Question 1: A block of 2 kg is placed on a horizontal surface. A force of 10 N is applied to the block horizontally. The coefficient of friction between the block and the surface is 0.3. Determine the acceleration of the block.
Answer: The force acting on the block is F = 10 N, and the mass of the block is m = 2 kg. The coefficient of friction between the block and the surface is μ = 0.3. The normal force acting on the block is N = mg = 2 kg x 9.81 m/s^2 = 19.62 N.
The force of friction acting on the block is Ff = μN = 0.3 x 19.62 N = 5.886 N.
The net force acting on the block is Fnet = F - Ff = 10 N - 5.886 N = 4.114 N.
The acceleration of the block is a = Fnet/m = 4.114 N / 2 kg = 2.057 m/s^2.
Question 2: A spring of spring constant 100 N/m is attached to a block of mass 2 kg. The block is placed on a horizontal surface with no friction. The spring is compressed by 0.2 m and then released. Find the maximum velocity of the block.
Answer : The spring constant of the spring is k = 100 N/m. The mass of the block is m = 2 kg, and the compression of the spring is x = 0.2 m.
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The potential energy stored in the spring when it is compressed is U = (1/2) kx^2 = (1/2) x 100 N/m x (0.2 m)^2 = 2 J. When the spring is released, this potential energy is converted into kinetic energy. The maximum velocity of the block can be found using the conservation of energy:
(1/2)mv^2 = U v^2 = 2U/m v = sqrt(2U/m) = sqrt(2 x 2 J / 2 kg) = 1 m/s.
Question 3: A car of mass 1000 kg is moving at a velocity of 20 m/s. The car hits a wall and comes to a stop in 0.5 seconds. Calculate the average force acting on the car during this time.
Answer: The initial velocity of the car is u = 20 m/s, and the final velocity is v = 0 m/s. The time taken to come to a stop is t = 0.5 s. The mass of the car is m = 1000 kg. Using the equation of motion: v = u + at 0 = 20 + a x 0.5 a = -40 m/s^2
The negative sign indicates that the acceleration is in the opposite direction to the initial velocity.
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The force acting on the car can be calculated using Newton's second law:
F = ma F = 1000 kg x (-40 m/s^2) F = -40000 N
The negative sign indicates that the force is acting in the opposite direction to the initial velocity.
The average force acting on the car during the 0.5 seconds can be calculated using:
Favg = Δp/Δt Δp = mvf - mui = 0 - 1000 kg x 20 m/s = -20000 kg m/s Favg = -20000 kg m/s / 0.5 s = -40000 N
The average force acting on the car during the collision is -40000 N.
Question 4: A box is placed on an inclined plane which makes an angle of 30 degrees with the horizontal. The mass of the box is 10 kg, and the coefficient of friction between the box and the inclined plane is 0.2. Calculate the force required to move the box up the inclined plane with a constant velocity.
Answer: The weight of the box acts vertically downwards and can be resolved into two components: one perpendicular to the inclined plane and one parallel to the inclined plane. The perpendicular component is given by:
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N = mg cos(θ) = 10 kg x 9.81 m/s^2 x cos(30) = 84.848 N
The force of friction acting on the box is given by:
Ff = μN = 0.2 x 84.848 N = 16.97 N
The force required to move the box up the inclined plane with a constant velocity is equal to the component of the weight of the box parallel to the inclined plane plus the force of friction acting in the opposite direction. This force can be calculated using:
F = mg sin(θ) + Ff = 10 kg x 9.81 m/s^2 x sin(30) + 16.97 N = 54.03 N
Therefore, a force of 54.03 N is required to move the box up the inclined plane with a constant velocity.
Question 5: A projectile is fired from a cannon at an angle of 30 degrees with the horizontal. The initial velocity of the projectile is 100 m/s. Neglecting air resistance, calculate the time taken for the projectile to reach its maximum height and the maximum height reached by the projectile.
Answer: The initial velocity of the projectile can be resolved into two components: one parallel to the horizontal and one perpendicular to the horizontal. The horizontal component of velocity remains constant throughout the motion, while the vertical component of velocity changes due to gravity.
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The time taken for the projectile to reach its maximum height can be calculated using: vy = u sin(θ) - gt 0 = 100 m/s x sin(30) - 9.81 m/s^2 x t t = 5.09 s
The maximum height reached by the projectile can be calculated using: y = uyt + (1/2)gt^2 y = 100 m/s x sin(30) x 5.09 s - (1/2) x 9.81 m/s^2 x (5.09 s)^2 y = 127.3 m
Therefore, the time taken for the projectile to reach its maximum height is 5.09 s and the maximum height reached by the projectile is 127.3 m.
Question 6: A block of mass 2 kg is attached to a spring with spring constant 100 N/m. The block is pulled 10 cm to the right and released. Calculate the maximum velocity and acceleration of the block.
Answer: The displacement of the block from its equilibrium position is x = 10 cm = 0.1 m. The mass of the block is m = 2 kg and the spring constant is k = 100 N/m.
The maximum displacement of the block occurs when the block is at its equilibrium position, so the initial velocity of the block is zero. The maximum velocity of the block occurs when the block has moved a distance equal to the amplitude of the motion. The amplitude of the motion is equal to the maximum displacement, which is given by:
A = x = 0.1 m
The maximum velocity of the block can be calculated using:
vmax = ±ωA ω = √(k/m) = √(100 N/m / 2 kg) = 7.07 rad/s vmax = ±7.07 rad/s x 0.1 m = ±0.707 m/s
The maximum velocity of the block is ±0.707 m/s, where the positive sign indicates that the block is moving to the right and the negative sign indicates that the block is moving to the left.
The maximum acceleration of the block occurs when the block is at its equilibrium position and the spring is at its maximum compression or extension. The maximum compression or extension of the spring is equal to the amplitude of the motion, which is given by:
A = x = 0.1 m
The maximum acceleration of the block can be calculated using: amax = ±ω^2A amax = ±(7.07 rad/s)^2 x 0.1 m = ±4.99 m/s^2
The maximum acceleration of the block is ±4.99 m/s^2, where the positive sign indicates that the block is accelerating to the right and the negative sign indicates that the block is accelerating to the left. For