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Work-Kinetic Energy

  1. Oct 24, 2007 #1
    1. The problem statement, all variables and given/known data
    With brakes fully applied, a 1540 kg car decelerates from a speed of 94.0 km/hr. What is the work done by the braking force in bringing the car to a stop? What is the change in the kinetic energy of the car?


    2. Relevant equations



    3. The attempt at a solution
    W = -Fd
    Not sure how to start this problem?
     
  2. jcsd
  3. Oct 24, 2007 #2
    This problem is, I think, most easily solved by finding the initial kinetic energy of the car.
    Calculations are simplified if you convert 94.0 km/hr to m/s ==> 94.0 km/hr ~ 26.111111 m/s.
    Now, K = 1/2 m v^2 , so the initial kinetic energy of the car is given by K = (1/2)x(1540 kg)x(26.111111 m/s)^2 = approx. 525 kJ.
    Since the final velocity of the car is zero, the final kinetic energy is zero as well. Thus, the change in kinetic energy is 0 - 525 kJ = -525 kJ.
    The work is equal to the total change in energy, so W = -525 kJ as well.
    ===> change in KE = W = -525 kJ <===
     
    Last edited: Oct 24, 2007
  4. Oct 24, 2007 #3
    I understand...but its saying its incorrect?
     
  5. Oct 24, 2007 #4
    well, i may have rounded a bit too much... also, it's common just to write the magnitude of the work, so try W = 525 kJ if -525 kJ doesn't work. A more exact answer is 524978.3951 J
     
  6. Oct 24, 2007 #5
    I got the change in kinetic energy right but, is the work is wrong when entered as positive. So do I enter is negative?
     
  7. Oct 24, 2007 #6
    yeah, negative was my first suggestion... in my original post i had it as positive but then edited it shortly thereafter to be negative... just a typo.
     
  8. Oct 25, 2007 #7
    Oh ok...that was right then, thank you much!!
     
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