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Homework Help: Work Done Question

  1. Mar 20, 2010 #1

    Paymemoney

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    1. The problem statement, all variables and given/known data
    A 40.0kg box initially at rest is pushed 5.00m along a rough, horizontal floor with a constant applied horizontal force of 130N. If the coefficient of friction between box and floor is 0.300, find the work done by the normal force?


    2. Relevant equations
    W=F*d


    3. The attempt at a solution
    The answer is 0N, however i got 1260(which i know does not seem logically correct).
    So can someone explain to me why does work done by the normal force equal to 0N.
     
    Paymemoney, Mar 20, 2010
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  3. Mar 20, 2010 #2

    thebigstar25

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    because the work = F.d, dont forget that F and d (the force and displacment) are both vectors , F.d is a dot product, then you have work=F.d=Fdcostheta , where theta is the angle between the force(in your case is the normal force) and the displacement .. and since theta is 90 degree then w=Fdcos90=0J (the unit of work is J not N)...
     
    thebigstar25, Mar 20, 2010
  4. Mar 20, 2010 #3

    Paymemoney

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    ok thanks i understand, however another question i would like to ask, Then how would you explain how i got the work done by friction to be -588. Because to find the Frictional force i times the (coefficient of friction by the Normal Force ). However if you say that the Normal Force is mgcostheta then i don't understand why the answer would be -588, because W=-392cos90 * 5 then it would be 0 Joules?
     
    Paymemoney, Mar 20, 2010
  5. Mar 20, 2010 #4

    thebigstar25

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    who said that the normal force is mgcostheta???? ...
    If u carefully draw the freebody diagram u can see that u have 4 forces acting on the box ..
    1- the force that horizontally applied which is 130 N ..
    2- the force of friction which is opposing the force you applied and = coeffient of friction * normal force..
    3-the weight of the box which is downward and = m*g ..
    4-the normal force which is upward..

    Sice the resulant motion is in the horizontal axes, then according to Newtons 2nd law .. The net force in the vertical axes = zero .. Which implies that : the normal force - the weight of the box = zero ..
    Then, N = mg (and not mgcostheta) ...
    So, the friction force = coeffient of friction* mg ..

    Hopefully this was clear enough :) .. If u still have question ask again ....
     
    thebigstar25, Mar 20, 2010
  6. Mar 20, 2010 #5

    Paymemoney

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    yep thanks i understand now :)
     
    Paymemoney, Mar 20, 2010
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