Calculating Work on a Frictionless Slope with Multiple Choice Help

[dominus96]

Homework Statement

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:

a) 61 J
b) 200 J
c) 140 J
d) -200 J

Homework Equations

F=ma

The Attempt at a Solution

I used sin 30 = y/50, so y (the height of the slope) equals 2.5. I don't know what to do from here.

 

[hotcommodity]

The net force on the block parallel to the incline is zero, as it moves with constant speed. This should tell you that the pushing force must be equal in magnitude to the component of the weight force that points along the incline. Newton's Second Law gives:

\Sigma F_{parallel} = F_{push} - Wsin 30 = 0

F_{push} = Wsin 30

Once you know the value of this force you can plug it into the work equation:

W_{F push} = F_{push}(\Delta x)

Does that make sense?

 

[kudoushinichi88]

Since the guy is moving the crate up a <b>frictionless</b> slope, the only work done is against gravity.

And since gravitational force is always directed vertically downwards, work done by the man;

W = mgh

where mg is the weight of the crate and h is the vertical distance that the crate has moved.