# Homework Help: Coefficient of kinetic friction between a car and the road

1. Jan 1, 2015

### discosucks

1. The problem statement, all variables and given/known data
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

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)

2. Relevant equations
(i) F = ma

Fk = μk Fn
(ii)

Dynamic equations

3. 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 .

2. Jan 1, 2015

### Staff: Mentor

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. Jan 1, 2015

### discosucks

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. Jan 1, 2015

### Staff: Mentor

Sure you can. Use your second equation, the one that describes Fk.

5. Jan 1, 2015

### discosucks

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. Jan 1, 2015

### Staff: Mentor

Good. Now it's time to apply Newton's 2nd law. (Your first equation.)

7. Jan 1, 2015

### discosucks

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. Jan 1, 2015

### Staff: Mentor

Yes.

Good.

You haven't used the velocity yet. You'll need that to figure out the time.

9. Jan 1, 2015

### discosucks

well using v = u + at

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

I got 20.60 s

10. Jan 1, 2015

### Staff: Mentor

The acceleration will be negative, since it's opposite to the velocity.

Show how you got that.

11. Jan 1, 2015

### discosucks

Ah ok I left that minus out

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

t = 4.393 .

12. Jan 1, 2015

### Staff: Mentor

Good!

13. Jan 1, 2015

### discosucks

Great thanks for your help :-)