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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.

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- Thread starter Aaron Wong
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- #1

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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

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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 d^{2}U/dx^{2} = 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 d^{2}U/dx^{2} = 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.

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

ie. The expression you started with is a 1st order

Notice that F=mg means that dU/dx = -mg and d

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.

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