# Finding the work done by friction when sliding on parabola [duplicate]

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This is not exactly a homework question, because I myself came up with the problem. I hope that this is the right place to ask this :)

So, given a "ramp" that follows some function $f(x)=x^n$+ c how to find the work done by friction when some object "slides" from $f(b)$ to $f(a)$, if

The friction coefficient is tx. $0.1$

Mass of the object is $1kg$

I did not yet study integrals in school, but this is what I got:

I assume that the function is $0.5x²$, and that i want to know the work done by friction from $f(4)$ to $f(0)$

You can get the "slope" at any given point by the derivative $f'(x)=x$ and the angle between $G$ and $Gy$ is $tan⁻¹(x)$

at that point force of friction is

$F=0.1*g*m*cos$(tan⁻¹(x))

work done by friction $W=F*d$

and distance traveled is $dx\over cos(tan⁻¹(x))$ (the length of the hypotenuse of tanget slope if one cathetus is dx)

so by putting the values in we get $W=0.1*g*m*cos(tan⁻¹(x))*$$(dx)\over cos(tan⁻¹(x))$ But if you start to integrate that, we get $G*m*\int (dx)dx$ because the cosines cancel out and we get a constant. Can someone tell how this should be done?

ps. Sorry for my bad English :)

## marked as duplicate by sammy gerbil, stafusa, Jon Custer, Kyle Kanos, John Rennie homework-and-exercises StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Nov 15 '17 at 9:36

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

• It is an exercise. – sammy gerbil Nov 14 '17 at 19:52

## 2 Answers

Your calculation is correct. The integral gives you $W=\mu mg (b-a)$, where $b-a$ is the horizontal distance moved.

However, integration is not required.

Consider a slope with angle $\theta$. The normal reaction force, and therefore also the friction force, is proportional to $\cos\theta$. The work done against friction in pushing an object a distance $L$ up or down the slope is proportional to $L\cos\theta$. But $L\cos\theta$ is just the horizontal distance moved by the object.

So for any surface, regardless of the shape, the work done against friction when moving an object along the surface is the same as when the object is moved the same horizontal distance on level ground with the same coefficient of friction. The work done against friction is independent of the change in height, it depends only on the change in horizontal position.

This result assumes that the pushing force is always parallel to the slope, so that the normal force (and therefore also the friction force and the work done) depends only on gravity. If the applied force has a component normal to the slope, this solution would not give you the correct answer.

You have to integrate over the (dx) , no need to add another dx more.