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.
Given that the mass of block is 15 kg , the coefficient of static friction between the block and trolley is 0.18 and the constant acceleration that acts on the trolley is 0.5 m / s 2 for 20 s .
a)
The equation to determine the force experienced by the block is,
F = m a
Substitute the values in the above equation.
F = ( 15 ) ( 0.5 ) = 7.5 N
The equation to determine the friction force experienced by the block is,
F f = μ m g
Substitute the value in the above equation.
F f = ( 0.18 ) ( 15 ) ( 10 ) = 27 N
As the value of the force experienced by the block is less than that of the friction force experienced by the block, so the block will not move.
Thus, the block will remain stationary with respect to the trolley for a stationary observer on the ground.
b)
The observer when moving with the trolley has an accelerated motion. So the system forms a non-inertial frame and the Newton’s law of motion is not applicable.
As both the frames are moving with the same acceleration, so they will be at rest with respect to each other.
Thus, the box would be at rest relative to the observer.
Given that the mass of block is
a)
The equation to determine the force experienced by the block is,
Substitute the values in the above equation.
The equation to determine the friction force experienced by the block is,
Substitute the value in the above equation.
As the value of the force experienced by the block is less than that of the friction force experienced by the block, so the block will not move.
Thus, the block will remain stationary with respect to the trolley for a stationary observer on the ground.
b)
The observer when moving with the trolley has an accelerated motion. So the system forms a non-inertial frame and the Newton’s law of motion is not applicable.
As both the frames are moving with the same acceleration, so they will be at rest with respect to each other.
Thus, the box would be at rest relative to the observer.
