Question

A block lying on a long horizontal conveyor belt moving at a constant velocity receives a velocity 5 m/s relative to the ground in the direction opposite to the conveyor.; After t = 4 sec, the velocity of the block becomes equal to the velocity of the belt. If the coefficient of friction between the block and the belt is 0.2. Then the velocity of conveyor belt is $(g=10m/s^2)$

• A. 13 m/s
• B. -13 m/s
• C. 3 m/s
• D. 6 m/s

Answer 1

The correct option is C 3 m/s

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 $a=-\mu g$. For the instant of time t when the velocity of the block vanishes, we obtain the equation 0 = v0 + v - $\mu gt$.
Here, t = 4 s, g = $10m/s^2$, $\mu$ = 0.2.
Hence, the velocity of the conveyor belt is v = $\mu gt$ - v0 = 3 m/s.