Coefficient of Sliding Friction

  • Thread starter camjam73
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  • #1
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Golf Cart Questions

Homework Statement



Ms. Boylan on her 1750-kg cart is traveling at a constant speed of 8.5-m/s on a wet, straight, and slippery road. A mindless student steps out in from of the cart. The wheels lock and she begins skidding and slides to a halt in 12.0-m. What is the coefficient of sliding friction between the tires and the slippery road?

Homework Equations



vf^2 = vi^2 + 2ad

a = (Fapplied - Fopposed) / mass

u = Ffriction / weight

The Attempt at a Solution



Her acceleration is found with:
(0)^2 = (8.5)^2 + 2(a)(12)
a = -3.01 m/s^2

So, -3.01 equals applied force minus Ffriction over 1750.

Then I get stuck. If I could just figure that out, I could divide it by (1750 * 9.80) to get the coefficient. Thanks for your help.
 
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Answers and Replies

  • #2
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There is no force applied.
Initially, she is going at constant speed so the forces are balanced. Force net = 0.
Afterwards, she is skidding so the only force acting when she is slowing down is the force of friction.
Ff=ma
And you know from here on. :)
 
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  • #3
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Thanks a bunch! Just one other golf cart question (not friction related).

Homework Statement



Two campus supervisors in golf carts approach each other, one from the east (Cart A) and the other from the south (Cart B). Cart A has a mass of 1750-kg and cart B has a mass of 1275-kg and is traveling at 6.7056 meters per second. They collide, become locked, and move together at an angle of 30.0 degrees. What was the speed of cart A?

Homework Equations



(m1)(v1) + (m2)(v2) = (m1+m2)(v)

The Attempt at a Solution



(1750)(v1) + (1275)(6.7056) = (3025)(v)

I'm stuck here trying to solve for v1. I obviously need to find the velocity of the combined mass, but I'm not sure how to use 30 degrees.
 
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  • #4
79
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Momentum is a vector, not a scalar. It also has direction.
P1 + p2 = pf
f=final
This only works as a vector sum and if the collision is in one dimension. Your collision is in two dimensions. So in two dimensions, if you draw a diagram, you have that p1 + p2 = pf (as vectors).
You then have a right angle triangle with the hypotenuse pf and the other two sides are p1 and p2.
p1^2 + p2^2 = pf^2
You also know the angle, which can help you determine another relationship.
 
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  • #5
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Oh, so it's just 6.7 tan 30 = 3.87 ?
 
  • #6
79
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Well actually I don't know. It doesn't say if the angle of 30 is formed with the horizontal or with the vertical. But that's how you would do it. I just don't know where the 30 degree angle is.
 
  • #7
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Thanks for your help, Husky.
 

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