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I am facing a problem in Physics.

Problem: What will be the work done by the frictional force over a polynomial curve if a body is sliding on this polynomial($a+bx+cx^2+dx^3+\ldots$) curve from rest from the height $h_1$ to height $h_2$ (where $h_1 > h_2$).

I tried to solve this as follows:

frictional force $F = k mg \cos\theta$, where $mg \cos\theta$ is normal force at that point. $k$ is coefficient of friction

Total work done=Line Integration over the polynomial(dot product of F and displacement).

But to go ahead from this point,i do not know.

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Is there any more information available, such as initial and final speeds of the body? The problem needs to be defined better. –  DarenW Aug 7 '12 at 6:39
    
"Total work done=Line Integration over the polynomial(dot product of F and displacement)." I think that is the solution already...If you're looking for total work done, and you've already found it, what more do you want to do? :) –  Rody Oldenhuis Aug 7 '12 at 7:56
    
@DarenW he states that the body starts "from rest". Anyway, how does the velocity matter in the calculation of the work? –  Rody Oldenhuis Aug 7 '12 at 7:57
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@user1220376 The normal force you wrote down is for a body that is at rest. When a body moves on a curve the normal force is different because there is acceleration in the normal direction (as you get in circular motion). Beyond that, see Qmechanic's answer below. –  Guy Gur-Ari Aug 7 '12 at 20:22

3 Answers 3

I) The easy way to calculate the work $W_{\rm fric}$ done by friction (if one also knows initial and final speeds of the body, cf. DarenW's comment), is to use energy conservation

$$ W_{\rm fric}~=~ -\Delta E_{\rm kin} -\Delta E_{\rm pot}. $$

II) Else one would have to set up Newton's 2nd law along the curve, which is a second order vector-valued ODE, and solve it.

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The stuff below doesn't help you with the problem--- that's Qmechanic's answer. You're supposed to use conservation of energy to infer the work done. But you asked what is the work done by friction for sliding on a polynomial curve:

But for a given polynomial, you know the height is y(x), so the speed, ignoring friction, would be

$$ v = \sqrt{2g (h_0 - y(x)) } $$

The centripetal force to keep you on the curve is

$$ F_c = {v^2 \over R} $$

Where R is the radius of curvature:

$$ {1\over R(x)} = {y'' \sqrt{(1+y'^2)} \over (1+y')^2} $$

while the normal force is by the cosine of the slope angle

$$ N = {mg\over \sqrt{1+y'^2}} $$

The work done by friction is the coefficient of friction $\mu$ times the total of these two forces, integrated over the curve:

$$ {dW\over \mu} = ({ 2g v^2 \over R} + mg {1\over \sqrt{1+y'^2}}) ds $$

Where $ds = \sqrt{1+y'^2} dx $, so that this is

$$ {W\over \mu} = \int { v^2 y'' (1+y'^2)\over (1+y')^2} + mg dx $$

Notably, the square roots cancel, and the second part, the friction for slow velocities, is just $\mu mg\Delta x $ for any curve, it's how far in X you moved.

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The problem is too general for being solved this way and there is a condition of contact you have not used. The normal force must never be zero for the mobile not to take off. –  Shaktyai Aug 7 '12 at 14:19
    
@Shaktyai: I have implicitly used the condition by saying that $v = sqrt{2g(h_0 - y) - W}$. You don't get any information about your problem from this, it is just a general interesting form of the work done by friction. You can't solve this equation, it's just an interesting fact that the slow velocity friction is proportional to the horizontal distance travelled, while the other part is not too bad for small $y'$ either. –  Ron Maimon Aug 7 '12 at 19:59

I would use the conservation of energy: Ec(1)-Ec(2)+mg(h1-h2)=W At least you know that without knowing initial and final velocities nothing much can be said.

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