Source: Principles of Physics
Problem
A jar of mass $m$ moves up an incline of angle $\theta$. At a distance $d_0$ from the bottom, its speed is $v_0$. The coefficient of kinetic friction is $\mu_k$.
- How much farther, $d_{up}$, will the jar move up the incline?
- What is its speed, $v_f$, when it returns to the bottom of the incline?
- How do the answers to Q1 and Q2 change if $\mu_k$ decreases?
[Q1] $d_{up} = \frac{v_0^2}{2g(\sin\theta + \mu_k \cos\theta)}$ [Q2] $v_f = \sqrt{2g(d_0 + d_{up})(\sin\theta - \mu_k \cos\theta)} = \sqrt{2gd_0(\sin\theta - \mu_k \cos\theta) + v_0^2\frac{\sin\theta - \mu_k \cos\theta}{\sin\theta + \mu_k \cos\theta}}$ [Q3] If $\mu_k$ decreases, $d_{up}$ increases and $v_f$ increases.
We apply the work-energy theorem, $\Delta K = W_{net}$. For the upward motion to the highest point, the change in kinetic energy is $\Delta K = - \frac{1}{2}mv_0^2$. The work is done by gravity, $W_g = -mg d_{up} \sin\theta$, and friction, $W_f = -(\mu_k mg \cos\theta) d_{up}$. Setting $W_g + W_f = \Delta K$ and solving for $d_{up}$ answers Q1. For the downward motion over the total distance $d_{total} = d_0 + d_{up}$, the jar starts from rest. The change in kinetic energy is $\Delta K = \frac{1}{2}mv_f^2$. The work is done by gravity, $W_g = mg d_{total} \sin\theta$, and friction, $W_f = -(\mu_k mg \cos\theta) d_{total}$. Setting $W_g + W_f = \Delta K$ yields $v_f$. For Q3, we inspect the dependence of the derived expressions for $d_{up}$ and $v_f$ on $\mu_k$.