An 1,100 kg car is traveling 46 km/h on a level road. The brakes are applied long enough to remove 51 kJ of energy.
A. What is the final speed of the car?
B. How much more kinetic energy must be removed by the brakes to stop the car?
C. Is this energy really being "removed?" Where does it "go?"
Work = [Final kinetic energy] - [Initial kinetic energy]
Kinetic energy = (1/2)mv^2
The Attempt at a Solution
Part A Attempt:
I started out by using the first equation, abbreviated as:
W = KE(f) - KE(i)
I then converted 51 kJ into 51,000 J and put that value in for "work." After that, I substituted the KE values for (1/2)mv^2 according to the second equation I listed.
51,000 J = [(1/2)mv^2] - [(1/2)mv^2]
I plugged in some more values. In order to get the units to come out as Joules, I had to convert 46 km/h for the initial veloctity to 12.7777... m/s.
51,000 J = [(1/2)*(1,100 kg)*(v^2)] - [(1/2)*(1,100 kg)*(12.777777... m/s)^2]
51,000 J = [550 kg*v^2] - [7027.77777... J]
58027.777... J = [550 kg*v^2]
This is the value I got for the final speed of the car. However, I can't shake off the feeling that I did something wrong here. I still don't fully understand what work exactly is. Was I right in inserting 51,000 J into that equation for W?
Part B Attempt:
This is a little easier. Using the 10 m/s value I obtained from the previous problem:
KE = (1/2)mv^2
KE = (1/2)(1,100 kg)(10 m/s)2
KE = 55,000 J
Part C Attempt:
I am confused on where the energy actually goes. I think it gets converted into thermal energy because the energy used for the brake gets released as gas into the atmosphere. Again, I'm not 100% sure on this.
Thanks for any help you can provide.