Answer
(a) W = 1.89 J
(b) The speed of the coin just as it hits the water is 26.5 m/s
Work Step by Step
(a) We can find the work done by gravity.
$W = F\cdot d$
$W = F ~d~cos(\theta)$
$W = (mg) ~d~cos(0^{\circ})$
$W = (0.0055~kg)(9.80~m/s^2)(35~m)(1)$
$W = 1.89~J$
(b) We can use the work-energy theorem to find the speed when the coin hits the water.
$KE_2 = KE_1+W$
$\frac{1}{2}mv_2^2 = \frac{1}{2}mv_1^2+W$
$v_2^2 = \frac{mv_1^2+2W}{m}$
$v_2 = \sqrt{\frac{mv_1^2+2W}{m}}$
$v_2 = \sqrt{\frac{(0.0055~kg)(4.0~m/s)^2+(2)(1.89~J)}{0.0055~kg}}$
$v_2 = 26.5~m/s$
The speed of the coin just as it hits the water is 26.5 m/s
