Solving a Physics Work Problem: Calculating Work on a Moving Car

[misyg]

While driving my 1,500kg car at 18 m/s, I unexpectedly see a stop sign and must brake hard. How much work must be done on the car to stop it?

I used KE=1/2(mv2)
I came up with the answer of 243,000 J, but my professor gave me only 5 out of 10 points for the answer...any one see where I went wrong?

 

[method_man]

Perhaps you didn't detail your answer enough. The question is, what do you have to do to stop your car? Change it's kinetic energy. So, work done is equal to the difference between final and initial kinetic energy:

W=E_KF-E_KI

 

[misyg]

So W=0J-243,000J
So it would take -243,000J of work to stop the car

 

[method_man]

Yes. You have a nice example http://www.fsas.upm.edu.my/~zainalas/PHY2001/2006/example4.pdf" on page 6.

 

[Pengwuino]

So W=0J-243,000J
So it would take -243,000J of work to stop the car

Yes, there is a very important aspect of the wording that must be given attention. Doing positive work on something, such as lifting an object off the ground, stores energy in the object that will later allow it to perform work of its own. Since in this problem you removed the ability of the car to do work (if it slows to a stop, its energy has been taken away), negative work had to be done on it.

Now, if the the question asks how much work is done by the car (which would have to be onto the ground or atmosphere or whatever heats up due to stopping the car), it would have been a positive number.