Work Done to Stop 1590 kg Car from 82 km/hr

[chaotixmonjuish]

With brakes fully applied, a 1590 kg car decelerates from a speed of 82.0 km/hr. What is the work done by the braking force in bringing the car to a stop?

I'm not really sure what to do here. I tried the whole (1590*82)/2. That didn't work.

 

[Mentz114]

Work = energy, so the work done to stop the car is the same as its energy when the braking starts.

 

[Link-]

The work done by the brakes is the change in kinetic energy.

W=1/2*m(Vf^2-Vi^2)

That's all. Notice that final velocity is zero.

I tried the whole (1590*82)/2.

You didn't use the kinetic energy equation. First of all you need to change units, velocity should be in m/s not on km/hr and this velocity should be square.

 

[chaotixmonjuish]

I got 409655 Joules, I'm not sure if this is right.

The velocity I used was 22.77

 

[cristo]

I got 409655 Joules, I'm not sure if this is right.

The velocity I used was 22.77

That's correct (well, more or less; I used 22.78, but it depends upon your rounding).

 

[chaotixmonjuish]

According to my lon capa it's wrong.

 

[chaotixmonjuish]

I got the answer wrong

 

[cristo]

What's a "lon capa." Like I said, it depends upon your rounding and how many significant figures the answer needs to be to.

 

[chaotixmonjuish]

It's some computer program that tolerates +/- 10%

I used 22.77, and I got 412.54 kJ