Question

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 67 kg player begins sliding 3.40 m from the base with a speed of 4.35 m/s. a.) If the player comes to rest at third base, how much work was done on the player by friction? b.) If the player comes to rest at third base, What was the coefficient of kinetic friction between the player and the ground?

Answer 1

Answer:

a) 633.39 J

b) 0.28

Explanation:

a)The kinetic energy of the player = $\frac{1}{2} mv^{2}$

Work done by friction = energy change of the player

                                    = $\frac{1}{2} 67(4.35)^{2}$ = 633.9 J

b) Assuming the frictional force stays constant,

Work done by friction = Frictional force×distance

Frictional force = kinetic friction(μ)×normal reaction

Normal reaction = weight = mass×gravitational acceleration ( g=10m/s2 )

Combining these equations

633.9  =  F×3.4 ⇒ F = 186.44 N

F = μmg ⇒ μ = F/mg

                     = 186.44/670

                      = 0.28