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Energy Problems.

  1. Jul 26, 2006 #1
    A 29.4 kg box initially at rest is pushed 3.5 m along a rough, horizontal floor with a constant applied horizontal force of 132.853 N. The acceleration of gravity is 9.8 m/s^2 . If the coefficient of friction between box and floor is 0.408, find the work done by the applied force. Answer in units of J.

    My understanding that it is asking for the applied force WITHOUT the friction? So just 3.5m * (29.4kg)(9.8m/s^2) = the applied force?
  2. jcsd
  3. Jul 26, 2006 #2


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    You may wish to reconsider your answer there, what is the definition of work done?
  4. Jul 26, 2006 #3
    Expanding Hootenanny's hint:

    [tex]W = \Sigma \vec{F} \times \Delta s[/tex]


    [tex]\Sigma \vec{F} = F - F_{a}[/tex]

    [tex]F[/tex] - constant applied horizontal force of 132.853 N
  5. Jul 26, 2006 #4

    But that didn't help me :(
  6. Jul 26, 2006 #5
    OK. No problem. :smile:

    [tex]\Sigma \vec{F} = F - f_k[/tex]

    [tex]F[/tex] - constant applied horizontal force of 132.853 N
    [tex]f_k[/tex] - force of kinetic friction

    To find the kinetic friction force:

    [tex]f_k =\mu_k.n[/tex]
    where [itex]n[/itex] is the normal force, and since the box is along an horizontal floor it is equal to the weight of the box.

    , you calculate [itex]\Sigma \vec{F}[/itex], using [itex]\Sigma \vec{F} = F - f_k[/itex]

    And finally, you compute the work done, using [itex]W = \Sigma \vec{F} \times \Delta s[/itex], since you now know [itex]\Sigma \vec{F}[/itex] and [itex]\Delta s[/itex].

    I hope this time you understand. If not, we are here to provide more help. :wink:
  7. Jul 26, 2006 #6
    Alright I got the correct answer.

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