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Find missing force for accelerating electron

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  1. Jan 29, 2015 #1
    • Moved from general physics forum
    [Mentor's note: no template because this thread was moved into the homework forum after it had already attracted several useful replies]
    [Mentor's note: Thread title changed to reflect question content]

    The question is:
    An electron is a subatomic particle (m = 9.11 10-31 kg) that is subject to electric forces. An electron moving in the +x direction accelerates from an initial velocity of +5.71 105 m/s to a final velocity of +2.03 106 m/s while traveling a distance of 0.037 m. The electron's acceleration is due to two electric forces parallel to the x axis: vector F 1 = +8.08 10-17 N, and vector F 2, which points in the -x direction. Find the magnitudes of the net force acting on the electron and the electric force vector F 2.

    Vf= final velocity Vi = initial velocity
    Fn= net force d= distance
    F1= forces parallel to x-axis

    to get acceleration I did: (Vf)^2- (Vi)^2 / 2d and I got 5.128 m/s^2
    I then calculated Fn= ma = 4.671 N
    To get F2 I did F1- Fn and got -4.671N

    but Webassign told me these are wrong, and I've already done other things and got other (wrong) answers... what am I doing wrong? This seems fairly straightforward.

    Thanks!
     
    Last edited by a moderator: Jan 30, 2015
  2. jcsd
  3. Jan 29, 2015 #2

    Stephen Tashi

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    Is there a factor with a power of ten associated with that result?
     
  4. Jan 29, 2015 #3
    Oh yes, sorry. 5.128 X 10 ^13
     
  5. Jan 29, 2015 #4

    Stephen Tashi

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    Why do think that calculation computes acceleration? Have you studied such a formula?
     
  6. Jan 29, 2015 #5
    After some searching on the web, someone had used that to find it. I'm guessing that is wrong? Should I do F1/ m to get it? I just wasn't sure if I should use that formula since its only the force in the +x direction and not net force.

    Or well, I guess using F1/m won't work, cause then the find Fn it'd be Fn=ma, which would get me F1 , and then F2 would be 0...
     
  7. Jan 29, 2015 #6

    Stephen Tashi

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    There is an equation that relates force and distance to a change in kinetic energy, but kinetic energy is [itex] \frac{mv^2}{2} [/itex] not [itex] \frac{v^2}{2} [/itex]. Perhaps you saw a problem where [itex] m [/itex] happened to be 1 .
     
  8. Jan 29, 2015 #7
    Hmm okay, so then should I be using (mv^2)/ 2 to find the acceleration? ... does that v stand for initial or final velocity?

    Sorry for all these questions, I probably seem so clueless... I've just never had a physics class before so I'm slightly overwhelmed. Thanks for you help, though.
     
  9. Jan 29, 2015 #8

    Stephen Tashi

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    Yes

    Almost all physics formulas have several interpretations. One interpretation will be the "change in" interpretation.

    For example the formula: Work = (Force)(distance) can be applied without reference to a "change in"
    However, when some initial amount of Work has already been done.
    Total Work = (Work already done) + (change in work)
    = (Work already done) + (Force)(change in distance) [ assuming Force is constant with respect to distance.]

    The formula (Work done) = kinetic energy likewise has a "change in" interpretation.
    In your problem the mass has an initial velocity so it has an initial kinetic energy. The Force in the problem acts to create a change in the kinetic energy. So you need the "change in" interpretation and that would be (change in work done) = (change in kinetic energy). The change in kinetic energy involves the initial and final kinetic energies, so it's like the formula that you used, except you need the [itex] m [/itex] term. Notice the 0.037 meters is a "change in distance", so you don't worry about finding the difference beween an initial and final distance.
     
  10. Jan 29, 2015 #9
    Okay, I think I'm beginning to get this... my original equation had the divisor of 2*d... are you saying it should just be everything divided by 2, or should it be divided by 2*d ?

    Thanks!
     
  11. Jan 29, 2015 #10

    Stephen Tashi

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    Solve for [itex] F [/itex] in the equation [itex] (F)(d) = \frac{m v^2}{2} [/itex] using the "change in" interpretation. [itex] F = [/itex] your original equation but with an [itex] m [/itex] in the numerator of the fraction.
     
  12. Jan 29, 2015 #11

    PeterDonis

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    Which specific answers did it say were wrong? All of them? Or only some?
     
  13. Jan 30, 2015 #12

    gneill

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    It is fine. It follows from the standard kinematics equation: ##{v_f}^2 - {v_i}^2 = 2 a d##.
     
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