How do I find work done by friction over a curve represented by a polynomial?

Question

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.

Answer 1

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.

Answer 2

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.

Answer 3

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.