How Is Work Calculated on a Slope with Deceleration?

[MathewsMD]

A man pushes an 80-N crate a distance of 5.0 m upward along a frictionless slope that makes an angle of 30◦ with the horizontal. The force he exerts is parallel to the slope. If the speed of the crate is constant, then the work done by the man is:

Attempt: Since gravity is a conservative force, and there is no nonconservative forces, then mgh = U. Therefore, since h = 5sin30 m, then the work done is the change in potential energy, right? Since you are working against gravity, with the direction of motion, work is positive. Also, I am not sure how to specifically work with a varying force (changing acceleration) in this case so I chose to use conservative forces.

W = (80 N)(5sin30 m)
W = 200 J

Although this is incorrect. I am unsure if I have made any errors, so any pointers would be helpful. Also, and guidance with this question would be greatly appreciated! :)

 

[HallsofIvy]

It is odd that you titled this "Work with changing velocity" while the problem specifies the "If the speed of the crate is constant"! It is, however, true that the work required to lift a mass is just the weight of the mass times the height lifted. Now, why do you say "this is incorrect"?

 

[MathewsMD]

Sorry, here's the correct question.

A man pushes an 80-N crate a distance of 5.0 m upward along a frictionless slope that makes an angle of 30◦ with the horizontal. His force is parallel to the slope. If the speed of the crate decreases at a rate of 1.5m/s2, then the work done by the man is:

My attempt at the solution is still the same

 

[physics&math]

Work = force * distance. If distance = r, then Work is the integral of Force*dr.

 

[Doc Al]

Sorry, here's the correct question.

A man pushes an 80-N crate a distance of 5.0 m upward along a frictionless slope that makes an angle of 30◦ with the horizontal. His force is parallel to the slope. If the speed of the crate decreases at a rate of 1.5m/s2, then the work done by the man is:

My attempt at the solution is still the same

First figure out the force with which the man pushes.