Q.
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
A) Zero
B) $+ 600\ J$
C) $- 600\ J$
D) $+ 1600\ J$
Explanation
(c) $F = \frac{3}{10}mg$
$W = - F\ s\ or\ \ W = - \frac{3}{10}mgs$
or $W = - \frac{3}{10} \times 200 \times 10\ J = - 600\ J$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.