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Silly work done on a car question.

  1. Nov 30, 2011 #1
    1. The problem statement, all variables and given/known data

    It takes 139 kJ of work to accelerate a car from 22.8 m/s to 27.7 m/s. What is the car's mass?

    2. Relevant equations

    Well, we know that work is also known as J and J is = to kg*m^2/s^2 so we can use the equation J = m(vf - vi) to solve this I think.

    3. The attempt at a solution

    So We know that 139 kJ is equal to 139x10^3 J. Also m(27.7 m/s - 22.8 m/s) = m(4.9m/s), so combing these we assume that 139000 J = ?kg(4.9m/s) But from here I'm getting confused badly. I think I'm misusing the equation somehow as I'm not sure how to manipulate the 4.9m/s in order to get rid of the kg*m^2/s^2 units on the other side.

    Any help will be much appreciated. :)
     
  2. jcsd
  3. Nov 30, 2011 #2

    Doc Al

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    Staff: Mentor

    The work done will equal the change in the car's kinetic energy--not momentum. (That's why your units are not working out.) How would you calculate that?
     
  4. Nov 30, 2011 #3
    KE = 1/2mv^2. I had pondered this, so in other words it's possible to use my change in velocity in place of V in the KE formula?
     
  5. Nov 30, 2011 #4

    Doc Al

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    What you need is ΔKE = KEf - KEi.
     
  6. Nov 30, 2011 #5
    I'm sorry but how can that be usable if you don't have the mass? Isn't it essentially equal to m = 2K / (delta V)^2. Sorry, I'm getting confused I just don't see this.

    Edit: Never mind. I think I'm misunderstanding stuff because some notes I took in my lecture I miscopied. :/
     
  7. Nov 30, 2011 #6

    Doc Al

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    Staff: Mentor

    You'll be using this to solve for the mass. (As you've probably figured out by now.)
     
  8. Nov 30, 2011 #7
    Ahh never mind. I figure out what the problem was, I had used the formula you mentioned before but divided it by 2 as opposed to multiplying it. I've got it now the answer is 1.123x10^3 kg. Thanks very much for your help. :)
     
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