Question

A block lying on a long horizontal conveyor belt moving at a constant velocity receives a velocity 0
= 5 m/s relative to the ground in the direction opposite to the direction of motion of the conveyor.
After t = 4 s, the velocity of the block becomes equal to the velocity of the belt. The coefficient of
friction between the block and the belt is  = 0.2. The velocity of the conveyor belt is :
a) 3 m/s b) 5 m/s c) 4 m/s d) 7 m/s

Answer 1

HEY MATE

In order to describe the motion of the block, we choose a reference frame fixed to the conveyor belt. Then the velocity of the block at the initial moment is vb = v0 + v, and the block moves with a constant acceleration a=−μg. For the instant of time t when the velocity of the block vanishes, we obtain the equation 0 = v0 + v - μgt.

Here, t = 4 s, g = 10m/s

2

, μ = 0.2.

Hence, the velocity of the conveyor belt is v = μgt - v0 = 3 m/s.

HOPE IT HELPS