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Work Done by Friction

  1. Apr 7, 2013 #1
    1. The problem statement, all variables and given/known data
    A constant external force P = 160 N (diagonally with an angle of 30) 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. The work done by friction is closest to:

    A) 1040 J B) -1040 J C) -1170 J D) 1110 J E) 1170 J


    2. Relevant equations
    W = F*d


    3. The attempt at a solution
    I made a free-body diagram, but I am confused on how to find the frictional force to plug into the work equation.
    So far, I have F_P = 160cos(30°), and F_P - f = 20a.
    a = (2.6-0.5)/4 = 0.525 m/s^2, so 112.58 - f = 20(0.525) and f = 102.08.
    W = (102.08)(8) = 816.64 J, but none of those are on the answers, so where did I go terribly wrong?
     
    Last edited: Apr 7, 2013
  2. jcsd
  3. Apr 7, 2013 #2
    First, where did you get the 30 degrees?

    Second, since you are not given a coeffiecient of friction, you won't be able to calculate the frictional force to plug into the work equation. YOu can, however, calculate the net work done, and the work done by the force.
     
  4. Apr 7, 2013 #3
    Sorry, there is a figure given with the force P acting on the box diagonally with an angle of 30°. So what you are saying is that I can find the work done by the force, but not by friction?
     
  5. Apr 7, 2013 #4
    There are two forces doing work in this problem. The force, p, and the the force of friction. There is also a net work. If you can find 2 of the 3 you can find the third.
     
  6. Apr 7, 2013 #5
    Okay, so I have the force of p (correct I hope), so could you guide me on how to find the net work? And how exactly would I use the two to find the frictional force? I am sorry, but my Physics lecturer is terrible, and I am really struggling in Physics.
     
  7. Apr 7, 2013 #6
    Net work is defined by a change in Kinectic energy.

    Net work would also be the sum of the all of the work done.

    Also, I don't believe you need the time in this problem. There maybe a way to use it, you shouldn't NEED it.
     
  8. Apr 7, 2013 #7
    Okay, so now I have Wnet = ΔKE = 0.5(20)(2.6^2-0.5^2) = 65.1 J.

    Using your second statement, I assumed Wnet = F + f, so 65.1 = 112.58 + f and f = -47.48 N.

    W = F*d = (-47.48)(8) = -379.84 J

    I am still doing something wrong hmm...
     
  9. Apr 7, 2013 #8
    Wnet is the sum of the work done, not the forces. ##W_{net}=W_p+W_{friction}##

    Otherwise, you're on the right track.
     
  10. Apr 7, 2013 #9
    You wrote:
    The question says 160 N as the force
     
  11. Apr 7, 2013 #10
    Using that I got ##65.1 = 900.67 + W_f##
    ##W_f = -835.57 J##
    I am still doing something wrong...
    I found ##W_P## by doing this: $$W_P = 130\cos 30\times 8 = 900.67 J$$
     
  12. Apr 7, 2013 #11
    As phiz pointed out, you're using 130N as the force, where in the question it stated, 160N.
     
  13. Apr 7, 2013 #12
    My mistake I fixed it and got the correct answer! Thank you for all your help!
     
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