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

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

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

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