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Homework Help: Mechanical equivalent of heat

  1. Mar 30, 2010 #1
    1. The problem statement, all variables and given/known data
    A 5850kg truck decelerates uniformly from 85.0km/h to 50.0 km/h. if 95.0% of the energy reduction is converted into heat, how many kilocalories are produced?

    2. Relevant equations
    kinetic energy (Ek)= 1/2mv2
    1 Cal= 4.18J

    3. The attempt at a solution
    So first I converted 85.0km/h and 50.0km/h to m/s. 85.0km/h=26.4m/s, 50.0km/h=13.9m/s.
    Since 26.4m/s is initial velocity and 13.9m/s is final velocity, so the uniform velocity is v1-vo,: 13.9m/s-23.6m/s=-9.7m/s. (Although I'm not quite sure it's right)

    so I converted it to cal which 1 cal= 4.18J. so answer came 6.25*104cal. and it was incorrect. someone help me?
  2. jcsd
  3. Mar 30, 2010 #2
    You just messed up the algebra. In particular this step:


    You can't do that.

    It should be:

    0.5*(5850kg)*(26.4m/s)2 - 0.5*(5850kg)*(13.9m/s)2
  4. Mar 30, 2010 #3
    hmm so speed not suppose to get subtracted?
    sorry, my algebra is my most weakest spot.
  5. Mar 30, 2010 #4
    Well the speeds are subtracted but you need to square them first before you subtract.
  6. Mar 30, 2010 #5
    umm shouldn't final speed go first then initial speed go last on your solution?
    how do you get the kcal part? I'm not sure what to do from there

    Also I made mistake, initial speed should be 23.6m/s
  7. Mar 30, 2010 #6
    Ok so firstly we are looking for a positive energy amount (they ask for energy produced) so we place the larger speed first.

    The equation I gave you solves for the energy reduction of the truck in Joules. You need to figure out how much of that energy goes into heat and then convert from Joules to Kilocalories.
  8. Mar 30, 2010 #7
    yay I got it. thank you.
  9. Mar 30, 2010 #8
    No problem. Good job.
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