Elastic Potential Energy (College Board AP® Physics 1: Algebra-Based)

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Leander Oates

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Physics

Elastic potential energy

The potential energy of common physical systems can be described using the physical properties of that system
Elastic potential energy is defined as:

The energy stored within a material (e.g. a spring) when it is stretched or compressed

Therefore, for a material obeying Hooke’s Law, the elastic potential energy of an ideal spring can be calculated using:

$U_{s} = \frac{1}{2} k {(∆ x)}^{2}$

Where:
- $U_{s}$ = elastic potential energy, measured in $J$
- $k$ = spring constant, measured in $N / m$
- $∆ x$ = distance the spring has been stretched or compressed from it's equilibrium length, measured in $m$
Elastic potential energy is a scalar quantity with magnitude only

Diagram of three springs with attached masses, showing unstretched, stretched, and compressed states, with energy equations and displacement denoted as Δx. — **A spring that is stretched or compressed has elastic potential energy**

It is very dangerous if a wire under large stress suddenly breaks because the elastic potential energy of the strained wire is converted into kinetic energy

$elastic potential energy = kinetic energy$

$U_{s} = K$

$\frac{1}{2} k {(∆ x)}^{2} = \frac{1}{2} m v^{2}$

$v \propto ∆ x$

This equation shows
- The greater the extension of a wire $∆ x$ the greater the speed $v$ it will have when it breaks

Worked Example

Cars are built with shock absorbers to make a ride more comfortable. Shock absorbers are strong springs that absorb energy when the car goes over a bump.

A shock absorber spring has a spring constant of $52 kN / m$ is fixed next to a wheel and compressed a distance of $10 cm$ . Which of the following is the correct value for the elastic potential energy of the spring?

A: $26 N \cdot m$

B: $260 N \cdot m$

C: $2600 N \cdot m$

D: $26 000 N \cdot m$

The correct answer is B

Step 1: List the known values

Spring constant, $k = 52 kN / m = 52 \times 10^{3} N / m$
Compression, $∆ x = 10 cm = 10 \times 10^{- 2} m$

Step 2: Substitute the values into the elastic potential energy equation

$U_{s} = \frac{1}{2} k {(∆ x)}^{2}$

$U_{s} = \frac{1}{2} (52 \times 10^{3}) {(10 \times 10^{- 2})}^{2}$

$U_{s} = 260 N \cdot m$

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