A block A of mass $45 k g$ is placed on another block B of mass $123 k g$ . Now block B is displaced by external agent by $50 c m$ horizontally towards right. During the same time block A just reaches to the left end of block B. Initial and final positions are shown in figures. The work done on block A in ground frame is
Assume $μ = 0.2$

- 18 J

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18 J

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36 J

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- 36 J

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Solution

The correct option is C $18 J$
The force acting on the block A in horizontal direction is friction towards right, $f = μ m_{A} g = 0.2 \times 45 \times 10 = 90 N$
Net displacement of the block A with respect to the ground = $50 c m - (40 c m - 10 c m) = 20 c m = 0.2 m$
Thus work done in the ground frame = $f d = + 18 J$

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Q. A block A of mass 45 kg is placed on another block B of mass 123 kg. Now block B is displaced by external agent by 50 cm horizontally towards right. During the same time block A just reaches to the left end of block B. Initial and final positions are shown in figures. The work done on block A in ground frame is

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