A body of mass m is launched up an incline of angle θ with…

A body of mass $m$ is launched up an incline of angle $\theta$ with speed $v_0$; the coefficient of friction is $\mu$. (a) How far up does it go? (b) How much work is done against friction on the way?

Explanation of Method

A body launched up a rough incline.

Step 1: (a) Distance

\[ \tfrac12 mv_0^2 = mgs\sin\theta + \mu mgs\cos\theta. \]

\[s = \frac{v_0^2}{2g\,(\sin\theta + \mu\cos\theta)}\]

Step 2: (b) Friction work

\[W_f = \mu mg\cos\theta \cdot s = \frac{\mu\cos\theta}{\sin\theta + \mu\cos\theta} \cdot \frac{mv_0^2}{2}\]

$\theta$ $v_0$ $\mu mg\cos\theta$ $s$ $\tfrac12 mv_0^2 = mgs\sin\theta + \mu mg s\cos\theta$

On the way up both gravity's slope component and friction act against the motion. The starting kinetic energy pays for the height gained and for the friction work.

Result

\[\boxed{\,s = \frac{v_0^2}{2g\,(\sin\theta + \mu\cos\theta)}\,}\]

\[\boxed{\,W_f = \mu mg\cos\theta \cdot s = \frac{\mu\cos\theta}{\sin\theta + \mu\cos\theta} \cdot \frac{mv_0^2}{2}\,}\]

Remark

Check. The original results are correct. The friction work is the fraction $\mu/(\tan\theta + \mu)$ of the starting kinetic energy. Whether the body then slides back down depends on whether $\tan\theta \gt \mu$.