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Homework Help: Kinetic Energy help please

  1. Mar 3, 2006 #1
    I've been trying this for an hour, please have mercy and help!

    You push a 67 kg box accross a floor where the coef. Kfriction is .55. The force you exert is horizontal. How much power is needed to push the box at a speed of .5m/s?

    I've tried w=Ki-Wnc, but I guess that isn't it.

    Thank you!!
  2. jcsd
  3. Mar 3, 2006 #2
    To solve this problem sum the forces in the x and y directions. You should get
    F_{x} = \frac{dp}{dt} + f_{s} and F_{y} = N= m g....(1)
    calling this eq (1). So we may write from eq (1)
    f_{s} = \mu_{s} m g ...(2)
    Now we wish to know the power, which we may relate to the force exerted by the equation
    P= F v ...(3)
    So substitute eq (2) and eq (1) into eq (3), with the numerical values given in the problem. Note that the quantity
    is the change in momentum with respect to time, if the force is a constant force (which I am assuming it is) this term is zero, which yields the result for eq (3) is
    P = \mu_{s} m g v
    which are all numerical values in problem statement. I hope this helps.
    sincerely, x
  4. Mar 4, 2006 #3


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    Staff Emeritus
    Science Advisor
    Gold Member

    Work done = force x distance.
    The force in this case is [itex] F = \mu R[/itex] where [itex]\mu[/itex] is the coefficent of friction.
    Let the distance moved be 0.5m.
    Power is [itex]\frac{workdone}{time}[/itex], the speed is 0.5m/s, therefore the time is 1 and in this case, power = workdone.
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