Asked by Anonymous

A 28 kg ball initially at rest rolls down a 159 m hill. When the ball reaches the bottom it is traveling at 31.3 m/s.
How much energy is dissipated by friction on the ball?
Answered by Henry
PEmax=mg*hmax = 28*9.8 * 159=43,630 J.
= KEmax @ bottom of hill with no friction.

KE = 14*31.3^2 = 13,716 J.

Energy lost = 43,630 - 13,716=29,914 J.

Answered by John
497.67
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