A frightened child is restrained by her mother as the child slides down a frictionless playground slide. If the force on the child from the mother is$\mathbf{100}\mathbf{}\mathbf{N}$up the slide, the child’s kinetic energy increases by $\mathbf{30}$$\mathbf{J}$as she moves down the slide a distance of$\mathbf{1}\mathbf{.}\mathbf{8}\mathbf{}\mathbf{m}$. (a) How much work is done on the child by the gravitational force during the$\mathbf{1}\mathbf{.}\mathbf{8}\mathbf{}\mathbf{m}$ descent? (b) If the child is not restrained by her mother, how much will the child’s kinetic energy increase as she comes down the slide that same distance of$\mathbf{1}\mathbf{.}\mathbf{8}\mathbf{}\mathbf{m}$?

It is given that,

1. Force given by mother is$F_m=100\mathrm{N}$
2. Increase in child's kinetic energy is$∆\mathrm{KE}=30\mathrm{J}$
3. Distance of sliding is$\mathrm{d}=1.8\mathrm{m}$

This problem deals with the work-energy theorem which states that the net work done by the forces applied on object is equal to the change in its kinetic energy. Using the work-energy theorem, find the work done by gravity. Also, find the increase in the kinetic energy using the concept of conservation of energy and total work done.

Formulae are as follow:

$W=F.d$

Where,

F is force, dis displacement and Wis the work done.

Total work done is

$W_{total}=W_g+W_m$

Work done by the force of mother is,

$$\begin{gathered} W_m=F_m\times d \\ {W}_m=100\times 1.8 \\ {W}_m=-180\mathrm{J} \end{gathered}$$

It is negative because displacement is opposite to force.

$W_g$is work done by gravity.

$W_{total}=W_g+W_m$

By work-energy theorem, the total work is the change in kinetic energy,

$$\begin{gathered} W_{total}=W_g+W_m=∆\mathrm{KE} \\ ∆\mathrm{KE}=W_g+W_m \\ 30=W_g-180 \\ {W}_g=210\mathrm{J} \end{gathered}$$

Hence, the work done on the child by gravity during 1.8m descent is$210\mathrm{J}$

If her mother had not restrained the child, then only gravitational force is applied on the child, so the work done is only due to gravity and that is,

$W_g=210\mathrm{J}$.

So, the increase in kinetic energy is equal to $∆\mathrm{KE}=210\mathrm{J}$