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

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SUMMARY

The discussion focuses on calculating the work done on a 16 kg crate being pushed up a 15° frictionless incline. The worker applies a force of 52.8 N, resulting in 79.1 J of work done by the applied force. The confusion arises in calculating the work done by the weight of the crate, where the correct approach involves determining the component of gravitational force acting parallel to the incline. The correct calculation for the work done by gravity is 117.6 J, considering the weight's direction relative to the crate's motion.

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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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