Coefficient of kinetic friction between a car and the road

  • Thread starter discosucks
  • Start date
  • #1
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Homework Statement


Just found this site ,looks really helpful , so i am doing some revision using past exam papers and have a few head scratches you guys maybe able to help with .

A car of mass 1000 kg is travelling at a speed of 25 m/s when it applies its brakes suddenly.
The car immediately skids. If the coefficient of kinetic friction between the car and the
road is 0.58, calculate:

i. the magnitude of the force of friction acting on the car during the skid

ii. the time taken from when the brakes are applied for the car to come to a complete
stop.

(Use g = 9.81 m/s2. You may neglect air resistance and assume level ground)

Homework Equations


(i) F = ma

Fk = μk Fn
(ii)

Dynamic equations


The Attempt at a Solution



so first i worked out that the force of the car is its mass (1000kg) x its acceleration (25 m/s) and that this is x the coefficient of 0.58 = 14500n

but when starting to work on the 2nd i became confused on what information i actually have . If i wanted to use one of the dynamic equations for solve the time taken to stop do i have enough to do so?

a= ?
v = 0 m/s
u =25m/s
t = ?
s=?

this led me to think that 25 m/s is not actually the acceleration like I have used it in the first part and now im very confused .
 

Answers and Replies

  • #2
Doc Al
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this led me to think that 25 m/s is not actually the acceleration
That's right. 25 m/s is the velocity of the car, not its acceleration. (Acceleration would be in units of ##m/s^2##.)

First step is to calculate the friction force on the car. (You'll then use that to figure out the acceleration.)
 
  • #3
23
3
thanks for the reply ,

see this is what confused me , i know that i need force and the coefficient to work out the friction force but i cant work out the force with out the acceleration ?
 
  • #4
Doc Al
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but i cant work out the force with out the acceleration ?
Sure you can. Use your second equation, the one that describes Fk.
 
  • #5
23
3
Ohh iv just noticed its the normal force I need to work out the friction .

so the normal force is 1000 x 9.81 and then this x 0.58 ( μk) = 5689.8 N

But how to i work out the acc from this?
 
  • #6
Doc Al
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Ohh iv just noticed its the normal force I need to work out the friction .

so the normal force is 1000 x 9.81 and then this x 0.58 ( μk) = 5689.8 N

But how to i work out the acc from this?
Good. Now it's time to apply Newton's 2nd law. (Your first equation.)
 
  • #7
23
3
AHH i see , so i can use my new fk as my force in terms of a = f/m ?

a = 5689.8/1000 = 5.6898 m/s2?

and was i right with my velocities ?

thanks again for your help .
 
  • #8
Doc Al
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AHH i see , so i can use my new fk as my force in terms of a = f/m ?
Yes.

a = 5689.8/1000 = 5.6898 m/s2?
Good.

and was i right with my velocities ?
You haven't used the velocity yet. You'll need that to figure out the time.
 
  • #9
23
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well using v = u + at

a=5.6898
v = 0 m/s
u =25m/s

I got 20.60 s
 
  • #10
Doc Al
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well using v = u + at

a=5.6898
v = 0 m/s
u =25m/s
The acceleration will be negative, since it's opposite to the velocity.

I got 20.60 s
Show how you got that.
 
  • #11
23
3
Ah ok I left that minus out

0 = 25 + (-5.698)t
-25 = -5.698t

t = 4.393 .
 
  • #12
Doc Al
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Ah ok I left that minus out

0 = 25 + (-5.698)t
-25 = -5.698t

t = 4.393 .
Good!
 
  • #13
23
3
Great thanks for your help :-)
 

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