019;&#x17D;đ?&#x2018;´ đ?&#x2018;šđ?&#x;?

B) đ?&#x2019;&#x17D;đ?&#x2018;Ž đ?&#x2018;´ C) đ?&#x2018;Ž đ?&#x;? đ?&#x2018;š đ?&#x2019;&#x17D; D) đ?&#x2018;Ž đ?&#x;? đ?&#x2018;š

(152 Major 1, Q8)

Solutions Exactly similar to previous questionâ&#x20AC;¦

3.1 Forces 16. Three masses are located as shown in the diagram below. The distance between m1 and m2 is x1 and the distance between m2 and m3 is x2. If m3=4m1 and x2=2x1, what is the net force on m2? A) đ??&#x2122;đ??&#x17E;đ??Ťđ??¨ B)

đ?&#x2019;&#x17D; đ?&#x;? đ?&#x2019;&#x17D;đ?&#x;? đ?&#x2018;Ž đ?&#x;? đ?&#x;?đ?&#x2019;&#x2122;đ?&#x;?

đ?&#x2019;&#x17D;đ?&#x;? đ?&#x2019;&#x17D;đ?&#x;? đ?&#x;?đ?&#x2018;Ž đ?&#x;? đ?&#x2019;&#x2122;đ?&#x;?

đ?&#x2019;&#x17D; đ?&#x;? đ?&#x2019;&#x17D;đ?&#x;? đ?&#x2018;Ž đ?&#x;? đ?&#x2019;&#x2122;đ?&#x;?

(152 Major 1, Q9)

đ?&#x2019;&#x17D; đ?&#x;? đ?&#x2019;&#x17D;đ?&#x;? đ?&#x2018;Ž đ?&#x;? đ?&#x;?đ?&#x2019;&#x2122;đ?&#x;?

đ?&#x2019;&#x17D;đ?&#x;? đ?&#x2019;&#x17D;đ?&#x;? đ?&#x;?đ?&#x2018;Ž đ?&#x;? đ?&#x2019;&#x2122;đ?&#x;?

đ?&#x2019;&#x17D; đ?&#x;? đ?&#x2019;&#x17D;đ?&#x;? đ?&#x2018;Ž đ?&#x;? đ?&#x2019;&#x2122;đ?&#x;?

(152 Major 1, Q9)

Solutions â&#x20AC;˘ Jđ??Žđ??Źđ?? đ??Žđ??Źđ??&#x17E; đ??&#x2026; =

đ?&#x2018;´đ?&#x2019;&#x17D; đ?&#x2018;Ž đ?&#x;? đ?&#x2018;š

two times:

(1) once to find the force by m1 on m2 (points left), and (2) another time to find the force by m3 on m2 (points right)

â&#x20AC;˘ You will get both forces as equal in magnitude and opposite in direction so they cancel each other and the net force is zero.

3.1 Forces 17. If the earth were 3 times as far from the sun as it is now, the gravitational force exerted on it by the sun would be A) 9 times as large as it is now. B) 3 times as large as it is now. C) one-ninth as large as it is now. D) one-third as large as it is now. E) the same as it is now.

(151 Major 1, Q11)

Solutions â&#x20AC;˘ Jđ??Žđ??Źđ?? đ??Žđ??Źđ??&#x17E; đ??&#x2026; =

đ?&#x2018;´đ?&#x2019;&#x17D; đ?&#x2018;Ž đ?&#x;? đ?&#x2018;š

â&#x20AC;˘ When R decreases to R/3, â&#x20AC;˘ Then F decreases to F/32 = F/9, due to squared relationship

â&#x20AC;˘ So force becomes 1/9

3.1 Forces 18. The gravitational potential energy of an appleEarth system changes when which factor changes? A) The apple’s size B) The apple’s speed

C) The apple’s temperature D) The apple’s mass

(151 Final, Q7)

3.1 Forces 18. The gravitational potential energy of an appleEarth system changes when which factor changes? A) The apple’s size B) The apple’s speed

C) The apple’s temperature D) The apple’s mass

(151 Final, Q7)

Solutions đ?&#x2018;´đ?&#x2019;&#x17D; đ?&#x2018;Ž đ?&#x;? đ?&#x2018;š

â&#x20AC;˘ A đ??Şđ??Žđ??˘đ??&#x153;đ??¤ đ??Ľđ??¨đ??¨đ??¤ đ??&#x161;đ?? đ??&#x2026; = tells us that Gravitational force only depends on mass and radius, and nothing elseâ&#x20AC;Ś â&#x20AC;˘ So the answer is massâ&#x20AC;Ś

3.1 Forces 19. The gravitational field strength (g) at the surface of Earth (which is at a distance R from the center of Earth) is approximately 9.80 N/kg. What would be the gravitational field strength at height (R) above the Earthâ&#x20AC;&#x2122;s surface? A) 4.90 N/kg. B) 9.80 N/kg. C) 19.6 N/kg. D) 39.2 N/kg.

E) 2.45 N/kg. (142 Major 1, Q7)

Solutions â&#x20AC;˘ Jđ??Žđ??Źđ?? đ??Žđ??Źđ??&#x17E; đ?? =

đ?&#x2018;´ đ?&#x2018;Ž đ?&#x;? đ?&#x2018;š

â&#x20AC;˘ And note that at earth surface, radius = R, g = 9.8 N/kg â&#x20AC;˘ If we increase R to 2R

â&#x20AC;˘ Then g decreases to g/22 = g/4, due to squared relationship â&#x20AC;˘ So g becomes g/4 = 2.45 N/kg

3.1 Forces 20. A person weighs 882 N at Earthâ&#x20AC;&#x2122;s surface and 270 N at the surface of another planet. What is the gravitational strength at the surface of that planet? A) 27.0 N/kg

B) 4.90 N/kg C) 3.00 N/kg

D) 2.00 N/kg (162 Major 1, Q12)

3.1 Forces 20. A person weighs 882 N at Earthâ&#x20AC;&#x2122;s surface and 270 N at the surface of another planet. What is the gravitational strength at the surface of that planet? A) 27.0 N/kg

B) 4.90 N/kg C) 3.00 N/kg

D) 2.00 N/kg (162 Major 1, Q12)

Solutions â&#x20AC;˘ Just use W = mg twice, Once for earth & once for planet We = mge = 882 N Wplanet = mgplanet = 270 N gplanet/ge= 270/882 gplanet = 270/882 x ge = 270/882 x 9.8 = 3.00 N/kg

3.1 Forces 21. You weigh yourself at the top of a high mountain and the scale reads 730 N. If your mass is 76.0 kg, what is the gravitational strength at your location A) 730 N/kg

B) 9.61 N/kg C) 9.80 N/kg

D) 76.0 N/kg (152 Final, Q4)

3.1 Forces 21. You weigh yourself at the top of a high mountain and the scale reads 730 N. If your mass is 76.0 kg, what is the gravitational strength at your location A) 730 N/kg

B) 9.61 N/kg C) 9.80 N/kg

D) 76.0 N/kg (152 Final, Q4)

Solutions â&#x20AC;˘ Just use W = mg at the mountain top Wmountain = mgmountain = 730 N gmountain = Wmountain/m = 730/76 = 9.61 N/kg

3.1 Forces 22. An astronaut has a mass of 100 kg and a weight of 327 N on another planet. The gravitational strength on that planet is A) one-half its value on earth. B) 3 times its value on earth. C) the same as on earth. D) 2 times its value on earth. E) one-third its value on earth.

(151 Major 1, Q13)

Solutions â&#x20AC;˘ Just use W = mg at the other planet Wplanet = m gplanet = 730 N gplanet = Wplanet /m = 327/100 = 3.27 N/kg ~ 9.81/3 N/kg ~ g/3 3.27 is roughly 1/3 of 9.81 = g/3

3.1 Forces 23. The magnitude of the acceleration of a stone thrown vertically upward is

A) zero until it reaches the highest point in its path. B) the same as that of a stone thrown vertically downward. C) greater than that of a stone thrown vertically downward.

D) less than that of a stone thrown vertically downward. (171 Major 1, Q18)

3.1 Forces 23. The magnitude of the acceleration of a stone thrown vertically upward is

A) zero until it reaches the highest point in its path. B) the same as that of a stone thrown vertically downward. C) greater than that of a stone thrown vertically downward.

D) less than that of a stone thrown vertically downward. (171 Major 1, Q18)

Solutions â&#x20AC;˘ As mentioned many times, the value of gravitational acceleration g = 9.8 m/s2 for objects thrown near the earth surface, and this g = 9.8 does not depend on which direction or velocity the object is thrown with â&#x20AC;˘ So Downward vertical acceleration is always the same

3.2 Newtonâ&#x20AC;&#x2122;s Laws of Motion

3.2 Newtonâ&#x20AC;&#x2122;s Laws of Motion A person is pushing horizontally on a box with a constant force, causing it to slide across a rough floor with a constant velocity . If the person suddenly stops pushing on the box, the box will 1.

A) Immediately come to a stop. B) Continue moving at a constant velocity for a while, then gradually slows down to a stop. C) Immediately change to a slower but constant velocity. D) Immediately begin slowing down and eventually stop. E) Continue moving at a constant velocity. (171 Final, Q8)

Solutions • Just push a book on your table at constant speed, • and then suddenly stop pushing. • you will notice that the book immediately comes to a stop

3.2 Newtonâ&#x20AC;&#x2122;s Laws of Motion 2. A ball on the end of a string is whirled with constant speed in a counterclockwise horizontal circle. At point A in the circle, the string breaks. Which of the curves sketched below (as seen from above) most accurately represents the path that the ball will take after the string breaks?

A) 3 B) 4 C) 2 D) 1

(161 Major 1, Q12)

A) 3 B) 4 C) 2 D) 1

(161 Major 1, Q12)

Solutions • This question is basically application of Newton's 1st law: an object moving in constant velocity (i.e. constant speed in straight line) will continue at same velocity if No net force act on it • at point a, when The string break, the ball was travelling in direction 3. • once broken string, no more net force is acting on the ball • So now, Newton's first law applies and it will continue travelling in direction 3 continuously

3.2 Newtonâ&#x20AC;&#x2122;s Laws of Motion 3. A ball on the end of a string is whirled with constant speed in a counterclockwise horizontal circle. At point A in the circle, the string breaks. Which of the curves sketched below (as seen from above) most accurately represents the path that the ball will take after the string breaks? A) 4 B) 2 C) 3

D) 1 (152 Major 1, Q10)

Solutions Same question!!!

3.2 Newtonâ&#x20AC;&#x2122;s Laws of Motion 4. What term refers to the tendency of an object to remain at rest or in uniform, straight-line motion?

A) Mass. B) Inertia.

C) Force. D) Momentum. E) Weight. (151 Major 1, Q9)

3.2 Newtonâ&#x20AC;&#x2122;s Laws of Motion 4. What term refers to the tendency of an object to remain at rest or in uniform, straight-line motion?

A) Mass. B) Inertia.

C) Force. D) Momentum. E) Weight. (151 Major 1, Q9)

Solutions

MEMORISE THIS! • Inertia is the tendency of an object: (1) if in uniform, straight-line motion, to remain in that uniform, straight-line motion (2) or if at rest, to remain at rest • unless acted upon by a force • Mass is a measure of the amount of inertia

3.2 Newtonâ&#x20AC;&#x2122;s Laws of Motion 5.

Inertia of an object depends on its _______.

A) speed. B) mass. C) momentum. D) weight. E) volume. (142 Major 1, Q9)

3.2 Newtonâ&#x20AC;&#x2122;s Laws of Motion 5.

Inertia of an object depends on its _______.

A) speed. B) mass. C) momentum. D) weight. E) volume. (142 Major 1, Q9)

Solutions • Mass (kg) is a measure of the amount of inertia • If you are confused about difference between mass and inertia,

• Think about the words wealth and money • Wealth is a general idea describing the state of a person, just like inertia describe the property of an object

• But money ($$$) is a numerical count of how much Wealth a person has, just like mass is a measurement of any body in kg. • So inertia = wealth • & Mass = money

3.2 Newtonâ&#x20AC;&#x2122;s Laws of Motion 6. A 4500-kg helicopter accelerates upward at 2.0 m/s2. What lift force is exerted by the air on the helicopter?

A) 13,500 N. B) 35,100 N.

C) 53,100 N. D) 9,000 N. (171 Major 1, Q9)

3.2 Newtonâ&#x20AC;&#x2122;s Laws of Motion 6. A 4500-kg helicopter accelerates upward at 2.0 m/s2. What lift force is exerted by the air on the helicopter?

A) 13,500 N. B) 35,100 N.

C) 53,100 N. D) 9,000 N. (171 Major 1, Q9)

Solutions • Fnet = Flift – W = Flift – mg • From 2nd Law, Fnet = ma

Lift force = F

• Combining these 2, • Flift – mg = ma • Flift = mg + ma = 4500(9.8) + 4500(2.0) • Flift = 53100 N W = mg

3.2 Newtonâ&#x20AC;&#x2122;s Laws of Motion 7. Two blocks of different masses (10 kg and 20 kg) are connected by a light rope on a frictionless surface as in the figure below. An applied force of 60 N [right] causes the blocks to accelerate. What is the acceleration of the 20-kg block? A) 3.0 m/s2

B) 6.0 m/s2 C) Zero

D) 2.0 m/s2 (162 Major 1, Q15)

B) 6.0 m/s2 C) Zero

D) 2.0 m/s2 (162 Major 1, Q15)

Solutions • First find acceleration of the whole system (both blocks) using F = ma • a = F/m = 60/(10 + 20) = 2.0 m/s2 • Now, both blocks are obviously accelerating at the same rate, otherwise the string will break

• So a10 kg = a20 kg = 2.0 m/s2

3.2 Newtonâ&#x20AC;&#x2122;s Laws of Motion 8. A 5-kg box is acted upon by four forces as shown in the figure below, what is the acceleration of the box? A) 6 m/s2, upward B) 10 m/s2, upward

C) Zero D) 4 m/s2, downward

(161 Major 1, Q7)

3.2 Newtonâ&#x20AC;&#x2122;s Laws of Motion 8. A 5-kg box is acted upon by four forces as shown in the figure below, what is the acceleration of the box? A) 6 m/s2, upward B) 10 m/s2, upward

C) Zero D) 4 m/s2, downward

(161 Major 1, Q7)

Solutions â&#x20AC;¢ EASY! Do it yourself!!!

3.2 Newtonâ&#x20AC;&#x2122;s Laws of Motion 9. According to Newtonâ&#x20AC;&#x2122;s second law of motion, when an object is acted upon by unbalanced forces, what can be said about its acceleration? A) It can have no acceleration. B) It is directly proportional to its mass. C) It is inversely proportional to the net force. D) It is directly proportional to its mass and inversely proportional to the net force. E) It is directly proportional to the net force and inversely proportional to its mass. (151 Major 1, Q14)

Solutions • Newton’s second law of motion says: Fnet = ma • So a = Fnet/m • a ∝ Fnet (acceleration directly proportional to the net force) • a ∝ 1/m (acceleration inversely proportional to mass)

3.2 Newtonâ&#x20AC;&#x2122;s Laws of Motion 10. You push yourself on a skateboard with a force of 30 N east and accelerate at 0.5 m/s2 east. What is the mass of the skateboard if your mass is 58 kg? A) 60 kg

B) 15 kg C) 2.0 kg

D) 4.0 kg (151 Final, Q3)

B) 15 kg C) 2.0 kg

D) 4.0 kg (151 Final, Q3)

Solutions â&#x20AC;¢ EASY! Do it yourself!!!

3.2 Newtonâ&#x20AC;&#x2122;s Laws of Motion 11. A 50-kg boy is standing on a skateboard near a wall. If the boy pushes on the wall by a 25-N force, what would be his acceleration away from the wall? A) Cannot be found. B) Zero.

C) 2.0 m/s2. D) 1250 m/s2. E) 0.50 m/s2. (142 Major 1, Q11)

Solutions â&#x20AC;¢ EASY! Do it yourself!!!

3.2 Newton’s Laws of Motion 12. The figure below shows a 5.0-kg block “A” being pushed with a 3.0-N force. In front of this block is a 10-kg block “B”; the two blocks move together. How much force does block “A” exerts on block “B”? A) 2.0 N B) 6.0 N C) 3.0 N D) 1.0 N

E) 4.0 N (171 Final, Q9)

Solutions • First find acceleration of the whole system (block A and B) using F = ma • a = F/m = 3/(5+10) = 0.2 m/s2 • Now find net force on block B • FB = maB = 10(0.2) = 2.0 N

• Now, the only force on block B in the horizontal Direction is coming from block A • So, Force on block B by block a = 2.0 N •

3.2 Newtonâ&#x20AC;&#x2122;s Laws of Motion 13. When a boy pulls a cart, the force that causes him to move forward is A) the force he exerts on the cart. B) the force the ground exerts on his feet.

C) the force the cart exerts on him. D) the force he exerts on the ground with his feet. (162 Major 1, Q19)

3.2 Newtonâ&#x20AC;&#x2122;s Laws of Motion 13. When a boy pulls a cart, the force that causes him to move forward is A) the force he exerts on the cart. B) the force the ground exerts on his feet.

C) the force the cart exerts on him. D) the force he exerts on the ground with his feet. (162 Major 1, Q19)

Solutions The force that causes HIM to move forward obviously must be acting on HIM. So (a) “on the cart” and (d) “on the ground” are out. As for (c) the force the cart exerts on him is acting backwards, so obviously it cannot make him move forward. Leaving (b) “the force the ground exerts on his feet” as the only correct answer, because he pushes the ground backwards, so the ground pushes HIM forward.

3.2 Newtonâ&#x20AC;&#x2122;s Laws of Motion 14. Which is true of the force pair of Newtonâ&#x20AC;&#x2122;s third law? A) The two forces never produce an acceleration. B) The two forces always cancel each other.

C) The two forces act on different objects. D) The two forces are in the same direction.

(161 Major 1, Q18)

C) The two forces act on different objects. D) The two forces are in the same direction.

(161 Major 1, Q18)

Solutions

MEMORISE THIS! â&#x20AC;˘ Newtonâ&#x20AC;&#x2122;s 3rd law states that forces occur in pairs that are: 1. Equal in magnitude

2. opposite in direction 3. always acting on different objects

3.3 Using Newtonâ&#x20AC;&#x2122;s Laws of Motion

3.3 Using Newtonâ&#x20AC;&#x2122;s Laws of Motion 1. Which of the following is true about an object in free fall? A) Its acceleration depends on its mass. B) It pulls on Earth, and Earth pulls on it.

C) It has no inertia. D) Its momentum is constant.

(151 Final, Q4)

C) It has no inertia. D) Its momentum is constant.

(151 Final, Q4)

Solutions • An object in free fall is • being pulled by the Earth (action), and • by 3rd law of motion, it is pulling the Earth (reaction). • Now we explore all the options…

Solutions A) Think of stone and feather, if no air resistance, they fall at same acceleration (g = 9.8 ms-2). Mathematically, F = ma, so a = F/m = N/kg.

A bigger object has more gravitational force (F) but also has more mass (m), so they cancel out each other and acceleration remains constant in free fall B) Remember Newton's third law, for every action (earth pull apple), there is equal and opposite reaction (apple pull earth) C) It has mass, so has inertia. D) In free fall, it is accelerating, so velocity increasing, so momentum, p = mv is also increasing

3.3 Using Newtonâ&#x20AC;&#x2122;s Laws of Motion 2. Which of the following is TRUE about an object in free fall?

A) Its momentum is constant. B) Its acceleration depends on its mass.

C) It has no inertia. D) It pulls on Earth, and Earth pulls on it. E) Its initial velocity is always zero. (142 Final, Q24)

3.3 Using Newtonâ&#x20AC;&#x2122;s Laws of Motion 2. Which of the following is TRUE about an object in free fall?

A) Its momentum is constant. B) Its acceleration depends on its mass.

C) It has no inertia. D) It pulls on Earth, and Earth pulls on it. E) Its initial velocity is always zero. (142 Final, Q24)

Solutions Same question!!!

3.3 Using Newtonâ&#x20AC;&#x2122;s Laws of Motion 3. Ahmad skydives and parachutes from a stationary helicopter. What is the acceleration of Ahmad when the air resistance equals twice his weight? A) 2.0 m/s2, upward

B) 9.8 m/s2, upward. C) 9.8 m/s2, downward.

D) 12 m/s2, downward (161 Major 1, Q10)

How to solve skydiving questions? For Skydiving questions, try to understand the story • when a person is falling from the sky to earth, there are two forces acting on him – gravity (down) and air resistance (up). • gravitational force = weight = mg

• air resistance depend on both area and speed.

How to solve skydiving questions? • air resistance depend on both area and speed. • when a person first jump from aeroplane, his speed is zero so air resistance = zero • gravitational force makes him accelerate down, his speed increases and so air resistance increases continuously until air resistance (up) = weight (down) • at this point, the net force on person = 0, so he falls at constant speed known as terminal velocity (v1)

How to solve skydiving questions? â&#x20AC;˘ after sometime, he opens his big parachute, so his area suddenly increases to very large â&#x20AC;˘ since air resistance depend on area, when the big parachute is open, his air resistance suddenly becomes very big and now air resistance (up) > gravitational force (down) â&#x20AC;˘ So, now the net force on him is upwards, and his downward speed will continue decreasing until the air resistance becomes smaller and smaller so that this equation

How to solve skydiving questions? • air resistance (up) = gravitational force (down)

• becomes balanced again. • at this point, he reaches a new terminal velocity (v2) with the parachute open, that is smaller than the old terminal velocity (v1) with the parachute closed.

• These are the main ideas of skydiving questions, so understand them well

B) 9.8 m/s2, upward. C) 9.8 m/s2, downward.

D) 12 m/s2, downward (161 Major 1, Q10)

B) 9.8 m/s2, upward. C) 9.8 m/s2, downward.

D) 12 m/s2, downward (161 Major 1, Q10)

Solutions Net force, Fnet = 2mg (up) – mg (down) = mg (up)

Air resistance = 2W = 2mg

Newton’s 2nd law: Fnet = ma So a = F/m = mg/m = g = 9.8 m/s2 Direction of acceleration = direction of Net force = up So a = 9.8 m/s2 up

W = mg

3.3 Using Newtonâ&#x20AC;&#x2122;s Laws of Motion 4. A sky diver with a mass of 80.0 kg jumps from an airplane. Five seconds after jumping, the force of air resistance on the sky diver is 300 N. What is the sky diverâ&#x20AC;&#x2122;s acceleration five seconds after jumping? A) 9.80 m/s2 downward.

B) 3.75 m/s2, downward. C) 2.75 m/s2, upward.

D) 6.05 m/s2 downward.

(152 Final, Q5)

B) 3.75 m/s2, downward. C) 2.75 m/s2, upward.

D) 6.05 m/s2 downward.

(152 Final, Q5)

Solutions Net force, Fnet = 784 N (down) – 300 N (up) = 484 N (down)

Air resistance = 300 N

Newton’s 2nd law: Fnet = ma So a = F/m = 484/30 m/s2 = 6.05 m/s2 Direction of acceleration = direction of Net force = down So a = 9.8 m/s2 down

W = mg = 80 x 9.8 N = 784 N

3.3 Using Newtonâ&#x20AC;&#x2122;s Laws of Motion 5. After a skydiver has reached his terminal velocity he opens his parachute. The air resistance is now 1180 N. If the mass of the skydiver and his parachute is 80.0 kg, his acceleration is A) 4.95 m/s2, upward. B) 14.75 m/s2, upward. C) 14.75 m/s2, downward. D) 9.80 m/s2, downward. E) 4.95 m/s2, downward. (151 Major 1, Q15)

Solutions Net force, Fnet = 1180 N (up) – 784 N (down) Air resistance = 1180 N

= 396 N (up) Newton’s 2nd law: Fnet = ma So a = F/m = (396/80) m/s2 = 4.95 m/s2 Direction of acceleration = direction of Net force = up So a = 4.95 m/s2 up

W = mg = (80 x 9.8) N = 784 N

3.3 Using Newtonâ&#x20AC;&#x2122;s Laws of Motion 6. A sky diver jumps from a helicopter and opens his parachute. If the mass of the sky diver and his parachute is m, what is his acceleration when the air resistance opposing his motion equals half of his weight? A) Zero. B) 4.9 m/s2. C) 9.8 m/s2. D) 20 m/s2.

E) 15 m/s2. (142 Major 1, Q10)

Solutions Net force, Fnet = mg (down) – 0.5mg (up) Air resistance = 0.5W = 0.5 mg = 0.5mg (down) Newton’s 2nd law: Fnet = ma So a = F/m = 0.5mg/m = 0.5g = 4.95 m/s2 Direction of acceleration = direction of Net force = down So a = 4.95 m/s2 down

W = mg

3.3 Using Newtonâ&#x20AC;&#x2122;s Laws of Motion 7. At a given instant in time, a 5.0-kg rock is observed to be falling with an acceleration of 7.0 m/s2. What is the magnitude of the force of air resistance acting upon the rock at this instant? A) 49 N B) 14 N C) 35 N D) Zero

(152 Major 1, Q15)

Solutions Net force, Fnet = 49 N (down) – R (up) = 49 – R (down) Newton’s 2nd law: Fnet = ma

Air resistance = R

–

49 – R = 5a = 5 x 7 = 35 So R = 14 N

W = mg = 5(9.8) N = 49 N

3.3 Using Newtonâ&#x20AC;&#x2122;s Laws of Motion 8. A skydiver jumps from a helicopter and opens his parachute. If the mass of the skydiver and his parachute is 100 kg, what is the air resistance acting on him when his acceleration is 4.9 m/s2 downward? A) 980N.

B) 490 N. C) 1470 N.

D) 4900 N. E) 4.9 N.

(142 Final, Q25)

B) 490 N. C) 1470 N.

D) 4900 N. E) 4.9 N.

(142 Final, Q25)

3.3 Using Newton’s Laws of Motion Net force, Fnet = 980 N (down) – R (up) = 980 – R (down)

Air resistance = R

Newton’s 2nd law: Fnet = ma = (100)(4.9) N 980 – R = 490 N So R = 490 N

W = mg = 100(9.8) N = 980 N

3.3 Using Newtonâ&#x20AC;&#x2122;s Laws of Motion 9. Assume you are standing inside an elevator, in which of the following cases your weight will equal zero?

A) When the elevator is in free fall. B) When the elevator accelerates upward.

C) When the elevator moves with constant velocity. D) None of these.

(171 Major 1, Q13)

How to solve elevator questions? For elevator questions, try to understand the story • Normally a person is in an elevator, that is accelerating upward or downward • If acceleration = 0, net force on person = 0 and the scale gives the correct weight of person/object • If acceleration is upward, the scale gives larger value than true weight • If acceleration is downward, the scale gives smaller than true weight

How to solve elevator questions? â&#x20AC;˘ In general there are two types of questions 1. either a person is standing on a weighing scale 2. or we have an object hanging on a spring balance by a hook â&#x20AC;˘ The information from previous page is summarised in this table, which applies to both of the above question types: Acceleration

Scale reading

Correct weight

Larger than correct

Down

Smaller than correct

How to solve elevator questions? Some important points to take note

â&#x20AC;˘ it is only the direction of acceleration that matter, not the direction that in the elevator is moving (velocity) â&#x20AC;˘ for example the elevator can be moving down but accelerating up (when you go from top floor and just reaching the bottom floor) or it can be moving up but accelerating down (opposite case - you go from bottom floor and just reaching at the top floor)

How to solve elevator questions? â&#x20AC;˘ First, we look at person standing on a weighing scale

Normal Reaction = R

â&#x20AC;˘ There are 2 forces acting on this person (1)The weight of person = gravitational pull on person by Earth (down) = W = mg (2) The Normal reaction/contact force on person by floor (up)

W = mg = is FIXED

How to solve elevator questions? • Now, think carefully: • When the elevator is not accelerating (at rest, or moving up or down with constant speed) • the weight of person (down) = normal contact/reaction force (up), so Fnet = 0 and the weighing scale reads the correct weight of the person.

• take note that the weighing scale reading is the normal contact/reaction force value.

Normal Reaction = R

W = mg = is FIXED

How to solve elevator questions? • However, when the elevator is accelerating upward, there is Fnet (up) on person

Normal Reaction = R

• because the ground is pushing up on the person • so normal contact/reaction force (up) > weight of person (down) • So the person accelerate upwards. • in this case the weighing scale (which read the normal contact reaction force) is giving greater than the correct weight of the person.

W = mg = is FIXED

How to solve elevator questions? • when the elevator is accelerating downwards, the exact opposite happens. • now the ground is falling downwards away from person

Normal Reaction = R

• hence there is less normal contact/reaction force on the person from the ground, • so normal contact/reaction force (up) < weight of person (down) • so Fnet is downward and person accelerates down. • weighing scale reads smaller than correct weight of person.

W = mg = is FIXED

How to solve elevator questions? To summarise everything in a few words • the weight is always, always fixed and acting on person downwards • the normal contact/reaction force is always up but its value will depend on the acceleration of the elevator • if acceleration = 0, normal contact/reaction force = weight and weighing scale gives correct weight • if acceleration is upward, the ground is pushing up on the person more than the weight is pulling down, so normal contact/reaction force > weight, so weighing scale gives larger than correct weight • if acceleration is downward, the ground is falling downward away from the person, so normal contact/reaction force < weight, so weighing scale reads smaller than correct weight

How to solve elevator questions? 2 types of elevator question • either a person standing on a weighing scale, or an object is hanging on a spring balance by the hook • they both have the exact same solution

• the only difference is: for the person on weighing scale, the upward force is the normal contact/reaction force but for object hanging on a Spring balance, the upward force is the spring force • weight reading is exactly the same in both cases, i.e. depending on the direction of acceleration (up = bigger, down = smaller).

3.3 Using Newtonâ&#x20AC;&#x2122;s Laws of Motion 9. Assume you are standing inside an elevator, in which of the following cases your weight will equal zero?

A) When the elevator is in free fall. B) When the elevator accelerates upward.

C) When the elevator moves with constant velocity. D) None of these.

(171 Major 1, Q13)

3.3 Using Newtonâ&#x20AC;&#x2122;s Laws of Motion 9. Assume you are standing inside an elevator, in which of the following cases your weight will equal zero?

A) When the elevator is in free fall. B) When the elevator accelerates upward.

C) When the elevator moves with constant velocity. D) None of these.

(171 Major 1, Q13)

Solutions • With all your background knowledge, you can now understand this question easily • Weight = W = mg will only be zero if (a) mass = 0 (and it's impossible not to have any mass!!!)

(b) g = 0 (far, far away in outer space, away from any planet or star) • Hence the weight will never, ever be zero on earth!

Solutions • Yes, if Elevator in free fall, then weighing scale will read zero • Just remember that the weighing scale tells you apparent weight (‫ )الوزن الظاهري‬not real weight (‫)الوزن الحقيقي‬ • And we have already learned that weighing scale gives normal contact/reaction force • So normal contact/reaction force = apparent weight = ‫الوزن الظاهري‬

• when question say weight, it is referring to real weight (‫)الوزن الحقيقي‬ and not apparent weight (‫ )الوزن الظاهري‬from weighing scale

3.3 Using Newtonâ&#x20AC;&#x2122;s Laws of Motion 10. Assume you are standing inside an elevator, in which of the following cases your weight will equal zero? A) When the elevator is in free fall. B) When the elevator accelerates upward. C) When the elevator moves with constant velocity. D) None of these. (161 Major 1, Q16)

Solutions â&#x20AC;¢ Same question!!!

3.3 Using Newtonâ&#x20AC;&#x2122;s Laws of Motion 11. Assume you are standing inside an elevator, in which of the following cases your weight will equal zero? A) When the elevator is in free fall. B) When the elevator accelerates upward.

C) When the elevator moves with constant velocity. D) None of these.

(152 Final, Q6)

C) When the elevator moves with constant velocity. D) None of these.

(152 Final, Q6)

Solutions â&#x20AC;¢ Same question!!!

3.3 Using Newtonâ&#x20AC;&#x2122;s Laws of Motion 12. Assume you are standing on a bathroom scale inside an elevator, in which of the following cases the reading of the scale will equal zero? A) When the elevator accelerates upward.

B) When the elevator is in free fall. C) When the elevator moves with constant velocity.

D) None of these (161 Final, Q27)

B) When the elevator is in free fall. C) When the elevator moves with constant velocity.

D) None of these (161 Final, Q27)

Solutions Net force, Fnet = W (down) – O (up) = W (down) = mg (down) Newton’s 2nd law: Fnet = ma

Normal contact, R=0N

So a = F/m = mg/m = g = 9.8 m/s2

Direction of acceleration = direction of Net force = down

So when a = 9.8 m/s2 down (free fall), R = O N, so weighing scale will read zero

W = mg

3.3 Using Newtonâ&#x20AC;&#x2122;s Laws of Motion 13. You are riding in an elevator holding a spring scale with a 1.0-kg mass suspended from it. You look at the scale and see that it reads 5.0 N. What can you conclude about the elevatorâ&#x20AC;&#x2122;s motion at that time? A) The elevator is accelerating upward.

B) The elevator is accelerating downward. C) The elevator is moving with constant velocity.

D) The elevator is at rest. (171 Major 1, Q14)

B) The elevator is accelerating downward. C) The elevator is moving with constant velocity.

D) The elevator is at rest. (171 Major 1, Q14)

Solutions • Using the rule (up = bigger, down = smaller) • Since weighing scale (5 N) < real weight (9.8 N), • We get acceleration = down

3.3 Using Newtonâ&#x20AC;&#x2122;s Laws of Motion 14. A 60.0-kg person is standing on a scale in an elevator that is accelerating downward at 1.70 m/s2. What is the reading of the scale? A) 102 N. B) 486 N.

C) Zero. D) 690 N.

E) 588 N. (171 Final, Q3)

C) Zero. D) 690 N.

E) 588 N. (171 Final, Q3)

Solutions Net force, Fnet = W (down) – R (up) = (60 x 9.8) N (down) – R (up) Newton’s 2nd law: Fnet = ma

Normal contact, R

So 588 – R = 60(1.7) R = 486 N, so weighing scale read 486 N W = mg

3.3 Using Newtonâ&#x20AC;&#x2122;s Laws of Motion 15. A person is on an elevator that moves downward with an acceleration of 1.8 m/s2. If the person weighs 686 N, what is the net force on the person? A) 560 N.

B) 686 N. C) 126 N.

D) 812 N. (162 Final, Q27)

B) 686 N. C) 126 N.

D) 812 N. (162 Final, Q27)

Solutions • This is an even easier Question • whenever question asked to find net force, always try to use Newton's second law if possible • Net force, Fnet = manet • From W = mg, m = W/g = 686/9.8 = 70 kg • Fnet = manet = 70(1.8) N = 126 N

Important: The question says “the person weighs 686 N” this is real weight, which is the one used in the formula W = mg

3.3 Using Newtonâ&#x20AC;&#x2122;s Laws of Motion 16. A student stands on a bathroom scale in an elevator at rest on the 2nd floor of a five-story building. The scale reads 833 N. As the elevator moves up, the scale reading increases to 936 N. What is the acceleration of the elevator? A) 11.0 m/s2

B) 103 m/s2 C) 1.21 m/s2

D) 1.07 m/s2

(162 Major 1, Q14)

B) 103 m/s2 C) 1.21 m/s2

D) 1.07 m/s2

(162 Major 1, Q14)

Solutions Important: The question says “at rest… The scale reads 833 N” this is real weight, which is the one used in the formula W = mg. Since elevator is accelerating upward, so R (up) > W (down), so net force, Fnet = R (up) – W (down)

= 936 – 833 = 103 N

Normal contact, R=0N

From W = mg, m = W/g = 833/9.8 = 85 kg

Newton’s 2nd law: Fnet = ma So a = Fnet /m = 103/85 = 1.21 m/s2

W = mg

3.3 Using Newtonâ&#x20AC;&#x2122;s Laws of Motion 17. A 100-kg man in an elevator is accelerating upward at a rate of 1.0 m/s2. What is the force pushing upward on the manâ&#x20AC;&#x2122;s feet? A) 980 N

B) 880 N C) 1080 N

D) 100 N (161 Major 1, Q13)

B) 880 N C) 1080 N

D) 100 N (161 Major 1, Q13)

Solutions force pushing upward on the man’s feet = normal contact/reaction force = weighing scale = R No elevator is accelerating upward, so R (up) > W (down), so net force, Fnet = R (up) – W (down) Normal contact, R

= R (up) – (100 x 9.8) N Newton’s 2nd law: Fnet = ma

= (100 kg)(1.0 m/s2) = 100 N (up) Combining 2 above equations, R – 980 = 100 R = 1080 N, so force pushing up on feet = 1080 N

W = mg

3.3 Using Newtonâ&#x20AC;&#x2122;s Laws of Motion 18. A 100-kg man in an elevator is accelerating upward at a rate of 1.0 m/s2. What is the force pushing upward on the manâ&#x20AC;&#x2122;s feet? A) 980 N

B) 880 N C) 100 N

D) 1080 N (152 Major 1, Q11)

B) 880 N C) 100 N

D) 1080 N (152 Major 1, Q11)

Solutions â&#x20AC;¢ Same question!!!

3.3 Using Newtonâ&#x20AC;&#x2122;s Laws of Motion 19. As seen from above, a car rounds a curved path as shown in the figure below at constant speed. Which direction best represents the direction of the net force acting on the car? A) (IV) B) (III)

C) (II) D) (V) E) (I)

(171 Final, Q2)

C) (II) D) (V) E) (I)

(171 Final, Q2)

Solutions For centripetal acceleration questions • just remember that any object moving in a circle at a constant speed, has: • centripetal acceleration of value ac = v2/r (magnitude)

• pointing towards centre of circle, and perpendicular to velocity (direction) • For a car, this centripetal force is provided by the friction between tires and road

Solutions For centripetal acceleration questions â&#x20AC;˘ For a car, the centripetal force is provided by the friction between tires and road (see textbook p90 last paragraph for details)

3.3 Using Newtonâ&#x20AC;&#x2122;s Laws of Motion 20. A car rounds a curve on a level road. The centripetal force on the car is provided by A) friction between the tires and the road. B) inertia.

C) gravity. D) the force applied to the steering wheel.

(161 Final, Q28)

3.3 Using Newtonâ&#x20AC;&#x2122;s Laws of Motion 20. A car rounds a curve on a level road. The centripetal force on the car is provided by A) friction between the tires and the road. B) inertia.

C) gravity. D) the force applied to the steering wheel.

(161 Final, Q28)

3.3 Using Newtonâ&#x20AC;&#x2122;s Laws of Motion 21. How does a safety belt or an air bag reduce injuries in car crashes? A) It decreases the time it takes to slow down the passenger. B) It increases the momentum of the passenger.

C) It increases the force acted on the passenger. D) It increases the time it takes to slow down the passenger.

E) It increases the acceleration of the passenger. (151 Major 1, Q17)

C) It increases the force acted on the passenger. D) It increases the time it takes to slow down the passenger.

E) It increases the acceleration of the passenger. (151 Major 1, Q17)

Solutions â&#x20AC;˘ This is an application of Newton's third law of motion. â&#x20AC;˘ Use this formula (textbook p91)

VERY IMPORTANT!!!

đ??&#x2026; â&#x2C6;&#x2020;đ?? = đ??Ś â&#x2C6;&#x2020;v â&#x20AC;˘ đ??Ś â&#x2C6;&#x2020;v obviously cannot change, because your mass is fixed, and v depends on your driving speed, and airbag cant change them â&#x20AC;˘ What air bag does is to make â&#x2C6;&#x2020;đ?? very long, so the force F from car on you becomes small and you donâ&#x20AC;&#x2122;t get badly injured in if accident happens â&#x20AC;˘ Remember â&#x20AC;&#x201C; bigger force causes worst injury!!!

Solutions

VERY IMPORTANT!!!

đ??&#x2026; â&#x2C6;&#x2020;đ?? = đ??Ś â&#x2C6;&#x2020;v â&#x20AC;˘ If you are still confused, think about catching a ball thrown at your hand. â&#x20AC;˘ You will move your hand backwards/downwards to make â&#x2C6;&#x2020;đ?? bigger so that F becomes smaller. â&#x20AC;˘ if You keep your hands tight and rigid, the â&#x2C6;&#x2020;đ?? will be very short leading to a large F, so very painful ball catching.

3.3 Using Newtonâ&#x20AC;&#x2122;s Laws of Motion 22. As a baseball is being caught, its speed goes from 30.0 m/s to 0.0 m/s in about 0.005 s. If the mass of the baseball is 0.145 kg, how much force acts on it? A) 6,000 N.

B) 4.35 N. C) 29.0 N.

D) 870 N. (171 Major 1, Q11)

Solutions â&#x20AC;˘ This is an application of Newton's third law of motion. â&#x20AC;˘ Use this formula (textbook p91)

VERY IMPORTANT!!!

đ??&#x2026; â&#x2C6;&#x2020;đ?? = đ??Ś â&#x2C6;&#x2020;v â&#x20AC;˘ đ??&#x2026; â&#x2C6;&#x2020;đ?? = đ??Ś (đ??Żđ??&#x; â&#x2C6;&#x2019; đ??Żđ??˘ ) â&#x20AC;˘ So đ??&#x2026; =

đ??Ś (đ??Żđ??&#x; â&#x2C6;&#x2019;đ??Żđ??˘ ) â&#x2C6;&#x2020;đ??

đ?&#x;&#x17D;.đ?&#x;?đ?&#x;&#x2019;đ?&#x;&#x201C; đ??¤đ?? đ?&#x;&#x2018;đ?&#x;&#x17D;.đ?&#x;&#x17D; â&#x2C6;&#x2019; đ?&#x;&#x17D;.đ?&#x;&#x17D; đ??Ś/đ??Ź đ?&#x;&#x17D;.đ?&#x;&#x17D;đ?&#x;&#x17D;đ?&#x;&#x201C; đ?&#x2019;&#x201D;

= đ?&#x;&#x2013;đ?&#x;&#x2022;đ?&#x;&#x17D; đ???

B) 4.35 N. C) 29.0 N.

D) 870 N. (171 Major 1, Q11)

3.3 Using Newtonâ&#x20AC;&#x2122;s Laws of Motion 23. A 60-kg front-seat passenger in a car moving initially with a speed of 18 m/s is brought to rest by an air bag in a time of 0.4 s. What is the average force acting on the passenger in this process? A) 1,080 N B) 150 N C) 45 N D) 2,700 N (161 Major 1, Q14)

Solutions • Again use

𝐅 ∆𝐭 = 𝐦 ∆v • 𝐅 ∆𝐭 = 𝐦 (𝐯𝐟 − 𝐯𝐢 ) • So 𝐅 =

𝐦 (𝐯𝐟 −𝐯𝐢 ) ∆𝐭

𝟔𝟎 𝐤𝐠 𝟏𝟖.𝟎 − 𝟎.𝟎 𝐦/𝐬 𝟎.𝟒 𝒔

= 𝟐𝟕𝟎𝟎 𝐍

3.3 Using Newtonâ&#x20AC;&#x2122;s Laws of Motion 24. A 60-kg front-seat passenger in a car moving initially with a speed of 18 m/s is brought to rest by an air bag in a time of 0.4 s. What is the average force acting on the passenger in this process? A) 1,080 N B) 150 N C) 2,700 N D) 45 N

(152 Major 1, Q12)

3.3 Using Newtonâ&#x20AC;&#x2122;s Laws of Motion 25. A 1-kg object is moving west at a constant speed of 10 m/s. A force of 100 N, along the east direction, is applied on the object for a time of 0.01 s. The momentum of the object at the end of the 0.01 s interval is A) 9.0 kg.m/s east.

B) 9.0 kg.m/s west. C) 1.0 kg.m/s west. D) 11 kg.m/s east.

(161 Major 1, Q19)

B) 9.0 kg.m/s west. C) 1.0 kg.m/s west. D) 11 kg.m/s east.

(161 Major 1, Q19)

Solutions â&#x20AC;˘ Again use đ??&#x2026; â&#x2C6;&#x2020;đ?? = đ??Ś â&#x2C6;&#x2020;v â&#x20AC;˘ But this time notice we can write mv as p: đ??&#x2026; â&#x2C6;&#x2020;đ?? = đ??Ś đ??Żđ??&#x; â&#x2C6;&#x2019; đ??Żđ??˘ = đ??Šđ??&#x; â&#x2C6;&#x2019; đ??Šđ??˘ â&#x20AC;˘ So đ??Šđ??&#x; â&#x2C6;&#x2019; đ??Šđ??˘ = đ??&#x2026; â&#x2C6;&#x2020;đ?? â&#x20AC;˘ đ??Šđ??&#x; = đ??Šđ??˘ + đ??&#x2026; â&#x2C6;&#x2020;đ?? = 1 kg x 10 m/s (west) + 100 N (east) x 0.01 s = 10 kg m/s (west) + 1 N s (east) = 10 kg m/s (west) + 1 kg m/s (east) (kg m/s = N s)

Solutions = 10 kg m/s (west) + 1 kg m/s (east) (kg m/s = N s are same units) = 10 kg m/s (west) – 1 kg m/s (west) = 9 kg m/s (west)

(change + to – when change direction)

3.3 Using Newtonâ&#x20AC;&#x2122;s Laws of Motion 26. A 1-kg object is moving west at a constant speed of 10 m/s. A force of 100 N, along the east direction, is applied on the object for a time of 0.01 s. The momentum of the object at the end of the 0.01 s interval is A) 9.0 kg.m/s east. B) 1.0 kg.m/s west.

C) 11 kg.m/s east. D) 11 kg.m/s west.

E) 9.0 kg.m/s west. (142 Major 1, Q13)

C) 11 kg.m/s east. D) 11 kg.m/s west.

E) 9.0 kg.m/s west. (142 Major 1, Q13)

Solutions Same question!!!

3.3 Using Newton’s Laws of Motion 27. A trolley “A” moves with 5.0 m/s toward a trolley “B” of equal mass which is at rest, as in the figure below. The trolleys stick together and move off as one. What is their speed after they stuck together? A) 5.0 m/s

B) Zero. C) 2.5 m/s.

D) 10 m/s (171 Major 1, Q15)

B) Zero. C) 2.5 m/s.

D) 10 m/s (171 Major 1, Q15)

Solutions â&#x20AC;˘ This is a simple conservation of momentum question đ??Š đ??&#x; = đ??Šđ??˘ â&#x20AC;˘ Which states, Total initial Momentum (đ??Šđ??˘ ) = total final Momentum (đ??Šđ??&#x; ) if no net external force acts on the system

â&#x20AC;˘ đ??Šđ??˘ =

đ??Ś đ?&#x;&#x201C; đ??Ź

(m kg) [only A moving at 5 m/s]

â&#x20AC;˘ đ??Šđ??&#x; = đ?&#x;&#x17D; = đ?&#x;?đ??Ś đ??¤đ??

đ??Ś đ??Ż đ??Ź

[both moving at v m/s]

â&#x20AC;˘ So 5m = 2mv ď&#x192; v = 2.5 m/s

3.3 Using Newtonâ&#x20AC;&#x2122;s Laws of Motion 28. Two frictionless trolleys A and B, of mass m and 3m respectively, are on a horizontal track as in the figure below. Initially they are clipped together by a device which incorporates a spring, compressed between the trolleys. At time=0 the clip is released, what is the acceleration of trolley A compared to the acceleration of trolley B? A) aA = 9aB B) aA = 3aB đ?&#x;? đ?&#x;&#x2018;

C) aA = aB

D) aA = aB (142 Final, Q15)

C) aA = aB

D) aA = aB (142 Final, Q15)

Solutions • This one uses both Newton's second and third law. • from 3rd law, we know that the forces acting on both trolley is equal in magnitude and opposite in direction • From 2nd law, F = ma, we get a = F/m. • So if mass is 3 times bigger for B, then acceleration will be 3 times smaller for B

Solutions 29.

Which of the following statements is FALSE?

A) When you drop a feather and a hammer from the same height, the feather will reach its terminal velocity first. B) If you push horizontally on a table and the table doesnâ&#x20AC;&#x2122;t move, then the static friction between the floor and the table is greater than your pushing force.

C) A bullet fired horizontally from a rifle is in free fall from the moment it leaves the rifle until it hits the ground below, if air resistance is negligible. D) The centripetal force that acts on a satellite revolving around Earth in a circular path is caused by its weight. (171 Major 1, Q17)

Solutions 29.

Which of the following statements is FALSE?

3.3 Using Newtonâ&#x20AC;&#x2122;s Laws of Motion 30. A rifle with a mass of 1.2 kg fires a bullet with a mass of 6.0 g (0.006 kg). The bullet moves with a velocity of 600 m/s after the rifle is fired. If external forces acting on the rifle can be ignored, what is the recoil velocity of the rifle? A) 600 m/s B) 3.0 m/s C) 3.6 m/s D) 0.0 m/s

(161 Major 1, Q15)

Solutions â&#x20AC;˘ This is a simple conservation of momentum question đ??Šđ??&#x; = đ??Šđ??˘

â&#x20AC;˘ Which states, Total initial Momentum (đ??Šđ??˘ ) = total final Momentum (đ??Šđ??&#x; ) if no net external force acts on the system â&#x20AC;˘ đ??Šđ??˘ = đ?&#x;&#x17D; â&#x20AC;˘ đ??Šđ??&#x; = đ?&#x;&#x17D; = đ?&#x;&#x17D;. đ?&#x;&#x17D;đ?&#x;&#x17D;đ?&#x;&#x201D; đ??¤đ??

đ?&#x;&#x201D;đ?&#x;&#x17D;đ?&#x;&#x17D;

đ??Ś đ??Ź

+ đ?&#x;?. đ?&#x;? đ??¤đ?? (đ??Ż )

â&#x20AC;˘ So v = â&#x2C6;&#x2019; 3.0 m/s

â&#x20AC;˘ Notice the answer is negative, So if we define forward direction as positive (bullet motion is forward), then negative answer means that rifle recoils backwards

3.3 Using Newtonâ&#x20AC;&#x2122;s Laws of Motion 31. An astronaut at rest in space fires a thruster pistol that expels 35.0 g of hot gas at 875 m/s as shown in the figure below. The combined mass of the astronaut and pistol is 84.0 kg. How fast is the astronaut moving after firing the pistol? A) 0.365 m/s

B) 30.6 m/s C) 875 m/s

D) Zero m/s (161 Final, Q29)

B) 30.6 m/s C) 875 m/s

D) Zero m/s (161 Final, Q29)

Solutions â&#x20AC;˘ This is a simple conservation of momentum question đ??Šđ??&#x; = đ??Šđ??˘

đ?&#x;&#x2013;đ?&#x;&#x2022;đ?&#x;&#x201C;

đ??Ś đ??Ź

+ đ?&#x;&#x2013;đ?&#x;&#x2019;. đ?&#x;&#x17D; đ??¤đ?? (đ??Ż )

â&#x20AC;˘ So v = - 0.365 m/s

â&#x20AC;˘ Notice the answer is negative, So if we define forward direction as positive (gas shoots forward), then negative answer means that astronaut moves backwards

3.3 Using Newtonâ&#x20AC;&#x2122;s Laws of Motion 32. A rifle with a mass of 1.2 kg fires a bullet with a mass of 6.0 g (0.006 kg). The bullet moves with a velocity of 600 m/s after the rifle is fired. If external forces acting on the rifle can be ignored, what is the recoil velocity of the rifle?

A) 600 m/s B) 3.0 m/s C) 3.6 m/s D) 0.0 m/s

(152 Major 1, Q13)

A) 600 m/s B) 3.0 m/s C) 3.6 m/s D) 0.0 m/s

(152 Major 1, Q13)

Solutions Same question!!!

3.3 Using Newtonâ&#x20AC;&#x2122;s Laws of Motion 33. A white ball of momentum 4.0 kg.m/s east, hits a black ball so that the total momentum of both balls after collision is 3.0 kg.m/s east. What was the momentum of the black ball before collision? A) 1.0 kg.m/s west B) 7.0 kg.m/s east

C) 3.0 kg.m/s west. D) 4.0 kg.m/s east.

E) 7.0 kg.m/s west. (142 Major 1, Q14)

C) 3.0 kg.m/s west. D) 4.0 kg.m/s east.

E) 7.0 kg.m/s west. (142 Major 1, Q14)

Solutions â&#x20AC;˘ đ??Šđ??˘ = đ?&#x;&#x2019;. đ?&#x;&#x17D; kg m/s east + pblack â&#x20AC;˘ đ??Šđ??&#x; = đ?&#x;&#x2018;. đ?&#x;&#x17D; kg m/s east â&#x20AC;˘ Using đ??Šđ??&#x; = đ??Šđ??˘ , â&#x20AC;˘ we get pblack = đ?&#x;&#x2018;. đ?&#x;&#x17D; kg m/s east â&#x2C6;&#x2019; đ?&#x;&#x2019;. đ?&#x;&#x17D; kg m/s east = â&#x2C6;&#x2019;đ?&#x;?. đ?&#x;&#x17D; kg m/s east = +đ?&#x;?. đ?&#x;&#x17D; kg m/s west

3.3 Using Newtonâ&#x20AC;&#x2122;s Laws of Motion 34.

Which of the following statements is FALSE?

A) When you push horizontally on a desk and the desk doesnâ&#x20AC;&#x2122;t move, the static friction between the desk and the floor is larger than your pushing force. B) Astronauts float inside a spacecraft orbiting Earth because the spacecraft and the astronauts are all in free fall. C) Sliding friction is larger than rolling friction. D) When a car rounds a level curve, friction between the tires and the road acts as the centripetal force. E) If you throw an object in any direction it will be in free fall if there is no air resistance. (151 Major 1, Q12)

3.3 Using Newtonâ&#x20AC;&#x2122;s Laws of Motion 34.

Which of the following statements is FALSE?

3.3 Using Newtonâ&#x20AC;&#x2122;s Laws of Motion 35.

Which of the following statements is FALSE?

A) When a car rounds a level curve, friction between the tires and the road acts as the centripetal force. B) When you push horizontally on a desk and the desk doesnâ&#x20AC;&#x2122;t move, the static friction between the desk and the floor is equal in size but opposite in direction to your pushing force. C) Rolling friction is smaller than sliding friction. D) Astronauts float inside a spacecraft orbiting Earth because the gravitational force acting on them is zero. E) When you throw an object downward, it will be in free fall if there is no air resistance. (142 Major 1, Q12)

3.3 Using Newtonâ&#x20AC;&#x2122;s Laws of Motion 35.

Which of the following statements is FALSE?

4.1 Work and Machines

4.1 Work and Machines 1. The brakes on a car do 240 kJ of work in stopping the car. If the car travels a distance of 40 m while the brakes are being applied, how large is the average force that the brakes exert on the car? A) 9,600 N

B) 6 N C) 6,000 N

D) 9,600,000 N

(151 Final, Q5)

B) 6 N C) 6,000 N

D) 9,600,000 N

(151 Final, Q5)

Solutions • Work done = Force x distance • W = Fd

• F = W/d = 240 kJ/40 m = 6000 N

4.1 Work and Machines 2.

Which of the following statements is TRUE?

A) Chemical potential energy is a mechanical energy. B) The rate at which energy is converted from one form to another is called acceleration.

C) Work doesnâ&#x20AC;&#x2122;t have direction. D) Bicycle is a simple machine.

(161 Final, Q26)

4.1 Work and Machines 2.

Which of the following statements is TRUE?

A) Chemical potential energy is a mechanical energy. B) The rate at which energy is converted from one form to another is called acceleration.

C) Work doesnâ&#x20AC;&#x2122;t have direction. D) Bicycle is a simple machine.

(161 Final, Q26)

4.1 Work and Machines 3. Equal forces are used to move blocks m1 and m2 across the floor. Block m1 has twice the mass of block m2, but block m2 moves twice the distance moved by block m1. Which block, if either, has the greater amount of work done on it? A) Block m2. B) The amount of work done on each block is equal. C) Block m1.

D) More information is needed. (152 Major 1, Q16)

Solutions • Work done = Force x distance; W = Fd • Key word: Equal forces so W ∝ d

• Since block m2 moves twice the distance (d) , therefore it has more work done on it

4.1 Work and Machines 4. A skater moves a horizontal distance of 100 m on a level surface while holding a 10.0 kg box as in the figure below. How much work is done in holding the box? A) 9,800 J.

B) 1000 J. C) 98.0 J.

D) Zero. (171 Major 1, Q21)

4.1 Work and Machines 4. A skater moves a horizontal distance of 100 m on a level surface while holding a 10.0 kg box as in the figure below. How much work is done in holding the box? A) 9,800 J.

B) 1000 J. C) 98.0 J.

D) Zero. (171 Major 1, Q21)

Solutions W = Fd cos q There are 3 cases when no work is done by a force on an object:

1. No force is exerted on the object (F = 0) 2. The object does not move (d = 0), like pushing a big camel!!!

3. The angle between force (F) and direction of movement (d) is 90Â° (so cos 90Â° = 0) For this question, the 3rd case applies

4.1 Work and Machines 5. A person is holding a box that has a mass of 10.0 kg. He walks a distance of 20 m at a constant speed on a level surface. How much work has this person done on the box? A) 200 J. B) 1,960 J.

C) Not enough information. We must know the speed of the person. D) Zero (162 Major 2, Q8)

Solutions W = Fd cos q There are 3 cases when no work is done by a force on an object:

1. No force is exerted on the object (F = 0) 2. The object does not move (d = 0), like pushing a big camel!!!

3. The angle between force (F) and direction of movement (d) is 90Â° (so cos 90Â° = 0) For this question, the 3rd case applies

4.1 Work and Machines 6. A worker slides a 60.0-kg crate up an inclined ramp that is attached to a platform 1.0 m above floor level, as shown in the figure below. A 400.0-N force, parallel to the ramp, is needed to slide the crate up the ramp at a constant speed. Assume that the friction is negligible, what is the length of the ramp (L)? A) 0.15 m. B) 0.68 m. C) 1.47 m.

D) 3.00 m (171 Major 1, Q22)

4.1 Work and Machines 7. A force of 50 N is used to drag a crate 4.0 m across a floor. The force is directed at an angle upward from the crate so that the vertical component of the force is 30 N and the horizontal component is 40 N as shown in the diagram. What is the total work done by the 50-N force? A) 160 J. B) 200 J. C) 120 J.

D) Zero. (152 Major 1, Q17)

Solutions W = Fd cos q Work done (W) = Force (F) x distance moved in direction of force (d cos q)

â&#x20AC;˘ There are 2 ways to solve, so think carefully: â&#x20AC;˘ Either just use the 50 N force alone, or separate it into its 2 components of 30 N and 40 N and find work done by each of them 1. W = Fd cos q = (50 N)(4 m)

đ?&#x;&#x2019;đ?&#x;&#x17D; đ?&#x;&#x201C;đ?&#x;&#x17D;

= 200 x

đ?&#x;&#x2019;đ?&#x;&#x17D; đ?&#x;&#x201C;đ?&#x;&#x17D;

= 160 J

2. Work done by 30 N force = Fd = (30 N)(0 m) = 0 J Work done by 40 N force = Fd = (40 N)(4 m) = 160 J So total = 160 + 0 = 160 J

4.1 Work and Machines 8.

Which of the following is not a simple machine?

A) A pulley. B) A knife. C) A nail clipper. D) A ramp. (162 Major 2, Q9)

4.1 Work and Machines 8.

Which of the following is not a simple machine?

A) A pulley. B) A knife. C) A nail clipper. D) A ramp. (162 Major 2, Q9)

Solutions There are 6 types of simple machines from your book: • Lever, pulley, inclined plane, wheel and axle, screw, wedge • Ramp is basically a inclined plane, they mean same thing • Knife is basically a wedge • Only nail clipper is something complicated, made of more than 1 simple machine, so it’s a compound machine

Solutions • Only nail clipper is something complicated, made of more than 1 simple machine, actually 2 levers as shown at the right • so it’s a compound machine

Earthâ&#x20AC;&#x2122;s Interior 4.1 23) Work and Machines 9. Which of the following cannot be done by a machine? A) Increase force. B) Change direction of a force.

C) Increase work. D) Increase speed.

(152 Major 1, Q19)

Earthâ&#x20AC;&#x2122;s Interior 4.1 23) Work and Machines 9. Which of the following cannot be done by a machine? A) Increase force. B) Change direction of a force.

C) Increase work. D) Increase speed.

(152 Major 1, Q19)

Solutions • See your text book page 111 for details • Machine cannot increase work because it will violate conservation of energy law • If you put in 100 J of energy, you cannot get more than 100 J of work done (conservation of energy)

4.1 Work and Machines 10.

Which of the following is a compound machine?

A) Bicycle. B) Wedge. C) Pulley. D) Screw. E) Inclined plane. (151 Major 1, Q20)

4.1 Work and Machines 10.

Which of the following is a compound machine?

A) Bicycle. B) Wedge. C) Pulley. D) Screw. E) Inclined plane. (151 Major 1, Q20)

Solutions There are 6 types of simple machines from your book: â&#x20AC;˘ Lever, pulley, inclined plane, wheel and axle, screw, wedge

â&#x20AC;˘ Even if you forget the 6 and are confused in the exam, just choose the most complicated machine as the compound machine

4.1 Work and Machines 11. A student exerts a force of 250 N on a lever, through a distance of 1.60 m, as he lifts a 150-kg crate. If the efficiency of the lever is 90.0 percent, how far is the crate lifted? A) 0.245 m

B) 2.40 m C) 0.00245 m

D) 3.31 m (171 Major 1, Q23)

B) 2.40 m C) 0.00245 m

D) 3.31 m (171 Major 1, Q23)

Solutions Work done = Force x distance x efficiency = 250 N x 1.6 m x 90%

= 360 J GPE = mgh = 150 kg x 9.8 N/kg x h = 1470h (height is unknown)

Since work done by student is converted into GPE, 360 = 1470h

h = 360/1470 = 0.245 m

4.1 Work and Machines 12. A force of 1.4 N is exerted through a distance of 40.0 cm on a rope in a pulley system to lift a 0.50 kg mass 10.0 cm. What is the efficiency of the pulley system? A) 87.5% B) 8.93% C) 11.2% D) 2.8%

(162 Major 2, Q2)

Solutions Work done = Force x distance x efficiency = 1.4 N x 0.4 m x X%

= 5.6X J GPE = mgh = 0.5 kg x 9.8 N/kg x 0.1 m = 0.49 J

Since work done is converted into GPE, 5.6X = 0.49

X = 0.49/5.6 = 8.75%

4.1 Work and Machines 13. The electric motor of an elevator uses 630 kJ of electrical energy when raising the elevator and passengers, of total weight 12500 N, through a vertical height of 29 m. What is the efficiency of the elevator? A) 100% B) 1.74% C) 57.5 % D) 19.8%

(161 Major 2, Q5)

Solutions Total electrical energy = 630 000 J GPE gained = mgh = 12500 N x 29 m = 362 500 J

Since a total of 630 000 J of work done, but only 362 500 J of this was usefully converted into GPE, the efficiency (X %) is just: X = energy converted into GPE/total electrical energy = 362 500/630 000 = 57.5%

4.1 Work and Machines 14. Workers do 10,000 J of work on a 2,000 N crate to push it up a ramp. If the ramp is 2 m high, then what is the efficiency of the ramp? A) 40%.

B) 0.4% C) 20%

D) 2.5% (152 Major 1, Q18)

4.1 Work and Machines 14. Workers do 10,000 J of work on a 2,000 N crate to push it up a ramp. If the ramp is 2 m high, then what is the efficiency of the ramp? A) 40%.

B) 0.4% C) 20%

D) 2.5% (152 Major 1, Q18)

Solutions Work done = 10 000 J GPE = mgh = 2000 N x 2 m = 4000 J

Since a total of 10 000 J of work done, but only 4 000 J was usefully converted into GPE, the efficiency (X %) is just: X = energy converted into GPE/total Work done = 4000/10 000 = 40%

4.1 Work and Machines 15. A machine has an efficiency of 20 percent. What is the input work if the output work is 140 J?

A) 7 J. B) 70,000 J.

C) 700 J. D) 2800 J. E) 28 J (142 Final, Q26)

4.1 Work and Machines 15. A machine has an efficiency of 20 percent. What is the input work if the output work is 140 J?

A) 7 J. B) 70,000 J.

C) 700 J. D) 2800 J. E) 28 J (142 Final, Q26)

Solutions Output = Input x efficiency = 140 J Input = Output/efficiency = 140 J/20% = 700 J

4.1 Work and Machines 16. A single, fixed pulley is used to lift a 50-kg object as shown in the figure below. What is the mechanical advantage of the pulley? A) 1.0

B) 0.5 C) 2.0

D) Not enough Information. (162 Major 2, Q7)

4.1 Work and Machines 16. A single, fixed pulley is used to lift a 50-kg object as shown in the figure below. What is the mechanical advantage of the pulley? A) 1.0

B) 0.5 C) 2.0

D) Not enough Information. (162 Major 2, Q7)

Solutions mechanical advantage = Output force/Input force = 50/50 = 1

50 N

4.1 Work and Machines 17. A certain lever has a mechanical advantage of 6.00. How heavy of a load could the lever lift with an input force of 20.0 N? A) 3.33 N.

A) 120 N. C) 0.300 N

D) 20.0 N (152 Major 1, Q20)

4.1 Work and Machines 17. A certain lever has a mechanical advantage of 6.00. How heavy of a load could the lever lift with an input force of 20.0 N? A) 3.33 N.

A) 120 N. C) 0.300 N

D) 20.0 N (152 Major 1, Q20)

Solutions mechanical advantage = Output force/Input force

4.1 Work and Machines 18. A crate weighs 950 N. If you can use a pulley system to lift that crate with a force of only 250 N, then what is the mechanical advantage of the pulley system? A) 30%

B) 50% C) 3.8

D) 0.3 (151 Final, Q6)

4.1 Work and Machines 18. A crate weighs 950 N. If you can use a pulley system to lift that crate with a force of only 250 N, then what is the mechanical advantage of the pulley system? A) 30%

B) 50% C) 3.8

D) 0.3 (151 Final, Q6)

Solutions mechanical advantage = Output force/Input force = 950/250 = 3.8

4.1 Work and Machines 19. An input force of 50.0 N is used to lift an object weighing 200 N with a system of pulleys. What is the mechanical advantage of the pulley system? A) 0.250 B) 4.00 C) 1000 D) 250

E) 150 (142 Major 1, Q16)

Solutions mechanical advantage = Output force/Input force = 200/50 = 4.0