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Differential equations and physics proof problem

  1. Mar 8, 2012 #1
    1. The problem statement, all variables and given/known data

    The kinetic energy K of an object of mass m and velocity v is given by K=1/2mv2. Suppose an object of mass m, moving in a straight line, is acted upon by force F=F(s) that depends on position s. According to Newton's Second Law F(s)=ma. A is acceleration of the object.

    Show that the work done from moving a position s0 to a position s1 is equal to the change in the object's kinetic energy; that is, show that

    Work=[itex]\int^{s1}_{s0}[/itex]F(s)ds=[itex]\frac{1}{2}[/itex]mv[itex]^{2}_{1}[/itex]-[itex]\frac{1}{2}[/itex]mv[itex]^{2}_{0}[/itex]

    where v0=v(s0) and v1=v(s1) are the velocities of the objects at positions s0 and s1.

    2. Relevant equations
    I think all of the relevant equations are listed above.


    3. The attempt at a solution

    Work=[itex]\int^{s1}_{s0}[/itex]F(s)ds=[itex]\frac{1}{2}[/itex]mv[itex]^{2}_{1}[/itex]-[itex]\frac{1}{2}[/itex]mv[itex]^{2}_{0}[/itex]

    Looking at the right side of the equation suggests that I take the integral and it splits in to two, since it's a definite integral. So I should start with this Work=[itex]\int^{s1}_{s0}[/itex]F(s)ds and mess with it until it's the same as the right side of the equation.

    This is where I get stuck. I asked a guy for help and he showed me
    ma=m[itex]\frac{dv}{dt}[/itex]
    and then said "s" is a displacement vector s=vdt

    He showed me the next step, which was =[itex]\int^{t1}_{t0}[/itex]Fvdt

    I'm very confused how he got there, but he showed me the whole problem. This is it in its entirety.

    Work=[itex]\int^{s1}_{s0}[/itex]F(s)ds

    =[itex]\int^{t1}_{t0}[/itex]Fvdt

    =[itex]\int^{t1}_{t0}[/itex]m[itex]\frac{dv}{dt}vdt[/itex]

    =m[itex]\int^{t1}_{t0}[/itex]vdv

    =m(v2/2) evaluated from t0 to t1

    =[itex]\frac{1}{2}[/itex]mv[itex]^{2}_{1}[/itex]-[itex]\frac{1}{2}[/itex]mv[itex]^{2}_{0}[/itex]

    If someone could explain what happened step-by-step, that would be amazing. I can follow some of the steps, but all the variable swapping is making this impossible for me to understand.
     
    Last edited: Mar 9, 2012
  2. jcsd
  3. Mar 8, 2012 #2
    You've got your spatial integral bounds in the wrong order - you should go from s0 to s1.

    So the first step is to note that F(s) = ma, as you have. Velocity is the derivative of position: v = ds/dt; a = dv/dt, so F(s) = m dv/dt. The first trick is that we need to convert this spatial integral into a time integral. Since v = ds/dt, v dt = ds, which is what allows you to make the change of variables that seems to have confused you.
     
  4. Mar 9, 2012 #3
    You were correct, my bounds were wrong and I fixed them. And thank you for responding. This little bit you told me helps so much. It's late here, but I'll definitely take another look at this tomorrow.
     
  5. Mar 9, 2012 #4
    I've pretty much gotten this figured out at this point. I still have a couple questions at the end, though.


    =m[itex]\int^{t1}_{t0}[/itex]vdv

    After the dt's cancel each other, the problem has a interval with the upper and lower bounds of t, but there's a dv now that the integral works with. Does that effect the integration in any way? Should I now be evaluating with the bounds of v instead of t?

    =m(v2/2) evaluated from t0 to t1

    At this point, how do the t's fit into the problem? Normally you'd replace a variable with the upper bound and subtract it by the replaced lower bound of the same equation, but what do I do with the upper and lower bound variables on my integral symbol?
     
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