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Energy and Work

  1. Apr 11, 2008 #1
    A 1000.0 kg car experiences a net force of 9500N while decelerating from 30.0m/s to 23.4m/s. How far does the car travel while slowing down?

    I do not even know how to start to setup the problem. I know that W=Fd but I do not know how to apply this with the velocities.

    Any help would be appreciated. Thank you.

  2. jcsd
  3. Apr 11, 2008 #2


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    You can calculate the car's acceleration from the given data using Newton's laws, then use the standard kinematic equations of motion to calculate the distance.
  4. Apr 11, 2008 #3
    since we are on the energy and momentum chapter is there any way to do it with KE and PE?
  5. Apr 11, 2008 #4


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    Oh, sure. Use the work energy theorem
    [tex]W_{total} = W_{net} = \Delta KE[/tex]. In general , you've got to be a bit careful when using this equation, because W_net includes work done by both conservative (like gravity) and non-conservative (friction, etc.) forces, but it this case, those forces are given as one net number, so you don't have to worry about PE change.
    Last edited: Apr 11, 2008
  6. Apr 11, 2008 #5
    Fd=1/2mvf^2 - 1/2mvi^2

  7. Apr 11, 2008 #6


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    yes, where the value you use for F is the given F_net. Check it out both ways using the energy method vs. Newton 2 and the kinematic equations.
  8. Apr 12, 2008 #7
    Would vi be 23.4m/s
    and vf= 30m/s

    If it is the other way around then F*D= a negative number

    any help would be appreciated.
  9. Apr 13, 2008 #8
    vf is 23.4m/s since you're slowing down, right? so the change in kinetic you get should be a negative number, and keep in mind that the force of friction is always negative, so the work done is negative too.
  10. Apr 13, 2008 #9
    so vi= 30.0m/s

    and the force of 9500N is negative???

    -9500N*d=(1/2)(1000kg)(23.4m/s)^2 - (1/2)(1000kg)(30m/s)^2

  11. Apr 13, 2008 #10
  12. Apr 13, 2008 #11
    d= 18.5m or 19m with sig figs.?
  13. Apr 13, 2008 #12
    A general rule of thumb is to use the least amount of sig figs as the values given to you in the question.
  14. Apr 13, 2008 #13
    which would be 2 from 9500N
  15. Apr 13, 2008 #14
    do i have the right number of sig figs
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