# Pushing Crate up Frictionless Slope Find Work

1. Oct 5, 2008

### Spartan Erik

1. The problem statement, all variables and given/known data

A man pushes a create which weighs 75 N 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 increases at a rate of 0.5m/s^2, then the work done by the man while moving the crate a distance of 1.5m is:

-62J, +62J, -41J, +41J, none of the above

2. Relevant equations

W = Fdcos(theta)
W(done by gravity)= mgdcos(theta)

3. The attempt at a solution

I'm not sure exactly how to approach this problem as they supply an increasing acceleration of 0.5m/s^2. Otherwise I would just apply W = Fdcos(theta). I also know that gravity has an effect on this, but I'm not sure how to include it in this situation

2. Oct 5, 2008

### Mattowander

Draw a free body diagram and resolve the weight of the box into components.

Weight pointing down the plane = mg sin theta
Weight normal to the plane (the normal force) = mg cos theta.

If the object is accelerating upward at .5m/s^2, what does that tell you about the net force on the object?

3. Oct 5, 2008

### Spartan Erik

Well the 75N is due to gravity, so the mass is 7.653kg

Breaking it into horizontal and vertical components:
mgsin(theta) = 7.653x9.8xsin(30) = 37.5
mgcos(theta) = 7.653x9.8xcos(30) = 64.951

If the object is accelerating upward at 0.5 m/s^2 and the mass is 7.653kg, then the force is 3.826N