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:
[tex]v^2 - u^2 = 2as[/tex]
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
[tex]s=1.0\cdot 10^2 m = 100 m[/tex] is the distance he travels before stopping
Solving for a, we find the acceleration of the skater:
[tex]a=\frac{v^2-u^2}{2s}=\frac{0-10.0^2}{2(100)}=-0.5 m/s^2[/tex]
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):
[tex]F= ma = -\mu mg[/tex]
where
[tex]-\mu mg[/tex] is the force of friction
m is the mass of the skater
[tex]\mu[/tex] is the coefficient of friction
[tex]a=-0.5 m/s^2[/tex] is the acceleration
[tex]g=9.8 m/s^2[/tex] is the acceleration of gravity
Solving for [tex]\mu[/tex], we find the coefficient of friction:
[tex]\mu = -\frac{a}{g}=-\frac{-0.5}{9.8}=0.051[/tex]
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