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Work done by a non-constant force

  1. Nov 23, 2009 #1
    1. The problem statement, all variables and given/known data
    Hey this is my first time on here and i'm kinda lost on what im doing. Any help would be great and the question is
    Two protons that are very far apart are hurled straight at each other, each with an initial kinetic energy of 0.18 MeV, where 1 mega electron volt is equal to 1*10^6 multiply (1.6*10^-19) joules. What is the separation of the protons from each other when they momentarily come to a stop?


    2. Relevant equations
    Dont think i can type the equation

    3. The attempt at a solution

    So i multiplied the initial kinetic energy by how much 1 electron vold is equal to and then multiplied it by the work

    0.18 * 1*10^6 * 1.6* 10^-19
    =2.88* 10^-14 J (kinetic energy of both protons)

    Next i try to find the distance and this is where i have the trouble

    2.88*10^-14 J = (9*10^9)(1.6*10^-19)^2/d
    d= (9*10^9)(1.6*10^-19)^2 / 2.88*10^-14 J
    d= 8 * 10^-15m
     
  2. jcsd
  3. Nov 23, 2009 #2

    rl.bhat

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    Hi clutch12, welcome to PF.
    The system consists of two protons.
    Initially what is the total kinetic energy of the system and what is the net potential energy of the system? What are these energies when they are momentarily at rest?
     
  4. Nov 23, 2009 #3
    oh so i now need to find the net potential energy of the system right?
     
  5. Nov 23, 2009 #4
    and thanks for the welcome
     
  6. Nov 23, 2009 #5

    Delphi51

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    It doesn't make sense to multiply energy by work. The answer would come out in Joules², which doesn't have any meaning for me.

    You need a formula for the electric potential energy of a charge near another charge. That is the same as the work done to push one charge close to another. It is the integral of dW = F*dr = kqq/r²*dr
    and the answer is kqq/r. No doubt you can look it up in your textbook.
    So you have .18 MeV = kqq/r
    and you just have to solve for r.
     
  7. Nov 23, 2009 #6
    oh never mind you mean kq1q2/r^2 *dr
     
    Last edited: Nov 23, 2009
  8. Nov 23, 2009 #7

    rl.bhat

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    Both the protons are moving with KE = 0.18 MeV each. So net KE of the system = 2*0.18 MeV.
     
  9. Nov 23, 2009 #8
    so basically from what i tried earlier all i would need to do is multiply 0.18 by 2 and then multiply everything all over to get my answer?
     
  10. Nov 23, 2009 #9
    and thanks you guys for your help i finally got the answer right
     
  11. Nov 23, 2009 #10

    Delphi51

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    And thanks to rl.bhat for catching the factor of 2!
     
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