A conveyor belt is moving with a constant velocity 2 m/s. A small block is projected along the belt with speed of 4 m/s in opposite direction of velocity of belt. If mass of block is ‘4 kg’ and coefficient of friction is 0.2, find work done (in J) against friction on the conveyor belt up to the instant when slipping between block and belt ceases. $(g = 10 m / s^{2})$

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answer is 48.

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Detailed Solution

Time taken to slipping ceased

$+ 2 = - 4 + μ g t \Rightarrow t = \frac{6}{(2)} = 3 s$

Work done against friction on conveyor belt

$\begin{array}{l} = μ (m g) \times (x) \\ = 0.2 \times 4 \times 10 \times (2 \times 3) = 48 J o u l e \end{array}$

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