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Finding the Work Done by Friction w/out Friction

  • #1
I feel like I'm missing some obvious or simple piece but I can't seem to figure out the first part. If I can get the first part, the second part is easy.

1. Homework Statement

An 80-kg baseball player running at 6.5 m/s goes into a slide 3.0 m from second base. He comes to rest just as he touches the bag. What is the work done by friction? What is the coefficient of kinetic friction between the player and the ground?

Homework Equations


W = F * d
W = ΔK
F = μN

The Attempt at a Solution


This seemed straight forward. I calculated the kinetic energy to get the overall work:
W = Kf - Ki = 1/2mv_f^2 - 1/2mv_i^2 = -169 J
This should be the same work as the friction but I've been told the correct answer should be -1700 J. Am I wrong about something here?
 
Last edited:

Answers and Replies

  • #2
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Check your arithmetic.
 
  • #3
Check your arithmetic.
m = 80kg
v_i = 6.5m/s
v_f = 0

1/2(80)(0)^2 - 1/2(80)(6.5)^2 =
0 - 1/2(80)(42.25) =
0 - 1/2(338) =
0 - 169 = -169

Same answer.
 
  • #4
Bystander
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Check it again --- it's glaring at me with big red eyes.
 
  • #5
Wow! How did I miss that twice!

1/2(80)(0)^2 - 1/2(80)(6.5)^2 =
0 - 1/2(80)(42.25) =
0 - 1/2(3380) =
0 - 1690 = -1690 ≈ -1700
 

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