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In a children’s park, there is a slide which has a total length of 10 m and a height of 8.0 m. A vertical ladder is provided to reach the top. A boy weighing 200 N climbs up the ladder to the top of the slide and slides down to the ground. The average friction offered by the slide is three-tenth of his weight. The work done by the slide on the boy as he comes down is:

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zero

+1600 J

+ 600 J

–600 J

answer is C.

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Detailed Solution

weight of the boy, $mg = 200 N$

Friction force by slide, $f = \frac{3}{10} mg$

As, work done by friction, $W = - f \times s$

or $W = - \frac{3}{10} mgs$

or $W = - \frac{3}{10} \times 200 \times 10 J = - 600 J$

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