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Work/ Kinetic Energy

  1. Sep 30, 2007 #1
    1. The problem statement, all variables and given/known data

    With brakes fully applied, a 1670 kg car decelerates from a speed of 78.0 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

    W=f*s, KE= .5(m)(v^2)

    3. The attempt at a solution

    I am not sure if for the first answer how to start the problem, I have the mass, but I do not know the acceleration to solve for F...how can I go about solving this problem? Also for the second question I thought I would just have to use the equation for KE to solve, but it doesnt seem to be working...any suggestions?
     
  2. jcsd
  3. Sep 30, 2007 #2

    dynamicsolo

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    Have you had the work-kinetic energy theorem yet? In any case, you can answer part 2 right off: what was the KE of the car at the start? what is the KE at the end?
     
  4. Sep 30, 2007 #3

    Dick

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    The answers to both questions are the same. Energy is conserved, so the KE of the car is equal to the energy consumed by the braking force. If you are getting a wrong answer for the second part it's likely because you've overlooked that the speed is in km/hr. Not m/sec. Convert km/hr to m/sec.
     
  5. Sep 30, 2007 #4

    dynamicsolo

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    I was resisting pointing this out. I wanted to find out if the point of the problem was to discover that this is the case. (If her course has covered the W-KE theorem, then the answer to one part would point to the answer for the other...)
     
  6. Sep 30, 2007 #5
    for the second part of the problem do I use the equation KE =.5m(v^2) ....m= 1670, v = 78000 cause I have to convert it to m/hr....isn't the KE at the end 0, since the car stops? so wouldn't my answer be the final minus the inital?
     
  7. Sep 30, 2007 #6

    Dick

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    Yes but v=78000 is completely meaningless. What units? The usual units of energy are joules. In that case you want the speed in m/sec. 78000 certainly isn't right. Hint, 1hr=3600sec.
     
    Last edited: Sep 30, 2007
  8. Sep 30, 2007 #7

    Dick

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    Apologies. You did say it was m/hr. But that's not the number you want to plug into (1/2)*mv^2 to get joules.
     
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