Question

What is the coefficient of friction between the skates and the ice?

Answer 1

The coefficient of friction is 0.051

Explanation:

The motion of the skater is a uniformly accelerated motion, therefore we can use the following suvat equation:

$v^2 - u^2 = 2as$

where:

v = 0 is the final velocity of the skater (he comes to a stop)

u = 10.0 m/s is his initial velocity

a is the acceleration

$s=1.0\cdot 10^2 m = 100 m$ is the distance he travels before stopping

Solving for a, we find the acceleration of the skater:

$a=\frac{v^2-u^2}{2s}=\frac{0-10.0^2}{2(100)}=-0.5 m/s^2$

We also know that the net force acting on the skater is the force of friction, therefore we can write (Newton's second law of motion):

$F= ma = -\mu mg$

where

$-\mu mg$ is the force of friction

m is the mass of the skater

$\mu$ is the coefficient of friction

$a=-0.5 m/s^2$ is the acceleration

$g=9.8 m/s^2$ is the acceleration of gravity

Solving for $\mu$, we find the coefficient of friction:

$\mu = -\frac{a}{g}=-\frac{-0.5}{9.8}=0.051$

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