In a children’s park, there is a slide which has a total length of 10 m and a height of 8 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. Then:
1) The work done by ladder on the boy as he goes up is zero
2) The work done by ladder on boy as he goes up is – 1600 J
3) The work done by slide on boy as he comes down is – 600 J
4) The work done by slide on boy as he comes down is 1600 J

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Detailed Solution

Frictional force $= \frac{3}{10} \times 200 = 60 N$
Work done by ladder on body is zero because while ladder applies force on boy, his point of application does not move. Work done by slide work done by friction
$= - 60 \times 10 = - 600 J$
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