Solution

Let $\alpha$ be the angle from the downward radius of the circle to the radius pointing to the sled measured counterclockwise. The centripetal force on the sled is

$$N-mg\cos\alpha=\frac{mv^2}{r}.$$

Consequently, the friction is

$$f=\mu N=\mu m\left(g\cos\alpha+\frac{v^2}{r}\right).$$

Thus, the work done by friction is

$$\int_0^{2\pi}-fr\,d\alpha=\int_0^{2\pi}-\mu m\left(g\cos\alpha+\frac{v^2}{r}\right)r\,d\alpha.$$

Since the average of $\cos\alpha$ for $0\le\alpha\le2\pi$ is zero, the integral simplies to $-2\pi\mu mv^2$.