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Work Question - Is something missing?

  1. Jun 26, 2008 #1
    1. The problem statement, all variables and given/known data
    Calculate the work done accelerating a 2.0 kg object from 2.0 m/s to 3.0 m/s.

    mass=2.0 kg
    Initial Velocity= 2.0 m/s
    Final Velocity= 3.0 m/s


    2. Relevant equations
    Work=applied force x displacement



    3. The attempt at a solution
    Correct me, if I am wrong, but don't I need a bit more information to answer this question? This was all the information given to me.
     
  2. jcsd
  3. Jun 26, 2008 #2

    alphysicist

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    Hi Sabellic,

    No, that's all you need. The equation you listed is true; however, there is another equation that involves work and velocity. What is it?
     
  4. Jun 26, 2008 #3
    Hmmmm. I don't know. The only thing that I can conjure up is that the work can be equivalent to (mass/acceleration) * (final velocity^2 - initial velocity^2)/2 * a^2

    But that doesnt work because i dont have the acceleration.
     
  5. Jun 26, 2008 #4

    alphysicist

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    That's so close! But it needs to be:

    (mass/acceleration) * (final velocity^2 - initial velocity^2)/2 * a

    and so the accelerations will cancel.

    Then notice that what you have left is:

    (1/2) m vf^2 - (1/2) m vi^2

    Does that look more familiar? What are those terms? Once you have that, you will have the important equation that you need to know.
     
  6. Jun 26, 2008 #5
    Oh God you're right. I wrote (mass/acceleration) instead of mass * acceleration. So yes, they would cancel out. Thanks a lot for this. So it will be 5 Joules, I think.
     
  7. Jun 26, 2008 #6

    alphysicist

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    That's good, but you don't want to have to derive that important formula every time. The way to think about it is

    total work = change in kinetic energy

    (where total work includes all conservative and non-conservative work).

    Since kinetic energy is (1/2) m v^2, you could write down the relation you need right away.
     
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