The cart starts from rest at the top of a hill with height, H1. It rolls down the hill and at the bottom of the hill it has a speed of 8 m/s. Next it is at the top of same hill but pushed so that it starts at the top of the hill with a speed of 6 m/s. How fast is it going at the bottom of the hill?
physics - drwls, Tuesday, October 11, 2011 at 2:05pm
The change in kinetic energy will be the same in both cases.
(M/2)8^2 = M/2[V^2 - 6^2]
cancel the M's
V = sqrt[6^2 + 8^2] = 10 m/s