K.E. @ top = K.E. @ bottom + work pushing up - P.E. at top
1/2 m vt^2 = 1/2 m vb^2 + f d - m g h
1/2 * 86.0 * v^2 = ...
... (1/2 * 86.0 * 2.40^2) + (591 * 2.95) - [86.0 * 9.8 * 2.95 * sin(35.0º)]
A Physics professor is pushed up on a ramp inclined at an angle of 35.0° above the horizontal as he sits in his desk chair that slides on frictionless rollers. The combined mass of the professor and the chair is 86.0 kg. He pushed at a distance 2.95 m along the inclined by a group of students who are exerting a constant force of 591N parallel to the ramp. The speed at the bottom of the ramp is 2.40 m/s. Find the speed at the top of the ramp.
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