Elastic Potential Energy

Elastic potential energy is defined as

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

Therefore, for a material obeying Hooke’s Law, elastic potential energy is equal to:

$E_{H} = \frac{1}{2} k ∆ x^{2}$

Where:
- k = spring constant of the spring (N m^–1)
- Δx = extension of the spring (m)

This can also be written as:

$E_{H} = \frac{1}{2} F ∆ x$

Where:
- F = restoring force (N)

This force is the same restoring force as in Hooke's law: $F = k ∆ x$

1-3-6-elastic-potential-energy-spring

A spring that is stretched or compressed has elastic potential energy

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

$E_{H} = E_{K}$

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

$v \propto ∆ x$

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

Worked example

A car’s shock absorbers make a ride more comfortable by using a spring that absorbs energy when the car goes over a bump. One of these springs, with a spring constant of 50 kN m^–1 is fixed next to a wheel and compressed a distance of 10 cm.

Calculate the energy stored by the compressed spring.

Answer:

Step 1: List the known values

Spring constant, k = 50 kN m^–1 = 50 × 10³ N m^–1
Compression, x = 10 cm = 10 × 10^–2 m

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

$E_{H} = \frac{1}{2} \times (50 \times 10^{3}) \times {(10 \times 10^{- 2})}^{2} = 250 J$

Elastic Potential Energy (HL IB Physics)

Revision Note

Elastic Potential Energy

Worked example

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Author: Ashika