A block of mass 15 kg is placed on a long trolley. The coefficient of static friction between the block and the trolley is 0.18.
A block of mass 15 kg is placed on a long trolley. The coefficient of static friction between the block and the trolley is 0.18. The trolley accelerates from rest with 0.5 m s–2 for 20 s and then moves with uniform velocity. Discuss the motion of the block as viewed by (a) a stationary observer on the ground, (b) an observer moving with the trolley.
3 Answers
Mass of the block, m = 15 kg
Coefficient of static friction, μ = 0.18
Acceleration of the trolley, a = 0.5 m/s2
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 × 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 = μmg
= 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 rest.
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.
An observer, moving with the trolley, has some acceleration. This is the case of noninertial 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.
Mass of the block, m = 15 kg
Coefficient of static friction, μ = 0.18
Acceleration of the trolley, a = 0.5 m/s2
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 × 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 = μmg
= 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 rest.
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.
An observer, moving with the trolley, has some acceleration. This is the case of noninertial 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.
1. Force experienced by block
F = ma = 15 × 0.5 = 7.5 N
Force of friction, Ff = µ mg = 0.18 × 15 × 10 = 27 N
Since the force experienced block is less than frictional force it wilt remain stationary with respect to trolley. For an observer on the ground block appears to move with same acceleration as trolley,
2. For an observer moving with trolley the block appears to he stationary as there is no relative motion between him and trolley and the block.