A small body starts sliding from rest down an incline of…

A small body starts sliding from rest down an incline of angle $\alpha$. The coefficient of friction grows with the distance travelled as $\mu = ks$ ($k$ constant). (a) How far does the body go before it stops? (b) What is its greatest speed?

Explanation of Method

Sliding down an incline with friction growing with distance.

Step 1: Speed against distance

\[ v\frac{dv}{ds} = g\sin\alpha - ks\,g\cos\alpha, \]

\[ v^2 = 2gs\sin\alpha - kg\,s^2\cos\alpha. \]

Step 2: (a) Stopping distance

\[v = 0 \ \text{at} \ s = \frac{2\tan\alpha}{k}\]

There $\mu = 2\tan\alpha \gt \tan\alpha$, so the body stays at rest.

Step 3: (b) Greatest speed

The speed is greatest where the acceleration is zero, $\mu = \tan\alpha$, i.e. at $s = \tan\alpha/k$: $v_{\max} = \sqrt{\frac{g\sin\alpha\tan\alpha}{k}}$.

$\frac{\tan\alpha}{k}$ $\frac{2\tan\alpha}{k}$ $s$ $v^2$ $v_{\max}^2 = \frac{g\sin\alpha\tan\alpha}{k}$ $\mu < \tan\alpha$ $\mu > \tan\alpha$

Square of the speed against distance: a parabola. The body speeds up while $\mu = ks$ is below $\tan\alpha$, is fastest where they are equal, and then brakes symmetrically to a stop.

Result

\[\boxed{\,v = 0 \ \text{at} \ s = \frac{2\tan\alpha}{k}\,}\]

\[\boxed{\,v_{\max} = \sqrt{\frac{g\sin\alpha\tan\alpha}{k}}\,}\]

Remark

Check. The original results are correct. The motion is symmetric: the body speeds up over the first half of the path and slows down over the second half in the same way.