The rider starts a height hi above the bottom of the pipe, slides down the side,

Question

The rider starts a height hi above the bottom of the pipe, slides down the side, across the bottom, and back up the other side. Assume the sides are approximately frictionless so that the only friction is across the bottom which has a width W = 3.5 m and the rider doesn't do anything to increase his energy ("jump"). If the rider, with inertia 68 kg starts at a height hi = 4 m and ends up after the first time across the pipe at a height that is 80% of the initial height, what is the force of friction across the bottom of the pipe? [variable equations along with guided answer would be much appreciated in order to further understand]

Explanation / Answer

Since he rises 20% = .8 m less than his initial height, he loses m*g*h kinetic energy over distance W, the force of friction = m*g*h/W = 68*9.8*.8/3.5 = 152 Newtons.

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