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Which strategy should I use to approach a problem? Sometimes I'll use $U= 1/2kx^2$ when I should use $F=-kx$ and vice versa. What are clues in a problem that would let me identify the correct approach to a given problem? And shouldn't both approaches give me the same answer in any scenario?

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Yes, both approaches should give you the same answer, provided you consider all forces / all expenditures of energy –  Pranav Hosangadi Oct 31 '13 at 4:51
    
In which scenario would it be inappropriate to use F=-kx? To calculate work done? –  TopDog Oct 31 '13 at 4:52
    
When you want to find the acceleration of the mass at the end of the spring, the force approach would be applied. This would also be the case if you were required to find the work required to compress a spring, etc –  Pranav Hosangadi Oct 31 '13 at 4:58
    
If there is time involved use balance of forces, otherwise you energy. –  ja72 Oct 31 '13 at 14:26
    
You can distinguish where to use which in the same way you distinguish between $F = mg$ and $U = mgh$ –  shortstheory Dec 1 '13 at 2:27
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2 Answers 2

Obviously both are equivalent, in terms of applicability (you get the second directly by taking the gradient of the first).

My "rule of thumb" is that for simple exercices,

  • the energy view is more suited for initial/final state problems (like, "find the end velocity of mass" or "find the maximal stretch of the spring") because you don't have to integrate the force,
  • the force view is more suited for trajectory problems (like "find the position of the mass as a function of time") because you don't have to derivate the energy.

Of course that's only a way to choose one way to start solving a given problem; most of the time both are perfectly usable.

When going to more complicated problems however this "rule" is not necessary true ; in any case you should choose the one that gives you as little unnecessary information as possible - while still answering the question.

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Force is just minus gradient of U. Gradient is just (spatial) derivative.

$F=-\frac{d}{dx}\frac{kx^2}{2}=-\frac{2xk}{2}=-kx$

So, if your task has "differential" nature, it will probably suitable for force method. And if your task has "integral" nature, then it will probably suitable for potential method.

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