Work done of a car braking and the power needed to stop it

In summary, the problem states that a car weighing 1500kg traveling at 20m/s brakes and takes 50m to stop. To calculate the work done, the formula W=Fs is used, where F is found using F=ma and a is found using the motion equation s=1/2(u+v)t. The power needed to stop the car can be calculated using the formula P=Fv, where v is the average velocity. However, the problem statement does not mention a constant rate of deceleration, so it is assumed. Part B is not clear, but it could be asking for the average power or for the formula for power versus time.
  • #1
FaraDazed
347
2

Homework Statement



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

Homework Equations


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

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 don't 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



Please help, my head hurts!
 
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  • #2
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
FaraDazed said:
Part B ... P=Fv
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?
 
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  • #4
rcgldr said:
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?

No I put the relevant equations there from everything we have learned 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
FaraDazed said:
I am not sure about part b
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.
 

FAQ: Work done of a car braking and the power needed to stop it

What is the work done by a car during braking?

The work done by a car during braking is the force applied by the brakes multiplied by the distance the car travels while braking. This work is equal to the change in kinetic energy of the car.

How is the work done by a car during braking related to the power needed to stop it?

The work done by a car during braking is directly related to the power needed to stop it. The power needed to stop a car is equal to the work done divided by the time it takes to come to a complete stop.

How does the weight of the car affect the work done and power needed during braking?

The weight of the car affects the work done and power needed during braking because it determines the force needed to stop the car. A heavier car will require more force to stop, resulting in a greater work done and power needed.

What factors besides weight can affect the work done and power needed during braking?

The speed of the car, the condition of the brakes, and the friction between the tires and the road are all factors that can affect the work done and power needed during braking. A higher speed will result in a greater work done and power needed, while worn brakes or low friction can increase the distance needed to stop and thus increase the work done and power needed.

How does the work done and power needed during braking change in different road conditions?

The work done and power needed during braking can vary depending on the road conditions. On a slippery road, the friction between the tires and the road will be lower, resulting in a longer distance needed to stop and a greater work done and power needed. On a rough or uneven road, the force needed to stop the car may be greater, resulting in a higher work done and power needed.

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