Calculating Work Done on a 16kg Crate by Force and Weight

AI Thread Summary
To calculate the work done by the weight of a 16 kg crate on a frictionless incline at a 15° angle, it's essential to consider the component of gravitational force acting parallel to the incline. The gravitational force (Fg) is 156.8 N, and only the portion acting down the incline contributes to the work done by gravity. This requires resolving the weight into components, with the parallel component calculated as Fg * sin(15°). The work done by gravity is negative since it opposes the motion of the crate. Understanding these components is crucial for accurately determining the work done by weight.
perfect_piccolo
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1. To push a 16 kg crate up a frictionless incline, angled at 15° to the horizontal, a worker exerts a force of 52.8 N, parallel to the incline. As the crate slides 1.50 m, how much work is done on the crate by the worker's applied force? b]

So I got the first part of this question without any problem. The answer is 79.1 J.

Its the next part of the question that I'm unsure about:

How much work is done on the crate by the weight of the crate?


I tried W=Fg(delta x)
W=156.8 N (1.5 m)
W = 235.2 J

But this obviously isn't right. Where am I going wrong?

Thanks!
 
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How much of the weight is acting parallel to the plane?
When you do your FBD mg gets split up into 2 parts, one is perpendicular to the incline, the other is pointed down the incline. You take that portion of the weight * the distance to get the work gravity does.
Also take direction into account, is gravity opposing the motion or helping move the crate?
 
Is anyone other than me bothered by this question?
 
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