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Homework Help: Interesting and annoying power problem

  1. Sep 29, 2004 #1
    I have a trouble with this problem:

    Suppose you have a particle with mass 30Kg, this particle is in influence of a force F that acts along the x axis and F=6+4x-3(x^2) and at x=0 the particle has 0m/s speed. Find the total work made by F between x=0 and x=3, then find the power given to the particle when x=3. (answers= 9 Joules and 22 Watts)

    Well the first thing is easy, only I have to integrate F between 0 and 3. but the second question I can't see what's to be done. I try to use the fact that if I have the work I must get the final time but I don't have an expression for the trayectory.

    Another way I have tried is to get the function of position by the fact that

    F=ma, a=F/m dv/dt=(6+4x-3(x^2))/m

    this is a partial differential equation involving the second derivative of a function u(x,t) and the general solution I get is:

    u(x,t)=a(x)+b(x)t+(t^2)((6+4x-3(x^2))/2m)

    but I don't have the sufficient conditions to get a good boundary problem or initial problem to separate a particular solution.


    My questions are: what conditions could I put, or what other way, easier, could I take to solve this?
     
  2. jcsd
  3. Sep 29, 2004 #2
    mdv/dt = 6 + 4x -3x^2 is a seperable second order differential equation.

    problem: there are no v terms in the force only x terms.
    solution: note that F is a position dependent force.

    problem: can't find a term for time.
    solution: is time really nescessary for finding power?

    this is a much better way to formulate the problem. nevermind about parioal derivatives and general solutions and whatever else. you don't need all that for this.
     
  4. Sep 29, 2004 #3

    krab

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    Science Advisor

    By conservation of energy, the work done is equal to the change in kinetic energy, which is 1/2mv^2. Use your answer to the first part and solve for v. Power is Fv.
     
  5. Sep 29, 2004 #4
    Thanks

    Thanks for the thinking. Another problem that is easier than i have wonder.
     
  6. Sep 29, 2004 #5
    power = work/time
    =((force*distance)/time
    = force* (distance/time)
    = force*velocity

    to find the velocity

    dv/dt = dv/dx (dx/dt) = dv/dx *v from the chain rule.

    vdv = Fdx

    perform the integral. done.
     
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