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B Relationship between potential energy and force

  1. Sep 9, 2016 #1
    Hi,
    I wonder which steps of my following deduce are wrong
    dU=-dW
    dU=mg dh
    dW=Fdx
    thus, mgdh=-Fdx (dx=dh)
    then, F=-mg which is a constant in most of situation
    However, F does change in some situation.

    I am confused about this.
     
  2. jcsd
  3. Sep 10, 2016 #2

    Simon Bridge

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    Sticking to 1D, if you have potential energy distribution U(x) so U is explicitly dependent of the position x ... then:
    F = -dU/dx ... so you see F will depend on x if dU/dx depends on x.

    Constraining motion to the x axis, the work done by force F is given by:
    W = Fx so dW/dx = x.dF/dx + F (by the product rule) so dW = x.dF + F.dx and substituting above gives:
    dW = x.dF - dU which gives a 2nd order DE in U.
    Thus dW = -dU everywhere only in the situation where d2U/dx2 = 0 ...
    ie. The expression you started with is a 1st order approximation only (expand U(x) as a power series and see.)

    Notice that F=mg means that dU/dx = -mg and d2U/dx2 = 0 so the relation is exact, and you will see that dU/dx does not depend on x (see first sentence above).

    Try for a harmonic oscillator ... ##U(x)=\frac{1}{2}kx^2 \implies F=?## ... here I have chosen U(x):U(0)=0 for you.
     
    Last edited: Sep 10, 2016
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