# physics

A 62.0-kg 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?

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1. (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)

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posted by drwls
2. After plugging in the values, I obtained the following results:

(a) W = K = -587J
(b) muk = 0.285

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posted by John

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