Answer
(a) The mass would have to fall a distance of 13.2 meters.
(b) $v = 8.04~m/s$
Work Step by Step
(a) If the initial potential energy on Mars is equal to the initial potential energy on Earth, then the drum will have a kinetic energy of 250.0 J.
$PE_{mars} = PE_{earth}$
$mg_mh_{m} = mg_eh_e$
$h_m = \frac{g_eh_e}{g_m}$
$h_m = \frac{(9.80~m/s^2)(5.00~m)}{3.71~m/s^2}$
$h_m = 13.2 ~m$
The mass would have to fall a distance of 13.2 meters.
(b) We can use conservation of energy to find the speed of the mass after it falls 13.2 meters.
$KE = PE$
$\frac{1}{2}mv^2 + 250.0~J = mgh$
$v = \sqrt{\frac{2mgh-500.0~J}{m}}$
$v = \sqrt{\frac{(2)(15.0~kg)(3.71~m/s^2)(13.2~m)-500~J}{15.0~kg}}$
$v = 8.04~m/s$
