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Simple problem in calculating kinetic energy

  1. Mar 22, 2009 #1
    1. The problem statement, all variables and given/known data
    If you push a crate horizontally with a force of 100 N across a 10 meter factory floor, and the friction between the crate the the floor is a steady 70 N, how much kinetic energy is gained by the crate?


    2. Relevant equations
    KE=1/2mv²
    Work=change in KE
    Work=force*distance traveled

    3. The attempt at a solution

    Im not sure how to get the net force. If all you have to do is add the friciton force and force from push than the solution should be:
    +100-(-70)=170 N
    170 N(10 meters)=1700 J

    is this correct?
     
  2. jcsd
  3. Mar 22, 2009 #2

    tiny-tim

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    Welcome to PF!

    Hi Laurlaur790! Welcome to PF! :smile:
    Nooo … what effect do you think friction has on KE? :wink:
     
  4. Mar 22, 2009 #3
    I think that the more friction there is, the less kinetic energy there is.
     
  5. Mar 22, 2009 #4

    tiny-tim

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    That's right! :biggrin:

    So what effect does that have on the numbers?
     
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