91Ó°ÊÓ

Moment of inertia of the merry-go-round (I) = 2100 kg m2.

Radius of the marry-go-round(R) = 2.40 m.

Force exerted by the child (F) = 18.0 N.

Time interval (t) =15.0 s.

The torque is given by,

$\mathbf{Ï„}\mathbf{}\mathbf{=}\mathbf{FR}$

$\mathbf{=}\mathbf{\pm ôÎ\pm }$ ……. (1)

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:

$\mathrm{Î\pm }=\frac{\mathrm{FR}}{\mathrm{I}}$ ……. (2)

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

Consider the formula for the angular speed:

$\mathrm{Ó\neg }=\mathrm{Î\pm Δ³Ù}$ …… (3)

Here, $\mathrm{Î\pm }$is angular acceleration and $\mathrm{Δ³Ù}$is the time interval.

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

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

Then using equation (2) in (3), angular speed is given by,

role="math" localid="1667982999190" $$\begin{gathered} \mathrm{Ó\neg }=\mathrm{Î\pm }\mathrm{Δ}t \\ ⇒\mathrm{Ó\neg }=\frac{FR\mathrm{Δ}t}{I} \\ ⇒\mathrm{Ó\neg }=\frac{(18.0)(2.40)(15.0)}{2100} \\ ⇒\mathrm{Ó\neg }=0.309 \frac{rad}{s} \end{gathered}$$

Hence, the angular speed is,.

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

Therefore,

$$\begin{gathered} \mathrm{W}={\mathrm{K}}_{\mathrm{rot}}=\frac{1}{2}{\mathrm{IÓ\neg }}^2 \\ ⇒\mathrm{W}=\frac{1}{2}\left(2100\right)(0.309{)}^2 \\ ⇒\mathrm{W}=100\mathrm{J} \end{gathered}$$

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

The average power supplied by the child is given by,

$\mathrm{P}=\frac{\mathrm{W}}{\mathrm{Δ³Ù}}$.

Substituting values, we get,

localid="1667983259762" $$\begin{gathered} \mathrm{P}=\frac{100}{15.0} \\ ⇒\mathrm{P}=6.67\mathrm{W} \end{gathered}$$

Hence, the average power supplied by the child is, $\mathrm{P}=6.67\mathrm{W}$.