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Work-Kinetic Energy Theorem

  1. Apr 27, 2010 #1
    1. The problem statement, all variables and given/known data
    In the neck of the picture tube of a certain black-and-white television set, an electron gun contains two charged metallic plates 2.80cm apart. An electric force accelerates each electron in the beam from rest to 9.60% of the speed of light over this distance. a) determine the kinetic energy of the electron as it leaves the electron gun. Electrons carry this energy to a phosphorescent material on the inner surface of the television screen material making it glow. For an electron passing between the plates in the electron gun, determine b) the magnitude of the constant electric force acting on the electron, c) the acceleration, and d) the time of flight.

    Electron mass = 9.10938188 * 10^-31 kg
    C = 299792458 m/s
    Electron velocity = 28780075.968 m/s (.0960 * C)
    Distance between plates = .028m

    2. Relevant equations

    KE = 1/2m* v^2
    F= mass * acceleration

    3. The attempt at a solution

    a) KE = 1/2m * v^2
    = 1/2(9.10938188*10^-31kg) * ( 28780075.968m/s)
    = 3.77261759 * 10^-16 Joules
    b) F=m*a
    = (9.10938188*10^-31kg) * [ (28780075.968m/s)^2 / (.028m) ]
    = 2.694722685 * 10^-14 N
    c) a = F/m
    = 2.694722685*10^-14N / 9.10938188*10^-31kg
    = 2.9581839*10-46 m/s^2
    d) ???

    and Im not sure if my parts b and c are correct
  2. jcsd
  3. Apr 27, 2010 #2


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    Homework Helper

    b is not correct, you should divide the KE by the distance.

  4. Apr 28, 2010 #3
    ah yea you're right it's N*m then when I divide it cancels out.

    w= f * d

    so ke = f * d
    ke/d = f?
  5. Apr 28, 2010 #4


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    Homework Helper

    The change of the KE is equal to the work. But initially, KE=0.

    Just divide your 3.7726 * 10^-16 Joules by 0.028 m.

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