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Homework Help: Work by a non constant force

  1. Jul 25, 2010 #1
    I don't know that if this is the right place to post this but it's safer than to post it in the general physics and get a warning
    1. The problem statement, all variables and given/known data
    let say we have an object moving with a non constant acceleration and has a non constant mass where we have a(t) as any function like t^3 or (7t^2-3)/2t and m(t) also is a function of time and of course position changes with respect to time I know the formula for work is W= [tex]\int m*a dx[/tex] but in this case its m(t) and a(t) and x is also a function of t so how would the integral work?

    2. Relevant equations
    F=m*a
    W=integral (ma)dx



    3. The attempt at a solution
     
  2. jcsd
  3. Jul 25, 2010 #2
    Use the fact that:

    [tex]
    a(t) = \frac{dv}{dt}
    [/tex]

    to find the velocity - time equation:

    [tex]
    v(t) = \int{a(t) \, dt} + C_{1}
    [/tex]

    where [itex]C_{1}[/itex] is an arbitrary integrating constant which can be determined only if you know the velocity at a particular instant in time [itex]v(t_{0}) = v_{0}[/itex]. Judging by what you had given in the statement of the problem, I do not see such kind of information.

    After you had found [itex]v(t)[/itex]. you simply use:

    [tex]
    v(t) = \frac{dx}{dt} \Rightarrow dx = v(t) \, dt
    [/tex]

    and substitute this into the integral for the total mechanical work. You integrate over t then.
     
  4. Jul 25, 2010 #3
    so in this case to calculate the work it would be necessary to know the velocity?
    ( I am assuming the information given is only a(t) and m(t) or F(t) and x(t))
     
  5. Jul 25, 2010 #4
    If you know [itex]a(t)[/itex] you need [itex]v(t_{0}) = v_{0}[/itex] to uniquely determine [itex]v(t)[/itex]. If you know [itex]x(t)[/itex], you can uniquely determine:

    [tex]
    dx = \dot{x}(t) \, dt
    [/tex]

    and you don't need additional information.
     
  6. Jul 25, 2010 #5
    ok but like this I would need to integrate from t1 to t2 not from x1 to x2 right?
     
  7. Jul 25, 2010 #6
    Yes! Your bounds must be with respect to whatever the [itex]d(\ldots)[/itex] quantity is in the integral.
     
  8. Jul 25, 2010 #7
    ok so the question maybe seem strange but if I have a ball of ice which is being exposed to the sun light so the ice melts and it's moving in a straight line but is being pushed with a non constant force and I have x(t)(so also v(t) and a(t) , F(t) ,and I want to know the work done by the force from the time it is whole to where it melts what would be the general interface of the solution? should I get when m(t) is zero as dF(t)/da(t) =0?
    (btw I am making the problem up)
     
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