After hitting a long fly ball that goes over the right fielder's head and lands in the outfield, the batter decides to keep going past second base and try for third base. The 60 kg player begins sliding 3.40 m from the base with a speed of 4.35 m/s. a) How much work was done on the player by friction?
b) What was the coefficient of kinetic friction between the player and the ground?

A) -567.675 J

B) 0.284

Explanation:

Parameters given:

Mass of player, m = 60 kg

Distance slid, d = 3.40 m

Speed = 4.35 m/s

A) We know that work done is equal to change in kinetic energy of a body:

W = ΔKE

Change in kinetic energy is given as:

where m = mass

v = velocity

=>W =  ΔKE = -0.5 * 60 * 4.35 * 4.35 = -567.675 J

The work is negative because the force of friction acts opposite the motion of the player.

B) We have that:

Fr = μmg

where Fr = Frictional force

μ = coefficient of kinetic friction

g = acceleration due to gravity

Work done is give as:

W = Fr * d

(d = distance moved)

=> Fr = W/d

Therefore:

W / d = μmg

=>μ = W / (m * g * d)

μ = (567.675) / (60 * 9.8 * 3.40)

μ = 0.284

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