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Calculus based physics - Energy

  1. Mar 6, 2012 #1
    1. The problem statement, all variables and given/known data
    As a particle moves along the x axis it is acted upon by a single conservative force given by Fx=(22-2.0x)N where x is in m. The potential energy associated with this force has the value +40 J at the origin (x=0). What is the value of the potential energy at x=4 m?


    2. Relevant equations



    3. The attempt at a solution

    Well, you have to move it so that it is Fx= 2xdx - 20dx ( -dU/dx ) so you have this equation. Then I integrated it and got x^2-20x+C... C is 40 in this case, so you have x^2-20x+40. Since it says potential energy at x=4, I plugged in the 4 and got 4^2-20*4+40= -24. That's not correct, though.


    1. The problem statement, all variables and given/known data
    A spring (k=300N/m) is placed in a vertical position with its lower end supported by a horizontal surface. A 1.5 kg block that is initially 0.10m above the upper end of the spring is dropped from rest onto the spring. What is the kinetic energy of the block at the instant it has fallen 0.20m ( compressing the spring 0.10m) ?


    2. Relevant equations

    1/2kxi2=Ei=mghf - energy conservation


    3. The attempt at a solution

    No idea. This equation above is a guess. I think it should be this one. If it is, you probably have to solve it for something, but I don't know. Solve for k ?
     
  2. jcsd
  3. Mar 6, 2012 #2
    For the second question:

    You're on the right track thinking about conservation of energy. What will define the energy of the system at the moment the block is dropped? What will define the energy of the system after it has dropped and compressed the spring?
     
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