A physics professor is pushed up a ramp inclined upward at an angle 35.0degrees above the horizontal as he sits in his desk chair that slides on frictionless rollers. The combined mass of the professor and chair is 89.0 kg. He is pushed a distance 2.05 m along the incline by a group of students who together exert a constant horizontal force of 606 N. The professor's speed at the bottom of the ramp is 2.30 m/s.

Use the work-energy theorem to find his speed at the top of the ramp

any takers? @Mathematics

Question

A physics professor is pushed up a ramp inclined upward at an angle 35.0degrees above the horizontal as he sits in his desk chair that slides on frictionless rollers. The combined mass of the professor and chair is 89.0 kg. He is pushed a distance 2.05 m along the incline by a group of students who together exert a constant horizontal force of 606 N. The professor's speed at the bottom of the ramp is 2.30 m/s.

Use the work-energy theorem to find his speed at the top of the ramp

any takers?   @Mathematics

phi · Answer

horizontal force opposing the students is 
fh=mgtan35=89⋅9.8⋅0.7=610.72N

The students apply a force of 606 N, so the net horizontal force is  -4.72 N
The force parallel to the slope is -4.72 cos35= -3.87 (downslope)
The work energy theorem is
$$F⋅x=0.5m(v_f^2−v_i^2) $$
plugging in m=89 kg, x= 2.05 m, vi= 2.3 m/s, F= -3.87 N, we get
-7.934/44.5 = vf^2 -  5.29 
or vf^2= 5.112, and vf= 2.26 m/s

anonymous · Answer

horizontal force opposing the students is 
fh=mgtan35=89⋅9.8⋅0.7=610.72N

USE FORMULA : F⋅x=0.5m(v_f^2−v_i^2)

anonymous · Answer

thanks for the input guys i got the correct answer i was calculating work done worng at teh beginning was trying to put a component of gravity into the force