5.3 Elastic potential energy — Practise with help

Warm-up: remember

Calculate the work done when a constant force of 8.0 N moves an object 0.50 m.
Work out $0.06^2$ .

Answers and feedback

$W = Fd = 8.0 \times 0.50 = 4.0\,\text{J}$ .
$0.0036$ . Squaring a decimal smaller than 1 makes it smaller, not larger — a trap worth knowing before you substitute into $\tfrac{1}{2}kx^2$ .

Both reappear in this section: work is what stretching a spring does, and the extension is squared.

Practise with help

1. A spring of spring constant 100 N/m is stretched by 0.20 m. Calculate the energy stored.

Hint

$E = \tfrac{1}{2}kx^2$ . Square the extension first.

Next step

$E = 0.5 \times 100 \times 0.20^2 = 0.5 \times 100 \times 0.040$ .

Full working

E = \tfrac{1}{2}kx^2 = 0.5 \times 100 \times 0.040 = 2.0\,\text{J}

The unit: $\text{N/m} \times \text{m}^2 = \text{N}\,\text{m} = \text{J}$ . Correct.

2. A spring of spring constant 400 N/m is stretched by 0.05 m. Calculate the energy stored.

Hint

$0.05^2 = 0.0025$ .

Next step

$E = 0.5 \times 400 \times 0.0025$ .

Full working

E = 0.5 \times 400 \times 0.0025 = 0.50\,\text{J}

A stiff spring stretched a little stores a modest amount. Compare with question 1: a softer spring stretched four times as far stores four times the energy.

3. A spring of spring constant 250 N/m is stretched by 0.12 m. Calculate the energy stored.

Hint

$0.12^2 = 0.0144$ .

Next step

$E = 0.5 \times 250 \times 0.0144$ .

Full working

E = 0.5 \times 250 \times 0.0144 = 1.8\,\text{J}

4. A spring stores 2.0 J when stretched by 0.10 m. Calculate the energy stored when it is stretched by 0.30 m.

Hint

The extension is three times bigger. The energy depends on $x^2$ .

Next step

The energy multiplies by $3^2 = 9$ .

Full working

E = 9 \times 2.0 = 18\,\text{J}

If you want to check it the long way: $E = \tfrac{1}{2}kx^2$ gives $2.0 = 0.5 \times k \times 0.010$ , so $k = 400\,\text{N/m}$ . Then $E = 0.5 \times 400 \times 0.09 = 18\,\text{J}$ . It agrees.

5. A spring stores 5.0 J. It is stretched to twice the extension. State the new stored energy and explain your reasoning in one sentence.

Hint

Write down how $E$ depends on $x$ .

Next step

$E \propto x^2$ , so doubling $x$ multiplies $E$ by 4.

Full working

$20\,\text{J}$ .

The stored energy is proportional to the square of the extension, so doubling the extension multiplies the energy by $2^2 = 4$ .

An answer of $10\,\text{J}$ comes from treating the relationship as proportional to $x$ rather than to $x^2$ . That is the mistake this question is designed to catch.

6. A force-extension graph for a spring is a straight line through the origin passing through $(0.08\,\text{m},\ 4.0\,\text{N})$ . Use the area under the graph to find the energy stored at 0.08 m, then check your answer with $E = \tfrac{1}{2}kx^2$ .

Hint

The area is a triangle: $\tfrac{1}{2} \times \text{base} \times \text{height}$ .

Next step

Area $= 0.5 \times 0.08 \times 4.0$ . And $k = \frac{4.0}{0.08} = 50\,\text{N/m}$ .

Full working

By the area: $\tfrac{1}{2} \times 0.08 \times 4.0 = 0.16\,\text{J}$ .

By the formula: $k = \frac{4.0}{0.08} = 50\,\text{N/m}$ , so $E = 0.5 \times 50 \times 0.08^2 = 0.5 \times 50 \times 0.0064 = 0.16\,\text{J}$ .

They agree. The triangle is where the $\tfrac{1}{2}$ comes from — the force rises steadily from zero, so the average force is half the final one.

A question from Section 5.2

7. Two springs of 30 N/m are joined in series. Find the spring constant of the combination, then find the energy stored when it is stretched by 0.20 m.

Hint

Series first, then energy.

Next step

$\frac{1}{k} = \frac{1}{30} + \frac{1}{30} = \frac{2}{30}$ , so $k = 15\,\text{N/m}$ .

Full working

$k = 15\,\text{N/m}$ , and

E = \tfrac{1}{2}kx^2 = 0.5 \times 15 \times 0.20^2 = 0.5 \times 15 \times 0.040 = 0.30\,\text{J}

A check on the series rule: one 30 N/m spring stretched by 0.20 m would store $0.5 \times 30 \times 0.04 = 0.60\,\text{J}$ . Two in series store half as much for the same extension, because the combination is easier to stretch — which is why the combined spring constant is half.