I've been trying to solve how to find the Work done against friction as a ball rolls down a curve, but haven't been able to find anything matching what I want online. The solution I have came up with is the following:
$$ W_{\mu} = mg\mu\int_{a}^{b}{\frac{\underline{c}'(t)\cdot\underline{i}}{||\underline{c}'(t)||}dt} $$
I derived this by first using the equation containing both $\cos(\theta)$ and the dot product of two vectors to find the $\cos(\theta)$ of the curve to the axis, as the normal force is $mg\cos(\theta)$.
I've verified that the formula works for a generalised linear slope, however I have not been able to verify it for all slopes.
Does this equation work? How would I go about verifying it for any general curve?
I am obtaining a slightly different result for the work done by the force of friction on the rolling ball.
I'm assuming $c(t)=\left(x(t),y(t)\right):a\leq t \leq b$ is the parametric curve which describes the motion of the ball on the time domain $[a,b]$.
Recall that any (force) vector can be decomposed into two components: a magnitude and a direction.
The magnitude of the force of friction at any time $t$ is $$\begin{eqnarray*}\mu\times \left(\text{magnitude of normal force}\right)&=&\mu \times \Big\|\text{Proj}_{\left(-y'(t),x'(t)\right)}\Big((0,-mg)\Big)\Big\| \\ &=& \frac{\mu mg|x'(t)|}{\|c'(t)\|}\end{eqnarray*}$$
The direction of the force of friction opposes the direction of motion which in this case is $-\frac{c'(t)}{\|c'(t)\|}$.
So, at any time $t\in [a,b]$ the force of friction is given by $F_f=-\frac{\mu mg|x'(t)|}{\|c'(t)\|^2} c'(t)$
Hence the work is $$\begin{eqnarray*}\int_{c\left([a,b]\right)}(F_f \cdot T)\mathrm{d}s&=&\int_a^b \left(-\frac{\mu mg|x'(t)|}{\|c'(t)\|^2} c'(t)\cdot \frac{c'(t)}{\|c'(t)\|}\right)\|c'(t)\|\mathrm{d}t \\ &=& -\mu mg\int_a^b |x'(t)|\mathrm{d}t\end{eqnarray*}$$