# Coefficient of Sliding Friction

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

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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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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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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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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Oh, so it's just 6.7 tan 30 = 3.87 ?

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.