A block lying on a long horizontal conveyor belt moving at a constant velocity receives a velocity 5 m/s at t = 0 sec. relative to the ground in the direction opposite to the direction of motion of th

Question

A block lying on a long horizontal conveyor belt moving at a constant velocity receives a velocity 5 m/s at $t = 0$ sec. relative to the ground in the direction opposite to the direction of motion of the conveyor. After $t = 4 sec$ , 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 . Then the velocity of the conveyor belt is : $(g = 10 m / s^{2})$

TestHub · Accepted Answer

Take the belt direction as positive.

Friction on the block (moving opposite to the belt) acts along +ve direction with $μg = 0.2 \times 10 = 2 m/ s^{2}$ .

v(t) = -5 + 2t.

At t = 4 s, v = -5 + 8 = +3 m/s = belt velocity.

Hence (C).

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