32
NEWTON'S LAWS
[CHAP. 3
ax
Fx 20 N 4:0 m=s2 m 5:0 kg
ay
Fy 30 N 6:0 m=s2 m 5:0 kg
These components of the acceleration are shown in Fig. 3-4. From the ®gure, we see that q a 4:0 2 6:0 2 m=s2 7:2 m=s2 and arctan 6:0=4:0 568:
Fig. 3-4
3.7
A 600 N object is to be given an acceleration of 0.70 m/s2 . How large an unbalanced force must act upon it? Notice that the weight, not the mass, of the object is given. Assuming the weight was measured on the Earth, we use FW mg to ®nd m
FW 600 N 61 kg g 9:81 m=s2
Now that we know the mass of the object (61 kg) and the desired acceleration (0.70 m/s2 ), we have F ma 61 kg 0:70 m=s2 43 N
3.8
A constant force acts on a 5.0 kg object and reduces its velocity from 7.0 m/s to 3.0 m/s in a time of 3.0 s. Find the force. We must ®rst ®nd the acceleration of the object, which is constant because the force is constant. Taking the direction of motion as positive, from Chapter 2 we have a
vf
vi t
4:0 m=s 3:0 s
1:33 m=s2
Now we can use F ma with m 5:0 kg: F 5:0 kg 1:33 m=s2
6:7 N
The minus sign indicates that the force is a retarding force, directed opposite to the motion.