Question

A person is pushed up a ramp inclined upward at an angle θ above the horizontal as he sits in his desk chair that slides on frictionless rollers. The combined mass of the person and chair is m. He is pushed a distance x along the incline by a group of students who together exert a constant horizontal force of FA. The person’s speed at the bottom of the ramp is v0 m/s. Use the work-energy theorem to find his speed at the top of the ramp.

Answer 1

Hi!

The work W done by force FA is W = FA*cos(θ)

The work-energy theorem say that the work W done by nonconservative forces is equal to the variation of mechanical energy:

`W = \Delta E_(potential) + \Delta E_(kinetic)`

If the distance moved is x, then the vertical displacement is x*sin(θ) Then,

`\Delta E_(potential) = mgx \sin(\theta)`

`\Delta E_(kinetic) = (m)/(2)v^2 =  W - \Delta E_(potential) = FA\cos(\theta) - mgx\sin(\theta)`

We can solve for the speed v at the top of the ramp:

`v = sqrt ((2)/(m) ( FA\cos(\theta) - mgx\sin(\theta)))`