physics
posted by Tina .
A 62.0kg baseball player slides 3.40 m from third base w/ a speed of 4.35 m/s. If the player comes to rest in third base , (a) how much work was done on the player by friction? (b) what was the coefficient of the kinetic friction between the player and the ground?

(a) The same as his initial kinetic energy
(b) The friction force F is
(Initial kinetic energy)/(sliding distance)
F = (1/2) M V^2/X
Divide that by the player's weight (M g) to get the kinetic friction coefficient.
mu,k = F/(M*g) = V^2/(2 g X) 
After plugging in the values, I obtained the following results:
(a) W = K = 587J
(b) muk = 0.285