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Another Work Problem

  1. Oct 22, 2008 #1
    1. The problem statement, all variables and given/known data


    A block with mass m = 10 kg slides along a horizontal floor. The kinetic friction force opposing the motion of the block is constant at a value 14 N. At clock reading t=0.0 s and position x=0.0 m, the block has an instantaneous velocity of v = 4 m/s to the right. A force P is applied to the block starting at the instant it is in the position x=0. The direction of P remains fixed and positive (to the right), but its magnitude varies according to the graph above. Use work and energy arguments to answer the questions below; do not make use of Newton's second law.
    a) Calculate the work done by the force P in the displacement from x=0 to x=4 m.

    b) Calculate the work done by the frictional force during the same displacement

    c) Calculate the work done by the net force acting on the block during the displacement.

    d) Calculate the change in kinetic energy of the block.

    e) Calculate the final kinetic energy of the block; that is, the kinetic energy it possesses on reaching the position x=4 m.

    f) Calculate the velocity of the block at the position x=4 m.

    g) Calculate how far the block will slide beyond position x=4 m if the force P abruptly drops to zero at this position and remains zero from there on.

    h) Calculate any potential energy changes of the earth-block system that take place over the course of the motion from x=0 to x=4 m.

    2. Relevant equations

    3. The attempt at a solution
    a) I found the area under the curve = 100 J
    b) 14N*4m*cos180 = -56 J
    c or d, I am not sure which) 100J+(-56J) = 44 J
    I am not sure about the rest. Can someone help me out??
    Last edited: Oct 22, 2008
  2. jcsd
  3. Oct 22, 2008 #2


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    d) what happened to the additional work? What did it go into?

    e) what's the total?

    f) You know m so what's v?

    g) how long until it stops then with friction slowing it by 14 J per meter?

    h) is it horizontal?
  4. Oct 22, 2008 #3
    Is the answer for c and d the same??
  5. Oct 22, 2008 #4


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    Where else would it go?
  6. Oct 22, 2008 #5
    I am not sure I know what part e is asking for...would initial kinetic energy be 1/2(10kg)(4m/s)^2=80 and then final would be 80-44=36??
  7. Oct 22, 2008 #6
    For part f, 44J=1/2(10kg)v^2-1/2(10kg)(4m/s)^2 = 4.98m/s ?
  8. Oct 22, 2008 #7
    Is the answer for h=0?
  9. Oct 22, 2008 #8
    g) 1/2(10kg)(4.98m/s)^2 = (14N)(x) x = 8.86m ?
    Last edited: Oct 22, 2008
  10. Oct 22, 2008 #9
    Sorry to be such a bother, but I applied for a tutor but they did not have one that could work around my school schedule :P
  11. Oct 22, 2008 #10


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    P was adding KE to the object, not subtracting it.
  12. Oct 22, 2008 #11
    You found the initial KE correctly, but not the final.

    Wnet = change in kinetic energy = KEfinal - KEinitial
    Wnet = 44J as you found out, so:
    44J = KEfinal - 80J
    KEfinal = 44J + 80J = 124J

    This looks good.

    for part h, I'm not sure, but I would guess that the answer is zero.

    For part g, there is still a frictional force of 14N slowing it down. But remember, you said you are not allowed to use newtons second law (F = ma) so you have to use the Wnet = KE2 - KE1 to solve for it.
  13. Oct 22, 2008 #12


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    No. The total KE at that point is equal to mV2/2

    Edit: OK. You got the right answer, but I was looking for 124 J = mV2/2
    Last edited: Oct 22, 2008
  14. Oct 22, 2008 #13
    So it would be 80+44?
  15. Oct 22, 2008 #14
    So 124J=1/2(10kg)(v^2) v=4.98m/s
  16. Oct 22, 2008 #15


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    This is the right idea, but the wrong total KE at x=4.

    Just redo it.

    Edit: OK so you did have the right total KE. I didn't recognize it.
  17. Oct 22, 2008 #16
    KE at x=4 is 124J
  18. Oct 22, 2008 #17
    That was a lot of great help. Thank you soo much!!
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