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Homework Help: Work Done Question

  1. Mar 20, 2010 #1
    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

    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.
  2. jcsd
  3. Mar 20, 2010 #2
    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)...
  4. Mar 20, 2010 #3
    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?
  5. Mar 20, 2010 #4
    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 ....
  6. Mar 20, 2010 #5
    yep thanks i understand now :)
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