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Finding velocity with changing distance and acceleration

  1. Apr 14, 2008 #1
    I have a physics problem I have been trying to figure out for a week and I'm now to the point of way past frustrated. The question is:

    An object with the mass 1.44*10^-27 kg is propelled at an initial speed of 3.84*10^5 m/s toward a fixed nucleus 4.80 m away. The object is repelled by the nucleus with the force F= a/x^2 where x is the seperation between the object and the nucleus and a = 2.59*10^-26 Nm^2. What is the speed of the object when it is 7.46*10^-10 from the nucleus?
    I have tried many ways to solve this to the point of I dont know what I'm doing anymore and just trying to plug in numbers into equations and still not getting the right answer of 3.15*10^5.
    equations:
    F=m*a
    final speed^2=initial speed^2+2*acceleration*distance
     
  2. jcsd
  3. Apr 14, 2008 #2
    Use conservation of energy.
     
  4. Apr 14, 2008 #3
    as in work= .5 mass* velocity^2-.5 mass*velocity^2?
     
  5. Apr 14, 2008 #4
    as in...
    energy = kinetic energy + potential energy = constant
     
  6. Apr 15, 2008 #5
    m = 1.44*10^-27 kg, initial speed v1 = 3.84*10^5 m/s, x1 = 4.80 m,
    force F= a/x^2 where a = 2.59*10^-26 Nm^2. x2 = 7.46*10^-10 m, find v2.
    note force is not a constant, but just like gravitational force f = GmM/x^2 except f is attractive but F is repulsive. Remember the corresponding potential energy for f is -GmM/x, then the corresponding potential energy for F is a/x,
    initial total energy = (1/2)mv1^2 + a/x1
    final total energy = (1/2)mv2^2 + a/x2
    energy conservation: (1/2)mv1^2 + a/x1 = (1/2)mv2^2 + a/x2
    then get v2 = 3.15*10^5 m/s
     
    Last edited: Apr 15, 2008
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