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Energy, force, acceleration, and work with time dependent position

  1. Oct 18, 2012 #1
    1. The problem statement, all variables and given/known data
    A 4.00-kg particle moves along the x axis. Its position
    varies with time according to x = t + 2.0t^3, where x is in
    meters and t is in seconds. Find (a) the kinetic energy of
    the particle at any time t, (b) the acceleration of the particle
    and the force acting on it at time t, and (c) the work done
    on the particle in the interval t = 0 to t = 2.00 s.


    2. Relevant equations

    K.E. = 1/2 m v^2 = 2 + 24t^2 + 72t^4
    acceleration = 12t
    force = 48t

    3. The attempt at a solution

    I got the answers as you can see for part a and b, and i actually got the right answer eventually for part c as well using W = ΔK.E....however, the first time i tried getting the work i used the formula W = ∫F dot dx, evaluated from 0 to 2. which lead me to:

    48t (t + 2t^3) from 0 to 2
    which gave me the answer W = 1728 J, but that is not the same answer i got for W = ΔK.E. and i can't figure out why using the force to find work did not yield the right answer...any idea?
     
  2. jcsd
  3. Oct 18, 2012 #2
    OK, this is just a thought, I also don't know for sure, but to calculate the work which is done, the force has to be applied along the path? If we multiply 48t (t + 2t^3), I'm not sure we achieve this.
     
  4. Oct 18, 2012 #3
    how? if F = 48t and x = (t+3t^3)
     
  5. Oct 18, 2012 #4
    While force is a straight line and x is a polynomial of third order?
    I hope someone comes with better answer.
     
  6. Oct 19, 2012 #5
    The equation is W=∫Fdx
    That requires F as a function of x, and to integrate over x. To get it as a function of x requires solving a cubic polynomial equation. Converting that to time is a less difficult.
    The first step would be to calculate dx=f(t)dt (hint, what is dx/dt?). Now you have W=∫F(t)f(t)dt. Now you can integrate over time.
     
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