A physics professor is pushed up a ramp inclined up-ward at 30.0o above the horizontal as she sits in her desk chair, which slides on frictionless rollers. The combined mass of the professor and chair is 85.0 kg. She is pushed 2.50 m along the incline by a group of students who together exert a constant horizontal force of 600 N. The professor’s speed at the bottom of the ramp is 2.00 m/s. Use the work–energy theorem to find her speed at the top of the ramp.

Solution 84P To solve this question, we shall have to take the condition that total mechanical energy at the bottom of the incline is equal to that of a given height. Mechanical energy at bottom = Mechanical energy at a height …..(1) Let us have a look at the figure below. The horizontal force is = 600 N Work done on the professor = 600 N × cos 30 × 2.50 J = 1300 J The professor’s speed at the bottom of the ramp is = 2.00 m/s Therefore, kinetic energy = 1 × 85.0 kg × (2.00) J = 170 J 2 So, total energy at the bottom of the ramp is = 1300 J + 170 J = 1470 J …..(2) Now, potential energy of the professor at the given height is 2 0 = 85.0 kg × 9.8 m/s × 2.50sin 30 = 1041 J Let, v be the speed of the professor at the top of the ramp. 1 2 Her kinetic energy at the top is = 2 × 85 kg × v 2 Her total energy at that height is = 42.5v J + 1041 J …..(3) From condition 1 and equating equations 2 and 3, we get 42.5v J + 1041 J = 1470 J 2 42.5v J = 429 J 2 2 2 v = 10.1 m /s v = 3.17 m/s Therefore, the approximate speed of the professor at the top of the ramp is 3.17 m/s.