Elastic potential energy

Elastic potential energy is the energy stored in a spring when it is stretched/compressed past its natural length. Elastic potential energy is also a type of mechanical energy so we use it with the work-energy principals to solve complex systems with springs.

The tension in the string is given by: $$T = \frac{\lambda}{l}e$$ where $T$ is the tension, $\lambda$ is the modulus of elasticity, $l$ is the natural length of the sprint and $e$ is the extension of the spring past its natural length.

To work of the potential energy we use $W = Fd$, since tension is not linear we can use integration i.e. $W = \int{F \text{ }dd}$ to derive the potential energy of a spring $$\begin{align} W &= \int^e_0{T \text{ } de}\\ &= \int^e_0{\frac{\lambda}{l}e \text{ } de}\\ &= \left[ \frac{\lambda}{l} \times \frac{1}{2}e^2 \right]^e_0\\ &= \frac{\lambda}{2l}e^2 \end{align}$$

Example

The following systems spring has a natural length of $1.25 \text{ m}$, a modulus of elasticity of $32 \text{ N}$. The particle has a mass of $0.2 \text{ kg}$ and a coefficent of friction with the surface of $0.3$. Find the speed of the particle when it is $0.6 \text{ m}$ from the wall.

$$\begin{align} \text{Initial M.E.} &= \text{G.P.E.} + \text{K.E.} + \text{E.P.E.}\\ &= 0 + \left(\frac{1}{2} \times 0.2 \times 0^2 \right) + \left(\frac{32}{2 \times 1.25} \times 0.75^2\right)\\ &= 7.2 \text{ J} \end{align}$$

$$\begin{align} \text{Final M.E.} &= \text{G.P.E.} + \text{K.E.} + \text{E.P.E.}\\ &= 0 + \left(\frac{1}{2} \times 0.2 \times v^2 \right) + \left(\frac{32}{2 \times 1.25} \times -0.65^2\right)\\ &= (0.1v^2 + 5.408) \text{ J} \end{align}$$

$$\begin{align} \text{Initial M.E.} - \text{Final M.E.} &= \text{Work done against friction}\\ 7.2 - (0.1v^2 + 5.408) &= (\mu R) d\\ 1.792 - 0.1v^2 &= 0.3 \times (0.2 \times 9.8) \times 1.4\\ &= 0.8232\\ -0.1v^2 &= -0.9688\\ v^2 &= 9.688\\ v &= 3.11\text{ms}^{-1}\\ \end{align}$$