Work done down an inclined plane

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SUMMARY

The discussion focuses on calculating the work done by a toy car of mass 20kg moving down a 35m inclined slope at an angle of arcsin(2/7) with a frictional force of 100N acting against it. The work done by friction is calculated as 100N multiplied by 35m, resulting in 3500J. Additionally, the work done by gravity must be considered, which is derived from the change in potential energy as the car descends 10m. The net work done on the car is the sum of the work done by gravity and the work done against friction.

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Dongorgon
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Homework Statement


A toy car of mass 20kg moves down a slope of length 35m, which is inclined at arcsin(2/7) to the horizontal. The height of the car initially is 10m, and a constant frictional force of 100N acts against the car through motion. Calculate the work done, power and efficiency of the system:


The Attempt at a Solution


My initial issue here is calculating the work done by the car traveling down the entirety of the slope. I'm not sure whether the work done is just the friction force x distance which would give 3500J, or whether you need to take into account the horizontal component of the cars weight when descending the slope, resolving horizontally?
Any advice here would be greatly appreciated! Thanks.
 
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Dongorgon,

It doesn't make sense to speak of the work done "by" the car. Work is force * distance, so work is done BY forces ON objects as those forces move those objects over distances. If you want to compute the total work done ON the car, you have to take into account the work done by all forces acting on the car. So you have to account for the work done by gravity as well as the work done by friction. Another way to look at it: the total (net) work done on the car will be the work done by the net force. I hope this helps.
 
You need to take into account both gravity and friction. The work done by friction is the force, 100N, times the distance traveled, 35m. The work done by gravity can be calculated by the decrease in potential energy, mg= (20)(9.81) times the vertical distance, (2/7)35= 10 m, the car has gone. Becareful- one is work done increasing the speed of the car, the other is work done decreasing the speed of the car.

(There is no "horizontal component of weight". Weight is a force vector pointing vertically.)
 
cepheid said:
the total (net) work done on the car will be the work done by the net force. I hope this helps.

Given the question about efficiency, I suspect they want the work done by gravity.
Dongorgon, have you quoted the problem exactly as given?
 

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