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](https://tex.z-dn.net/?f=v%5E2%20-%20u%5E2%20%3D%202as)
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
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](https://tex.z-dn.net/?f=a%3D%5Cfrac%7Bv%5E2-u%5E2%7D%7B2s%7D%3D%5Cfrac%7B0-10.0%5E2%7D%7B2%28100%29%7D%3D-0.5%20m%2Fs%5E2)
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](https://tex.z-dn.net/?f=F%3D%20ma%20%3D%20-%5Cmu%20mg)
where
is the force of friction
m is the mass of the skater
is the coefficient of friction
is the acceleration
is the acceleration of gravity
Solving for
, we find the coefficient of friction:
![\mu = -\frac{a}{g}=-\frac{-0.5}{9.8}=0.051](https://tex.z-dn.net/?f=%5Cmu%20%3D%20-%5Cfrac%7Ba%7D%7Bg%7D%3D-%5Cfrac%7B-0.5%7D%7B9.8%7D%3D0.051)
Learn more about friction:
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