Work and Friction Problem

[NYKnick]

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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

 

[andrewsmack05]

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

 

[mikexima]

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

 

[zhemnis]

. 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?

 

[DrknoSDN]

. 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?

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.

 

[DrknoSDN]

i don't think the above calculation is correct, maybe i'm wrong.

first of all think logically, going back to the original question where the car skids to a stop in 7 seconds, after getting a = -5m/s*s, plug that into this equation: Vf=Vo + 2ad, where (Vf=0, Vo=35, a=-5m/s*s), you would get d = 3.5m. and with a coefficient of .5 it skids for that long to come to a stop.

the question states it comes to a stop this time in 175(!!!) meters, that immediately tells me the coefficient of friction is tiny, that's why it took so long (both in time and distance) to stop.

my calculation:

given: Vf = 0, Vo= 35, X=175, a=?
now use this formula to find a: Vf^2 = Vo^2 + 2ad => a = -.1m/s*s

since F friction = F net (they are both in the negative direction btw so no need for negating signs here)
Muk * N = ma
Muk * mg = m(0.1)
Muk = 0.1 / 10
Muk = 0.01

this makes sense to me because since the friction force is tiny (due to the property of the road surface or tire), it's almost like breaking on ice (very low friction), where it would take you a long distance before you come to a stop. and your answer 0.35 is still relatively close to .5 for Muk, which is disproportional to the dramatic difference in distance needed to come to a stop (3.5m vs 175m)

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.

my calculation:

given: Vf = 0, Vo= 35, X=175, a=?
now use this formula to find a: Vf^2 = Vo^2 + 2ad => a = -.1m/s*s

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.

Vf^2 = Vo^2 + 2ad
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.

 

[ipmed]

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

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