Answer
(a) W = 847 J
(b) (i) KE = 847 J
(ii) v = 18.8 m/s
(c) The work done by gravity on the watermelon would not change. However, the amount of kinetic energy just before it hit the ground would decrease and the speed would also decrease.
Work Step by Step
(a) $W = mg~d = (4.80~kg)(9.80~m/s^2)(18.0~m)$
$W = 847~J$
(b) (i) Since gravity is the only force acting on the watermelon as it falls, the kinetic energy just before it hits the ground is $KE = 847~J$
(ii) $\frac{1}{2}mv^2 = 847~J$
$v^2 = \frac{(2)(847~J)}{4.80~kg}$
$v = \sqrt{\frac{(2)(847~J)}{4.80~kg}}$
$v = 18.8~m/s$
(c) The work done by gravity on the watermelon would not change. However, the air resistance would do negative work on the watermelon. Thus the amount of kinetic energy just before it hit the ground would decrease and the speed would also decrease.
