#### Answer

(a) $\omega = 0.309~rad/s$
(b) $Work = 100~J$
(c) The average power is 6.67 watts.

#### Work Step by Step

(a) The angular impulse provided by the child will be equal in magnitude to the angular momentum of the merry-go-round.
$I\omega = \tau ~t$
$\omega = \frac{R~F ~t}{I}$
$\omega = \frac{(2.40)(18.0~N)(15.0~s)}{2100~kg~m^2}$
$\omega = 0.309~rad/s$
(b) The work done by the child is equal to the rotational kinetic energy of the merry-go-round.
$Work = K_{rot}$
$Work = \frac{1}{2}I\omega^2$
$Work = \frac{1}{2}(2100~kg~m^2)(0.309~rad/s)^2$
$Work = 100~J$
(c) $P_{ave} = \frac{Work}{t}$
$P_{ave} = \frac{100~J}{15.0~s}$
$P_{ave} = 6.67~W$
The average power is 6.67 watts.