Question

In a children’s park, there is a slide which has a total length of 10 m and a height of 8 m (figure). 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 in three-tenths of his weight. Find (a) the work done by the ladder on the boy as he goes up. (b) the work done by the slide on the boy as he comes down. (c) Find the work done by forces inside the body of the boy.

Answer 1

As the boy climbs up with no effort, thus the frictional force will be zero while climbing the ladder. Whereas, there will be some frictional force offered by the slide at every point of the length of the slide. Work done is the product of friction and the displacement, thus can be used to solve the problem.

Formula used:
$w=fs$
$w=(mg)h$

Complete step by step answer:
From given, we have the data,
The length of the slide = 10 m
The height of the slide = 8 m
The weight of a boy = 200 N
The friction offered by the slide = $\dfrac{3}{10}\times (200\,N)=60\,N$

Part (a): the work done by the ladder on the boy as he goes up

The work done to climb up the ladder by the boy is the work done by the boy alone.
As the ladder does not do any work to help the boy to go up, thus, the work done by the ladder on the boy as he goes up is given as follows.
${{w}_{L}}=0$

Part (b): the work done by the slide on the boy as he comes down

The given data represents the ideal condition.
The slide offers friction to the motion of the boy of the amount given as follows.
$\dfrac{3}{10}\times (200\,N)=60\,N$

Therefore, the work done by the slide is equal to the work done by the friction.
Thus, the equations are given as follows.

& {{w}_{s}}={{w}_{f}} \\\ & {{w}_{s}}=-f\times s \\\ & {{w}_{s}}=-60\times 10 \\\ & {{w}_{s}}=-600\,J \\\ \end{aligned}$$ The negative sign indicates the downward direction, as the boy slides down. Part (c): the work done by forces inside the body of the boy $$\begin{aligned} & {{w}_{b}}=(mg)\times h \\\ & {{w}_{b}}=200\times 8 \\\ & {{w}_{b}}=1600\,J \\\ \end{aligned}$$ The answers are: (a) The work done by the ladder on the boy as he goes up is equal to 0 W. (b) The work done by the slide on the boy as he comes down is equal to -600 J. (c) The work done by forces inside the body of the boy is equal to 1600 J. **Note:** The things to be on your finger-tips for further information on solving these types of problems are: The units of the given parameters should be taken into consideration while solving the problem. If the internal force or work done will be opposite in direction to that of the external force or work done, the negative sign should be used.