#### Answer

(a) $W = 225~J$
(b) $F = 225~N$
(c) $P = 6750~watts$

#### Work Step by Step

(a) The work that the person does on the rock is equal to the kinetic energy of the rock.
$W = KE$
$W = \frac{1}{2}mv^2$
$W = \frac{1}{2}(0.50~kg)(30~m/s)^2$
$W = 225~J$
(b) We can use the work to find the force applied by the person.
$F~d = W$
$F = \frac{W}{d}$
$F = \frac{225~J}{1.0~m}$
$F = 225~N$
(c) The maximum power output occurs when the rock reaches its maximum speed of 30 m/s
$P = F~v$
$P = (225~N)(30~m/s)$
$P = 6750~watts$