# Homework Help: Work done of a car braking and the power needed to stop it

1. Nov 11, 2012

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

A car weighing 1500kg traveling at 20m/s brakes, and takes 50m to stop.

Part A) Calculate the work done
Part B) Calculate the power needed to stop the car

Known starting variables are..
u=20
v=0
s=50
m=1500

2. Relevant equations
P=Fv
W=Fs
F=ma
s=1/2(u+v)t
v=u+at

3. The attempt at a solution

Part A:

To find the work done I think I have to find out the force of the car going forward, and to find that out, I think I need to know the cars retardation (F=ma) first.

To find the retardation I used the motion equation below to find the time it took to stop.

s=1/2(u+v)t
50=0.5x20xt
50=10xt
t=5

So the time it took to stop is 5s.

Now I know the time, i used the motion equation below to find the retardation.

v=u+at
0=20+(ax5)
a=-20/5
a=-4

So the cars retardation is 4m/s^2.

Now I can find out the force going forward by using F=ma

F=ma
F=1500x4
F=6000

So the force going forward is 6000N

So therefore the work done is 6000x50 = 300000J ?

Part B:

Now I have worked out the power needed to stop the car two ways and I dont know which, if either, is correct.

The first way i just did work done divided by time it took to stop.

300000/5 = 60000W

But the second way I done it was force going forward times velocity.

6000x20 = 120000W

2. Nov 11, 2012

I think I may have worked out why I was getting two different answers for Part B, can anyone tell me if this is why?

P=Fv

power = force x velocity

the velocity in this equation is the average velocity, and I was using the initial velocity. The average velocity would be the dispalcement divided by the time, so it would be 50/5 = 10m/s. Then using the equation above it works, 6000 x 10 = 60000.

3. Nov 11, 2012

### rcgldr

The power at t = 0, v = 20 is 6000x20 = 120000w, the power at t = 5, v = 0 is 0 w. Should Part B be asking for the average power, or are you supposed to generate an equation for power?

P = F v = 6000 (20 - 4 t).

The relevant equations and your solution are for a constant rate of deceleration, but the problem statement doesn't mention a constant rate of deceleration, although it's probably assumed. Are the relevant equations given as part of the problem?

Last edited: Nov 11, 2012
4. Nov 11, 2012

No I put the relevant equations there from everything we have learnt so far. I assume the car brakes with a constant deceleration, although I don't know for sure, but everything we have done so far is with constant acceleration/deceleration.

I am not sure about part b, we have had only one lesson on power and work done. Its a very fast paced class and the teacher isn't exactly the best.

5. Nov 11, 2012

### rcgldr

Part B isn't worded very well, it should either ask to calculate the average power, or to show the formula for power versus time.

Part B could have asked to caculate the minimum (also constant) power to stop the car, but that involves an unrealistic force and acceleration that approach infinity as velocity approaches zero (something that a car could not do), and the math is much more complex, probably beyond where your class is at now.