91Ó°ÊÓ

Moment of inertia of the merry-go-round $\left(I\right)=2100kg{m}^2$.

Radius of the marry-go-round$\left(R\right)=2.40m$.

Force exerted by the child $\left(F\right)=18.0N$.

Time interval $\left(t\right)=15.0s$.

The torque is given by,

Here, F is the force, R is the radius, $\mathrm{Î\pm }$is angular acceleration and I is the moment of inertia.

The angular acceleration is:

$\mathrm{Î\pm }=\frac{FR}{I}$ ……. (2)

Here, F is the force, R is the radius, and I is the moment of inertia.

Consider the formula for the angular speed:

$\mathrm{Ó\neg }=\mathrm{Î\pm }\mathrm{Δ}t$ …… (3)

Here, $\mathrm{Î\pm }$is angular acceleration and $\mathrm{Δ}t $is the time interval.

The initial time is $t_i=0$s and final time is $t_f=15s$.

Therefore, the time interval is $\mathrm{Δ}t=15s$.

Then using equation$\left(2\right)$in $\left(3\right)$, angular speed is given by,

Hence, the angular speed is, $\mathrm{Ó\neg }=0.309\frac{\text{rad}}{\text{s}}$.

The work done on the merry-go-round by the child is equal to the rotational kinetic energy.

Therefore,

Hence, the work done on the merry-go-round by the child is$W=100 J$.

The average power supplied by the child is given by,

$P=\frac{W}{\mathrm{Δ}t}$.

Substituting values, we get,

Hence, the average power supplied by the child is, $P=6.67 W$.