Hooke's Law and Elastic Potential Energy Calculations

$$F = ke$$

Where:

• $F$ = Force in Newtons (N)
• $k$ = Spring constant in Newtons per metre (N/m)
• $e$ = Extension in metres (m)

A classic trap in exams is giving the extension in centimetres or millimetres. You must always convert this to metres (by dividing by 100 for cm, or 1,000 for mm) before calculating.

Worked Example: Calculating Force

Question: A spring with a spring constant of $25\text{ N/m}$ is stretched by $12\text{ cm}$. Calculate the force applied.

Step 1: Identify and convert units.

• $k = 25\text{ N/m}$
• $e = 12\text{ cm} \div 100 = 0.12\text{ m}$

Step 2: State the formula.

• $F = ke$

Step 3: Substitute the values.

• $F = 25 \times 0.12$

Step 4: Calculate the final answer.

• $F = 3.0\text{ N}$

Limits and Deformation

If you stretch a spring too far, it stops obeying Hooke's Law. The point where the relationship between force and extension becomes non-linear (where a graph of force against extension starts to curve) is called the limit of proportionality.

Before this point, the object undergoes elastic deformation, meaning it will return to its original shape and length after the forces are removed. If pulled beyond the limit of proportionality, it undergoes inelastic (plastic) deformation. When this happens, it does NOT return to its original shape or length after forces are removed, but instead remains permanently stretched.

Calculating Elastic Potential Energy

Stretching or compressing an object requires work to be done, which transfers energy into its elastic potential energy store. The energy stored when work is done to change an elastic object's shape is known as elastic potential energy.

As long as the spring has not been stretched beyond its limit of proportionality, you can calculate this energy using the equation (which is provided on the AQA Physics equation sheet):

$$E_e = \frac{1}{2}ke^2$$

Where:

• $E_e$ = Elastic potential energy in Joules (J)
• $k$ = Spring constant in Newtons per metre (N/m)
• $e$ = Extension in metres (m)

Because the extension is squared ($e^2$), the relationship is mathematical: doubling the extension increases the stored energy by four times ($2^2 = 4$).

Worked Example: Calculating Elastic Potential Energy

Question: A spring ($k = 250\text{ N/m}$) is stretched from an original length of $10.0\text{ cm}$ to $11.4\text{ cm}$. Calculate the energy stored. Give your answer to 2 significant figures.

Step 1: Find the extension ($e$) and convert to metres.

• $e = 11.4\text{ cm} - 10.0\text{ cm} = 1.4\text{ cm}$
• $1.4\text{ cm} \div 100 = 0.014\text{ m}$

Step 2: State the equation.

• $E_e = 0.5 \times k \times e^2$

Step 3: Substitute the values.

• $E_e = 0.5 \times 250 \times (0.014)^2$

Step 4: Explicitly square the extension.

• $e^2 = 0.000196$
• $E_e = 0.5 \times 250 \times 0.000196$

Step 5: Calculate the final answer.

• $E_e = 0.0245\text{ J}$
• Rounded to 2 significant figures: $0.025\text{ J}$