First slide

Kinetic friction

Question

A skier starts from rest at point A and slides down the hill without turning or breaking. The friction coefficient is $μ$ . 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 $μ$ on the velocity of the skier.)

Moderate

A

h = $μ$ s
B

$h = \frac{μ}{s}$
C

h = 2 $μ$ s
D

h = $μ$ s²

Solution

According to question, the given situation is shown in the figure below.
For a sufficiently safe-horizontal displacement, $∆$ s can be considered straight. If the corresponding length of path element is $∆$ L, the work done against frictional force is given by
$w_{∆ F} = μ m g (Δ L \cos θ)$
From $Δ O P Q, \cos θ = \frac{Δ s}{Δ L}$
So, $W_{∆ F} = μ m g Δ L (\frac{Δ s}{Δ L}) = μ m g Δ s$
On adding up, we find that along the whole path, the total work done by the friction force is $μ$ mgs. By law of conservation of energy, this must equals to the decrease in potential energy of skier.
$∴ μ m g s = m g h$
Hence, $h = μ s$

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