Pushing Crate up Frictionless Slope.. Find Work

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

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?

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