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How do you calculate potential energy in a spring?

The potential energy stored in a spring, commonly known as elastic potential energy, can be calculated using the formula:

PE=12kx2PE = \frac{1}{2} k x^2

In this equation, kk represents the spring constant, which quantifies the stiffness of the spring, while xx signifies the displacement of the spring from its equilibrium position. This formula is derived from Hooke’s Law, which states that the force exerted by a spring is directly proportional to its displacement from the equilibrium position.

To compute the potential energy, the first step is to determine the spring constant kk. This can be achieved by applying a known force to the spring and measuring the resulting displacement. The spring constant can then be calculated using the formula:

k=Fxk = \frac{F}{x}

where FF is the applied force and xx is the corresponding displacement. Once you have established the spring constant, you can calculate the potential energy for any given displacement.

The displacement xx refers to the distance the spring is either stretched or compressed from its equilibrium position. This measurement can be taken directly if the spring is at rest, or it can be inferred from the maximum amplitude if the spring is in motion.

The formula PE=12kx2PE = \frac{1}{2} k x^2 indicates that the potential energy is proportional to the square of the displacement. Consequently, if the displacement is doubled, the potential energy increases by a factor of four. Conversely, halving the displacement results in the potential energy decreasing to a quarter of its initial value.

It is important to note that this formula is valid under the assumption that the spring adheres to Hooke’s Law and that the displacement remains within a reasonable range. If the spring is stretched or compressed beyond its elastic limit, it may not return to its original shape, and the formula may no longer be applicable. In such cases, a more complex model may be necessary to accurately determine the potential energy.

Answered by: Prof. Emma Johnson
IB Physics Tutor
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