# Work and Friction Problem

Discussion in 'MCAT Study Question Q&A' started by NYKnick, Aug 8, 2011.

1. ### NYKnick 5+ Year Member

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A car moving at 35m/s on dry pavement, skids to a stop in 7 seconds. What is the coefficient of friction between the car's tires and the pavement?

a. .2
b. .5
c. 1
d. 2

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2. ### andrewsmack05 SDN Lifetime Donor Lifetime DonorVerified Account 7+ Year Member

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Hey,

Here is what I got:

Force = - mu(coefficient of friction) x Normal Force

simplifying:

mass x acceleration = -mu x (mass x gravity)

thus mass will cancel:

acceleration = -mu x gravity

since it is decelerating, acceleration is negative

plugging in the numbers we get:

-(35 m/s) / (7s) = -mu x (10 m/s^2)

(-5 m/s^2) = -mu x (10 m/s^2)
-0.5 = -mu

mu=0.5

3. ### mikexima 7+ Year Member

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yup that works. think this was a prolem in EK.

if you had the mass of the car you could solve it as a work problem. W = change in energy = change in momentum

4. ### zhemnis

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. A car moving at 35 m/s on dry pavement, skids to a stop over 175 m. What is the coefficient of friction between the car’s tires and the pavement?

5. ### DrknoSDN

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I would start by finding time by using average velocity = Vfinal + Vinitial /2 = 35+0/2 = 17.5 m/s. And if D=175m then Time=10 seconds

If you have 10 seconds you know the car is slowing down at 3.5 m/s.
Since Force(kinetic) = u(k)*Fnormal
Mass is constant, and if we are assuming the normal force on the car is (m*g) you can just say u(k) = 3.5/10 = 0.35 is the coefficient of kinetic friction.

6. ### DrknoSDN

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Lol, Xishen... Gotta read the question man.
I was responding to zhemnis, and when you change variables you can't assume that the acceleration is going to be constant between 2 problems.
Zhemnis asked a new question 3 years later, with entirely new variables.
You are not given acceleration in his problem so assuming -5m/s accel is going to give you a wrong answer. Acceleration is going to be different for these problems.

And,,, your kidding right? The equation is correct so I am not sure where the error occurred... but I'll help you with the arithmetic.

0 = 35^2 + 2(a)(175m)
0 = (1225) + 2(a)(175m)
-1225 = 2(a)(175)
-1225/(2*175) = a
-1225/350=a
... get this... A = -3.5m/s.

I just didn't use the equation cause it's faster to do simple math in your head. Just find average velocity (moving velocity/2), and then you can find how long it takes to travel D... Once you have time, then A=((deltaV)/T)

I'm not guaranteeing that I'm correct.. Just that your misunderstanding was probably from an incorrect assumption about acceleration.. followed by incorrect algebra. Sorry to be harsh sounding but that's just a lot of mistakes in one post.

7. ### ipmed ipmed 5+ Year Member

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You dont need mass to solve this problem. 1/2mv^2=MuGD and the M cancel

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