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Work done by friction on a box

  1. Oct 27, 2012 #1
    1. The problem statement, all variables and given/known data
    i've been stuck on this problem for about 45mins.

    A constant external force P = 160 N is applied to a 20 kg box,which is on a rough horizontal surface. The force pushes the box a distance of 8.0 m, in a time interval of 4.0 s, and the speed changes from v1 = 0.5 m/s to v2 = 2.6 m/s. Find the work done by friction. The work done by friction is closed to:
    There is a force applied at 30degree to the horizontal of the box.

    the answer is -1043 J but idk how to get that answer.

    2. Relevant equations
    Ff = µFn
    W=Fd
    W=mgd

    3. The attempt at a solution
    W=mgd
    W=20kg*9.8*cos(30)*8=1357.9J
     
    Last edited: Oct 27, 2012
  2. jcsd
  3. Oct 27, 2012 #2

    tiny-tim

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    welcome to pf!

    hi snoopyrawr! welcome to pf! :smile:
    (so the force is applied at 30° to the horizontal?)

    W = mgd is wrong

    W = Pd is right

    (and anyway, that's the work done by the force P … the question asks for the work done by the friction :wink:)

    try again :smile:
     
  4. Oct 27, 2012 #3
    yes there is a horizontal force of 30degrees applied on the box and how would you set up this problem if the friction coefficient is not given?
     
  5. Oct 27, 2012 #4

    tiny-tim

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    use the work energy theorem
     
  6. Oct 27, 2012 #5
    so it would be like this:
    W=KEf-KEi (KEf= kinetic final, KEi= kinetic initial)
    W=(1/2(20kg)(2.6m/s)^2)-(1/2(20kg)(0.5m/s)^2)
    W=65.1 J

    is there another step i have to do?
     
  7. Oct 28, 2012 #6

    tiny-tim

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    (just got up :zzz:)
    yup! :smile:
    that's the total work done …

    so what's the work done by friction? :wink:
     
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