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What is its speed in m/s when x = 8.00m

  1. Mar 23, 2015 #1
    1. The problem statement, all variables and given/known data
    The graph below shows how the force on a 0.500 kg particle varies with position. If the particle has speed https://www.physicsforums.com/x-apple-ql-id://92193C91-FD4D-4A2B-AE3A-A915237F9B1D/x-apple-ql-magic/4C0641E9-916A-4CAB-8E8F-A20E6CE7E040.pdf [Broken] at x = 2.00 m, what is its speed in m/s when x = 8.00 m?
    Have a picture
    2. Relevant equations
    KE = 0.5m*vf^2 - 0.5 m*vi^2
    W = F * d

    3. The attempt at a solution
    So what I basically did was to get an area of the graph because it will be F * d. After getting an area (work) I assumed that I could use KE equation like
    KE = 0.5m*vf^2 - 0.5 m*vi^2 = 51J
    Then I got Vf as 14.5m/s. However, the answer is 15.0 m/s. Can anyone help me what I did wrong here???
     

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    Last edited by a moderator: May 7, 2017
  2. jcsd
  3. Mar 23, 2015 #2

    Simon Bridge

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    Looks like rounding off to me.
     
  4. Mar 23, 2015 #3
    Are the answers reliable? I got somewhere around 14.5 as well
     
  5. Mar 23, 2015 #4
    You get the answer 15.0 if you take the work exerted on the particle until x=8 to be 55 (so including the x=0 till x=2 part) . But I think your calculation is actually the correct one and the answer given is wrong..
     
  6. Mar 23, 2015 #5

    BvU

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    Get the same result as you do. The 15 m/s comes out when you do ##\Delta{\rm KE} = 55 = {\tfrac 1 2} mv_f^2 - {\tfrac 1 2} m\;2.23^2, \ ## and I really see no reason to use 55 instead of 51 J.

    Especially since the particle can't have been released (with a small nudge) at x = 0, since the same calculation then gives 4 m/s at x = 2.

    I thought about starting off at some point with a negative speed, but the calculation you did is valid, no matter what happens before, so I'm puzzled o_O ?!
     
  7. Mar 23, 2015 #6
    Yeah, the work exerted on the particle before the velocity measurement cannot have contributed to the increase of velocity after the measurement. So the answer given is just wrong..
     
  8. Mar 23, 2015 #7
    I just found these questions from the internet so it's not 100 percent reliable haha
     
  9. Mar 23, 2015 #8
    Haha I think the answer is just wrong! Thank you for helping me out though! :)
     
  10. Mar 24, 2015 #9

    BvU

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    So 14.8 is wrong and 15 is correct ? We all end up at 14.5 !
     
  11. Mar 24, 2015 #10

    Simon Bridge

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    If you insist. But all things considered, it looks like the "correct" answer is in fact incorrect.
    Appearances can be deceiving.
     
  12. Jun 24, 2015 #11
    Actually the answer given is correct when rounding 14.999 to three sig fig. Work is area under curve of force vs position graph. Setting 55 = 1/2mVf^2 - 1/2mVi^2. When working with five sig figs for calculations (this comes from squaring the velocity) your answer will be 14.999 m/s => 15.0 m/s
     
  13. Jun 24, 2015 #12

    haruspex

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    Read the question carefully. It says from x=2, not x=0. The work done is only 51J.
     
  14. Jun 24, 2015 #13
    Haruspex, you are correct. I made a poor assumption that the graph continued to (0,0) for I had just seen and assigned this question. Thank you. The correct work is 51J as you stated, which results in a final velocity of 14.456 m/s ==> 14.5 m/s.
     
  15. Jun 25, 2015 #14

    SammyS

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    This this the given graph:

    image-jpg.80862.jpg
    At a position of about x = 1.660 m, the particle has a kinetic energy of zero. It will never be to to the left of that under the influence of this as the net force. To do so would imply negative kinetic energy, and thus an imaginary speed.
     
  16. Jun 25, 2015 #15

    haruspex

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    That helps narrow down the error in the question. Most likely it was supposed to say the given speed is at x=0. The only other I can think of is that the initial speed should have been given as 2.83, making the answer (c).
     
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