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 66 kg player begins sliding 3.40 mfrom the base with a speed of 4.35 m/s.

Part A

If the player comes to rest at third base, how much work was done on the player by friction?

W =

  J  

Part 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

Part A -

Use the expression for finding the acceleration, a -

V^2 = Vo^2 + 2ad
Substituting: 0^2 = (4.35)^2 + 2(a)(3.40)

=> a = - (4.35)^2 / (2*3.40) = -2.78 m/s^2

Now, F = ma
=> F = (66)*(2.78)
F = 183.5 N

So, the work done on the player by the friction, W = F*d

= 183.5 * 3.40 = 623.8 J

Part B -

W = uN *d
Where N is normal force

=> W = u[(66)(9.8)](3.40)

=> 623.8 =  u[(66)(9.8)](3.40)
=> u = 623.8 / (66*9.8*3.40) = 0.28

So, coeff. of kinetic friction = 0.28