Kathy Williams

2022-01-09

On an essentially frictionless, horizontal ice rink, a skater moving at 3.0 m/s encounters a rough patch that reduces her speed to 1.65 m/s due to a friction force that is $25\mathrm{%}$ of her weight. Use the work–energy theorem to find the length of this rough patch.

Mary Nicholson

Step 1
The ice-skater experiences three forces during his/her transition across the rough patch: gravity, the normal force, and the force of friction. The first two cancel out, so that the force of friction is the total force acting on the skater, equal to ${F}_{f}=\mu mg$. Since $\frac{{F}_{f}}{m}g=\frac{1}{4}$ , we have $\mu =\frac{1}{4}$. The friction is opposite to the displacement and so the work done by friction is ${W}_{f}=-\mu mgl$, where l is the length of the rough patch.

Now, if the initial and the final speeds of the skater were ${v}_{i}=3\frac{m}{s}$ and ${v}_{f}=1.65$, respectively, by the work energy theorem we have
${W}_{\to t}={W}_{f}=-\mu mgl={K}_{f\in al}-{K}_{\in itial}=\frac{1}{2}m\left({v}_{f}^{2}-{v}_{i}^{2}\right)⇒$
$l=\frac{1}{2\mu g}\left({v}_{i}^{2}-{v}_{f}^{2}\right)$
$=\left(\frac{1}{2\cdot 0.25\cdot 9.8}\left({3}^{2}-{1.65}^{2}\right)\right)m$
$=1.28m$

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