This is the problem that I was given and I have redone it at least 5 times alrea

Question

This is the problem that I was given and I have redone it at least 5 times already. Can anyone tell me what I am doing wrong?
"A physics professor is pushed up a ramp inclined upward at an angle 28.0 above the horizontal as he sits in his desk chair that slides on frictionless rollers. The combined mass of the professor and chair is 83.0 kg. He is pushed a distance 2.10 m along the incline by a group of students who together exert a constant horizontal force of 595 N. The professor's speed at the bottom of the ramp is 1.65 m/s. Use the work-energy theorem to find his speed at the top of the ramp."

W=595(2.1)-83(2.1)(9.8)sin(28)=447.577
K1=1/2mv2=[83(1.652)]/2=112.98

so...

v=[2(447.58+112.98)]/2)=3.68 m/s

Explanation / Answer

First of all, let me clarify the wrong in your solution. It is given that, the force is acting horizontally. So, you have to take the component of force in parallel to the plane i.e. F cos. That's it. All the remaining procedure done by you is correct. Here, I am doing the solution for you. Component of force along the plane, F' = F cos                                                             = 595 * cos 28                                                             = 525.35 N Net work done, W = ( F' - m g sin ) * L                               = [ 525.35 - ( 83 * 9.8 * sin 28 ) ] * 2.1                               = 301.31 J Change in kinetic energy, K = (1/2) m ( vf^2 - vi^2 )                                              = 0.5 * 83 * ( vf^2 - 1.65^2 )                                               = 41.5 * ( vf^2 - 1.65^2 ) From work-energy theorem, K = W 41.5 * ( vf^2 - 1.65^2 ) = 301.31 vf^2 - 2.72 = 301.31 / 41.5 vf^2 = 9.983 vf = 3.16 m/s

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