## Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition)

(a) $W = 225~J$ (b) $F = 225~N$ (c) $P = 6750~watts$
(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 $P = F~v$ $P = (225~N)(30~m/s)$ $P = 6750~watts$