A block of mass 2 kg is given a velocity $v = 10 m s^{- 1}$ towards left and is gently placed on a conveyor belt that is moving towards right with a constant velocity of $u = 10 m s^{- 1}$ . Coefficient of friction between the block and the belt is $μ = 0.5$ . A is an observer moving along with the conveyor belt and B is another observer who is at rest on ground. Both the observers record the event till the block stops slipping on the conveyor. $(g = 10 m / s^{2})$

Which of the following statements are correct?

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Work done by friction on the block as reported by B is zero.

Work done by friction on the block as reported by A is – 400J.

Observer B finds that the power of the motor driving the belt must be increased by $Δ P$ to keep the belt moving with a constant velocity after the block was placed and till it slipped on the belt. Value of $Δ P$ is 100 W.

Work done by friction on the block as reported by A is zero.

answer is B, C, A.

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Detailed Solution

In reference frame of ground, initial velocity of block is $10 m s^{- 1} (\leftarrow)$ and its final velocity, when slipping stops is, $10 m s^{- 1} (\to)$ .
A) $W_{B} = \frac{1}{2} m (v^{2} - u^{2})$ $= \frac{1}{2} (2) (10^{2} - 10^{2})$ = 0
B) In reference frame of belt,
$W_{A} = \frac{1}{2} m (0 - v_{r e l}^{2})$ $= \frac{1}{2} (2) (0 - 20^{2})$ = – 400 J
C) Friction force on the belt (as long as the block slips) is
$f = 0.5 \times 2 \times 10 = 10 N .$
$∴ Δ P = f . u = 10 \times 10 = 100 W .$