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Force v Displacement graph

  1. Feb 21, 2016 #1
    1. The problem statement, all variables and given/known data
    If the following Force v. Displacement graph is for a 9.9kg object and it's initial velocity is 28.6 m/s the what is its velocity at 30 m?
    0406768d-b992-4a57-9d87-a50f2b36ebc3.gif

    2. Relevant equations
    I am not certain that i am even using the right equations with this but:

    W = fd
    KE = 1/2mv(final)^2 - 1/2mv(initial)^2

    3. The attempt at a solution
    I thought since W=Fd and work is proportional to kinetic energy, which is
    KE=1/2mv(final)^2 - 1/2mv(initial)^2 and i then could substitute in the Fd, making the new equation:

    Fd = 1/2mv(final)^2 - 1/2mv(initial)^2

    This could then be rearranged to find final velocity:

    sqrt { (Fd + 1/2mv(initial)^2) / 1/2m } = v(final)

    Only the issue is that i do not know which F to use to find the final velocity at d=30meters. Because if i use the 0 the whole Fd cancels out, and i get the wrong answer. But if i use 100 then it comes out as larger than the initial velocity, which is not possible as it is losing speed due to lack of force after 20 meters.
     
  2. jcsd
  3. Feb 21, 2016 #2
    Actually work done is a scalar product of force and displacement vector and the amount of work done is sored with the body as kinetic energy so calculate the work done from the graph and use it to find the final velocity.
     
  4. Feb 21, 2016 #3
    thanks, haha so i was way off, isn't work then supposed to be the area under the line on the graph?
     
  5. Feb 21, 2016 #4
    As the force changes over time, The best way to find the work in graphs is by getting the area
    KhJXEzh.png

    You have a triangle and a rectangle. Find the area and sum it up. You will end you will the the work done
     
  6. Feb 21, 2016 #5
    in the graph the value of force and displacement is given so calculate the work done
     
  7. Feb 21, 2016 #6
    And do i use the numbers of the x and y axes to find area? in this case the rectangle being 100x20 and the triangle as 1/2x100x10 ?
     
  8. Feb 21, 2016 #7
    i keep getting an answer larger than the initial velocity of 28.6, is that at all possible?
     
  9. Feb 21, 2016 #8
    I got 2500 for the work, is that correct?
     
  10. Feb 21, 2016 #9
    I havent calculated the velocity but it should be larger than the initial. Work is adding energy to the system. You add more energy this energy will be kinetic. While the mass and 0.5 obviously dont change then the only thing that will change is the velocity.

    and the work is 2500 right :D

    So yes it should be larger.
     
  11. Feb 21, 2016 #10
    Yes. So, show us your detailed calculations for getting the final velocity. Incidentally, what are the units.
     
  12. Feb 21, 2016 #11
    W = 1/2mv(final)^2 - 1/2mv(initial)^2
    2500 = (0.5 x 9.9 x V^2) - (0.5 x 9.9 x 28.6^2)
    2500 = 4.95V^2 - 4048.9
    6548.9 = 4.95V^2
    1323.0 = V^2
    36.4 = v(final)

    Units:
    W is in Joules
    Mass is in kilograms
    Velocity is in meters per second
    Displacement is in meters
    Force is in newtons
     
  13. Feb 21, 2016 #12
    I didn't check your arithmetic, but your approach looks OK.
     
  14. Feb 21, 2016 #13
    Thank you, i got the right answer, just checked it! (:
     
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