Faulty conveyor belt

First, we observe that the box cannot move faster than the belt, i.e., the speed of the box cannot be greater than $v_{0}$ . Second, when the box reaches either $v_{0}$ or $-v_{0}$ it stops until the belt changes its direction of motion. Using Newton's second law, we find that the acceleration to the right and left are $a_{R}=\mu_{L}g =3.92~\text{m}/\text{s}^2$ and $a_{L}=\mu_{R}g=2.94~\text{m}/\text{s}^2$ . These number suggest that the steady-state motion of the box will be described by the graph shown below. Note that the motion as described in the graph is periodic and this is why we call it steady-state (or terminal). We note that the displacement of the box when it is moving relative to the belt is always zero (in a graph of $v$ vs. $t$ , the displacement equals the area under the curve). Thus, the displacement of the box occurs only when it is not moving relative to the belt (see the intervals $\Delta{t_{1}}$ and $\Delta{t_{2}}$ ). Let $\Delta{x_{1}}$ and $\Delta{x_{2}}$ be the displacements during these intervals. It is not hard to find that $\Delta{x_{1}}=-v_{0}(T-2\frac{v_{0}}{\mu_{R}g})<0 \quad \text{and} \quad \Delta{x_{2}}=v_{0}(T-2\frac{v_{0}}{\mu_{L}g}),$ where $T=1~\mbox{s}$ . The total displacement during the cycle is $\Delta{x}=\Delta{x_{2}}+\Delta{x_{1}}=\frac{2v_{0}^2}{g}(\frac{1}{\mu_{R}}-\frac{1}{\mu_{L}}).$ Finally, we find the average velocity $\bar{v}=\frac{\Delta{x}}{2T}=\frac{v_{0}^2}{gT}(\frac{1}{\mu_{R}}-\frac{1}{\mu_{L}})=0.085~\text{m/s}.$ One should note, that the above formula is not general. Note that for other values of $\mu_{R}$ and $\mu_{L}$ the graph shown below would be completely different. For example, if $\mu_{R}=0$ then the velocity of the box would be constant and equal to $v_{0}.$

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