Source: High school physics (Chinese)
Problem Sets:
Problem
A block is launched up an inclined plane with an initial speed $v_i$. The plane is inclined at an angle $\theta$ to the horizontal, and the coefficient of kinetic friction between the block and the plane is $\mu_k$. The block slides up to a maximum distance and then slides back down to its starting point. Assume the condition for sliding down, $\tan\theta > \mu_k$, is met.
- Find the maximum distance $d$ the block travels up the incline.
- Find the speed $v_f$ of the block when it returns to its starting point.
[Q1] $d = \frac{v_i^2}{2g(\sin\theta + \mu_k\cos\theta)}$ [Q2] $v_f = v_i \sqrt{\frac{\sin\theta - \mu_k\cos\theta}{\sin\theta + \mu_k\cos\theta}}$
We use the work-energy theorem for both parts of the motion. The normal force is $N = mg\cos\theta$, so the friction force is $F_f = \mu_k mg\cos\theta$.
[Q1] For the upward slide, the initial energy is $E_i = \frac{1}{2}mv_i^2$ (at $h=0$) and the final energy is $E_f = mgh = mgd\sin\theta$ (at $v=0$). The work done by friction is $W_f = -F_f d = -\mu_k mgd\cos\theta$.
$$W_f = E_f - E_i \implies -\mu_k mgd\cos\theta = mgd\sin\theta - \frac{1}{2}mv_i^2$$Solving for $d$:
$$\frac{1}{2}mv_i^2 = mgd(\sin\theta + \mu_k\cos\theta) \implies d = \frac{v_i^2}{2g(\sin\theta + \mu_k\cos\theta)}$$[Q2] For the downward slide, the initial energy at the top is $E_{top} = mgd\sin\theta$ (at $v=0$) and the final energy at the bottom is $E_{bot} = \frac{1}{2}mv_f^2$ (at $h=0$). The work done by friction is again $W_f = -\mu_k mgd\cos\theta$.
$$W_f = E_{bot} - E_{top} \implies -\mu_k mgd\cos\theta = \frac{1}{2}mv_f^2 - mgd\sin\theta$$Solving for $v_f^2$:
$$v_f^2 = 2gd(\sin\theta - \mu_k\cos\theta)$$Substitute the expression for $d$ from Q1:
$$v_f^2 = 2g \left( \frac{v_i^2}{2g(\sin\theta + \mu_k\cos\theta)} \right) (\sin\theta - \mu_k\cos\theta) = v_i^2 \frac{\sin\theta - \mu_k\cos\theta}{\sin\theta + \mu_k\cos\theta}$$