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Work-Energy theorem and kinetic energy pertaining to a car

  1. Dec 2, 2008 #1
    1. The problem statement, all variables and given/known data

    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?"

    2. Relevant equations

    Work = [Final kinetic energy] - [Initial kinetic energy]

    Kinetic energy = (1/2)mv^2

    3. 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]
    v=10 m/s

    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.
    Last edited: Dec 2, 2008
  2. jcsd
  3. Dec 2, 2008 #2


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    Homework Helper

    In this problem the energy is removed from the system. So the formula should be

    Work = [Initial kinetic energy] - [Final kinetic energy]
    -(51,000 J) = [(1/2)*(1,100 kg)*(v^2)] - [(1/2)*(1,100 kg)*(12.777777... m/s)^2]
  4. Dec 2, 2008 #3
    That would give me a negative value of 79.9 to take the square root of, which is imaginary. Or do I disregard the negative?

    EDIT: Never mind, this worked out perfectly for part a and I ended up getting 8.4 m/s. Now it's just part c that confuses me.

    EDIT Again: I figured out Part C. It's heat energy due to friction.

    Thanks rl.bhat!
    Last edited: Dec 2, 2008
  5. Dec 2, 2008 #4


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    Homework Helper

    Check your calculations.
    Initial KE = (1/2)*(1.100kg0*(12.7777)^2 J
    Energy removed = 51000 J
    Now find the final KE and final speed.
  6. Dec 2, 2008 #5
    My calculation mistake was a pretty sheepish one...not squaring the velocity.
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