How Much Work is Done Pushing a Crate Up an Incline?

[Jeann25]

To push a 25.0 kg crate up a frictionless incline, angled at 25.0 degrees to the horizontal, a worker exerts a force of 209N parallel to the incline. As the crate slides 1.50m, how much work is done on the crate by the worker's applied force?

I know that W=d x F(cos(angle))

Why doesn't it work when I use (1.50m)(209N)(Cos(25)) = 284 J

Answer is supposed to be 314 J. Is the angle used in the equation supposed to be 0 because the force is parallel to the incline? If the force was parallel to the x axis, would you then use Cos (25)? Just want to make sure I'm understanding this.

 

[willworkforfood]

Because the force is parallel to the incline, the angle of the incline is not relevant to solving the problem. The mass of the crate likewise has nothing to do with solving the problem because you are given the magnitude of the force.

Work in this case = F*d*cos(angle between the force and distance vectors, which is 0 in this case) = 209*1.5 * 1 = 313.5 which rounds to 314.

You always need to be on the lookout for extraneous information in problems like these :)

 

[quicknote]

Your understanding is correct. When calculating work, the angle used in the equation is the angle between the force and the displacement. In this case, the force is parallel to the incline, so the angle between the force and the displacement is 0 degrees. Therefore, the correct equation to use is W = d x F x cos(0) = 314 J. If the force was parallel to the x axis, then you would use cos(25) as the angle between the force and the displacement would be 25 degrees.