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anonymous

  • 5 years ago

Can someone please explain in a little more detail on the potential energy of a spring? I'm given the formula: U = 1/2kx^2 I understand that k represents the stiffness of the spring, but x (According to my text) says "it is the difference between the length of the stretched or compressed spring and the length of the spring when it is neither stretched nor compressed. "

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  1. anonymous
    • 5 years ago
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    \[U _{s} = 1/2kx ^{2}\]

  2. anonymous
    • 5 years ago
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    first of all potential energy of spring increases in both cases, compression and expansion. you are right, x is the difference between the stretched or compressed length to its natural length.

  3. anonymous
    • 5 years ago
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    This is derived from the Work-Energy Theorem, which states that\[W = \Delta U\]Let's take a look at the work done by a spring. From the definition of work, \[W = \int\limits {\bf \vec F} ~ dx\]where, for a spring, \[{\bf \vec F} = kx\]Therefore, \[W = \int\limits kx dx\]If we integrate, we obtain\[W = {1 \over 2} kx^2\]From the Work-Energy Theorem, we can see that\[{1 \over 2} kx^2 = \Delta U\]If we take the spring being at equilibrium when \(x=0\), then\[{1 \over 2} kx^2 = U\]

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