The player covers a distance of 7.1 m before coming to rest
Explanation:
The motion of the player is a uniformly accelerated motion, so we can use the following suvat equation to find the distance through which the player has slided:
[tex]v^2-u^2=2as[/tex] (1)
where
v = 0 is the final velocity (he comes to a stop)
u = 8.25 m/s is the initial velocity
a is the acceleration
s is the distance covered by the player
We need to find an expression for the acceleration. The only force acting on the player, which is the force responsible for his deceleration, is the force of kinetic friction:
[tex]F_f = -\mu_k m g[/tex]
where
[tex]\mu_k = 0.49[/tex] is the coefficient of kinetic friction
m is the mass of the player
[tex]g=9.8 m/s^2[/tex] is the acceleration of gravity
The negative sign is due to the fact that the direction of the force is opposite to the motion of the player
According to Newton's second law, therefore the acceleration of the player is
[tex]a=\frac{F}{m}=\frac{-\mu_k mg}{m}=-\mu_k g[/tex]
So now we can substitute into (1) and find the distance covered by the player:
[tex]v^2-u^2 = -2\mu_k g s\\s=\frac{v^2-u^2}{-2\mu_k g}=\frac{0-(8.25)^2}{-2(0.49)(9.8)}=7.1 m[/tex]
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