A movie stuntman (mass 81.8 kg) stands on a window ledge 7.33 m above the floor. Grabbing a rope attached to a chandelier, he swings down to grapple with the movie’s villain (mass 66.4 kg), who is standing directly under the chandelier. (Assume that the stuntman’s center of mass moves downward 7.33 m. He releases the rope just as he reaches the villain) If the coefficient of kinetic friction of their bodies with the floor is uk = 0.71, how far do they slide?
The Attempt at a Solution
So what I did is first solve for the potential of the stuntman ( 81.8 * 9.8 * 7.33). Then subtract out the friction caused by the two people ( total mass * 9.8 * .71). This is the energy left to the system. So set this equal to F * d where force is equal to total mass * 9.8. And that's how I get my d, which is wrong.