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Checking my work on a conservation of energy problem

  1. Oct 16, 2013 #1
    1. The problem statement, all variables and given/known data

    Given: [itex]\frac{1}{2}m (\dot x)^2 + mg x = E[/itex]

    Gravitational force is mg.

    We need to show that by solving this DE that we can confirm that the conservation of energy correctly describes one-dimensional motion (the motion in a uniform field). That is, that the same motion is obtained as predicted by Newton' equation of motion. I had this up as a thread before -- (thanks to tiny-tim!) but I discovered that there might have been some bad errors in my original set-up. So I wanted to see if I corrected those.

    3. The attempt at a solution

    My attempt was as follows:

    [itex]\frac{1}{2}m\dot x + mg x = E[/itex] so by moving things around a bit I can reduce this to

    [itex](\dot x)^2 = \frac{2(E - mgx)}{m}[/itex]

    and from there I can solve the DE:

    [itex]\dot x = \sqrt{\frac{2(E - mgx)}{m}} \Rightarrow \int dx =\int \sqrt{\frac{2(E - mgx)}{m}}dt[/itex]

    from there I can move some variables some more, in my initial attempt I left the [itex]\sqrt{2}[/itex] on the right but this time I will move it over:

    [itex]\frac{\sqrt{m}}{\sqrt{2(E-mgx)}} \int dx = \int dt \Rightarrow \frac{\sqrt{m}}{\sqrt{2}} \int \frac{dx}{{\sqrt{E-mgx}}} = \int dt[/itex]

    trying a u substitution where [itex]u=\sqrt{E-mgx}[/itex] and [itex]du = -mgdx[/itex] I should have [itex]\frac{\sqrt{m}}{-mg \sqrt{2}} \int \frac{du}{{\sqrt{u}}} = \int dt[/itex] which gets me to [itex]\frac{2\sqrt{m}}{mg\sqrt{2}} \sqrt{u} + c = t[/itex]

    going back to what I substituted for u I have:
    [itex]\frac{2 \sqrt{m}}{mg \sqrt{2}} \sqrt{E-mgx} + c = t[/itex]

    and when I do the algebra I get (after moving the c over and squaring both sides):

    [itex]\frac{4m}{2m^2g^2}({E-mgx}) = t^2 - 2ct+ c^2[/itex]
    [itex]\frac{2E}{mg^2}-\frac{2x}{g} = {2}t^2 - 2ct+ c^2[/itex]
    [itex]-\frac{2x}{g} =-\frac{2E}{mg^2}+ {2}t^2 - 2ct+ c^2[/itex]
    [itex]x(t) =\frac{E}{mg}- gt^2 + cgt- \frac{c^2g}{2}[/itex]

    From here I plug in the equation for kinetic energy. [itex]E = (1/2) mv^2[/itex] which in this case would be [itex]\frac{E}{mg} = \frac{(1/2) mg^2t^2}{mg} = \frac{1}{2}gt^2[/itex]

    which changes my x(t) expression to:

    [itex]x(t) =-\frac{1}{2}gt^2 + cgt- \frac{c^2g}{2}[/itex]

    That looks like an equation of motion to me. And if we take a derivative of the original expression we get:

    [itex]\frac{1}{2}m \dot x \ddot x + \frac{1}{2}m \dot x \ddot x + mg \dot x = 0 \Rightarrow m \dot x \ddot x + mg \dot x = 0[/itex]

    and that shows that [itex]g=\ddot x [/itex]. Maybe that's what we were shooting for in the first place.

    and looking at the x(t) expression I got, we can replace g with [itex]\ddot x[/itex] and that makes x(t) into:
    [itex]x(t) =-\frac{1}{2}x + c\dot x- \frac{c^2}{2}\ddot x[/itex]

    I feel like I am close here, but not quite there. So I am hoping someone can tell me where I messed up.
     
  2. jcsd
  3. Oct 16, 2013 #2

    arildno

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    You have complexified this too much!
    Take the energy equation, which holds for ALL times, and differentiate it.
    That is allowed, because it holds for ALL t's.
    We then get:
    [tex]m\ddot{x}\dot{x}+mg\dot{x}=0[/tex]
    That is:
    [tex]\dot{x}(m\ddot{x}+mg)=0[/tex]
    meaning that Newton's 2.law of motion is contained as one of two possible solutions.
    If the velocity is zero, then position is also unchanging, and the energy equation holds trivially. The other solution, Newton's 2.law, is the one governing a system in motion, and consistent with the energy law of conservation.
     
  4. Oct 16, 2013 #3
    OK, so it was enough to differentiate and have done with it? That seems too simple. (I thought I had to do the other stuff b/c the question said to solve the DE)
     
  5. Oct 16, 2013 #4

    arildno

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    Simplicity is what is called for!
    :smile:
     
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