Q. 35. A block of mass 15 kg is placed on a long trolly. The coefficient of friction between the block and the
trolly is 0-18. The trolly accelerates from rest with 0.5 m 5-2 for 20 s and then moves with a uniform
velocity. Discuss the motion of the block as viewed by (a) a stationary observer on the ground (b) an
observer moving with the trolly.
Answers
Answer
and
Explanation:
(Question a)
Mass of the block, m = 15 kg
Coefficient of static friction, = 0.18
Acceleration of the trolley, a = 0.5 m/s
As per Newton's second law of motion, the force (F) on the block caused by the motion of the trolley is given by the relation:
F = ma = 15 x 0.5 = 7.5 N
This force is acted in the direction of motion of the trolley.
Force of static friction between the block and the trolley:
f = mg
= 0.18×15× 10 = 27 N
The force of static friction between the block and the trolley is greater than the applied external force. Hence, for an observer on the ground, the block will appear to be at moving with constant acceleration.
When the trolley moves with uniform velocity there will be no applied external force. Only the force of friction will act on the block in this situation.
(Question B) An observer, moving with the trolley, has some acceleration. This is the case of a non-inertial frame of reference. The frictional force, acting on the trolley backward, is opposed by a pseudo force of the same magnitude. However, this force acts in the opposite direction. Thus, the trolley will appear to be at rest for the observer moving with the trolley.