physics question help

[pizza1994]

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A 5 kg box is pushed 3 m up a 60 degree incline at a constant speed where the coefficient of kinetic friction is 0.4. What is the work done by the person pushing the box?


the answer is 150 J.

anybody know how to do this? 🙂

 

[Skizye]

The work done on the box goes to its change in potential energy as well as heat lost due to friction. So W = dU + Wf = mgh + u(Fn)d
Where h is just the vertical distance (i.e. 3sin60) while d is the total distance it is pushed (i.e. 3)

 

[pizza1994]

The work done on the box goes to its change in potential energy as well as heat lost due to friction. So W = dU + Wf = mgh + u(Fn)d
Where h is just the vertical distance (i.e. 3sin60) while d is the total distance it is pushed (i.e. 3)

why is work equal force of friction and potential energy????

 

[Skizye]

http://faculty.wwu.edu/vawter/PhysicsNet/Topics/Work/WorkEngergyTheorem.html

When you do work on something you are putting energy into it. Friction is essentially taking energy out of the box as it is pushed so you have to account for that in the amount of work that is done on the box.

 

[el_duderino]

why is work equal force of friction and potential energy????

It's moving at a constant speed, so the pushing force is equal to the force of friction. Those are the only two forces present in the direction of motion.