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A skier starts from rest at point $A$ and slides down the hill without turning or breaking. The friction coefficient is $\mu$. When he stops at point $B$, his horizontal displacement is $S$. What is the height difference between points $A$ and $B ?$
(The velocity of the skier is small so that the additional pressure on the snow due to the curvature can be neglected. Neglect also the friction of air and the dependence of $\mu$ on the velocity of the skier.)
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(The velocity of the skier is small so that the additional pressure on the snow due to the curvature can be neglected. Neglect also the friction of air and the dependence of $\mu$ on the velocity of the skier.)
Solution:
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Verified Answer
The correct answer is:
$h=\mu S$
According to question, the situation is shown in the figure
For a sufficiently safe-horizontal displacement $\Delta S$ can be considered straight. If the corresponding length of path element is $\Delta L$, the friction force is given by $\mu_{m g} \frac{\Delta S}{\Delta L}$ and the work done by the friction force equals force times displacement
$\mu_{m g} \frac{\Delta S}{\Delta L} \Delta L=\mu_{m g} \Delta S$
Adding up, we find that along the whole path the total work done by the friction force is $\mu_{m g} s$. By energy conservation this must equals the decrease mgh in potential energy of skier.
Hence, $h=\mu S$
For a sufficiently safe-horizontal displacement $\Delta S$ can be considered straight. If the corresponding length of path element is $\Delta L$, the friction force is given by $\mu_{m g} \frac{\Delta S}{\Delta L}$ and the work done by the friction force equals force times displacement
$\mu_{m g} \frac{\Delta S}{\Delta L} \Delta L=\mu_{m g} \Delta S$
Adding up, we find that along the whole path the total work done by the friction force is $\mu_{m g} s$. By energy conservation this must equals the decrease mgh in potential energy of skier.
Hence, $h=\mu S$
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