4.2 Force-extension graphs and stored energy — Practise with help

Work each problem first. Open a hint only if you are stuck, and the full working only to check. For each graph, decide whether the line is straight before you choose a formula.

Warm-up: remember

Work out $\tfrac{1}{2} \times 12 \times 0.30$ .
Find the area of a triangle with base $0.040\,\text{m}$ and height $8.0\,\text{N}$ .
Find the gradient of the line through $(0, 0)$ and $(0.050\,\text{m},\, 15\,\text{N})$ .

Answers

$1.8$ .
$\tfrac{1}{2} \times 0.040 \times 8.0 = 0.16$ . The unit would be $\text{N}\,\text{m}$ , which is the joule.
$\dfrac{15}{0.050} = 300$ .

If question 1 or 2 goes wrong, work on the arithmetic before the physics. If question 3 goes wrong, the gradient is the change in the vertical value divided by the change in the horizontal value.

Practise with help

1. A spring extends by $0.060\,\text{m}$ under a force of $12\,\text{N}$ . Find the spring constant.

Hint

Hooke's law connects the force, the spring constant and the extension. Make the spring constant the subject.

Next step

Write $k = \dfrac{F}{\Delta L}$ , then substitute $F = 12$ and $\Delta L = 0.060$ .

Full working

k = \frac{F}{\Delta L} = \frac{12}{0.060} = 200\,\text{N}\,\text{m}^{-1}

The spring constant is $200\,\text{N}\,\text{m}^{-1}$ .

2. A spring of constant $300\,\text{N}\,\text{m}^{-1}$ is stretched by $0.050\,\text{m}$ . Find the force and the energy stored.

Hint

Find the force first with $F = k\Delta L$ . Then use the force in the energy formula.

Next step

Work out $F = 300 \times 0.050$ , then put that force into $E = \tfrac{1}{2}F\Delta L$ .

Full working

\begin{aligned} F &= k\Delta L = 300 \times 0.050 = 15\,\text{N} \\ E &= \frac{1}{2}F\Delta L = \frac{1}{2} \times 15 \times 0.050 = 0.375\,\text{J} \end{aligned}

The force is $15\,\text{N}$ and the energy stored is $0.375\,\text{J}$ .

3. A force-extension graph is a straight line through $(0, 0)$ and $(0.025\,\text{m},\, 10\,\text{N})$ . Find the gradient and the energy stored at $10\,\text{N}$ .

Hint

The gradient is the spring constant. The energy stored is the area under the line, and the line is straight, so the area is a triangle.

Next step

Use $k = \dfrac{10}{0.025}$ for the gradient, then use $E = \tfrac{1}{2}F\Delta L$ with the same two values.

Full working

k = \frac{10}{0.025} = 400\,\text{N}\,\text{m}^{-1}

E = \frac{1}{2} \times 10 \times 0.025 = 0.125\,\text{J}

The gradient is $400\,\text{N}\,\text{m}^{-1}$ and the energy stored is $0.125\,\text{J}$ .

4. A wire has a Young modulus of $2.0 \times 10^{11}\,\text{Pa}$ , a cross-sectional area of $1.0 \times 10^{-7}\,\text{m}^2$ and a length of $2.00\,\text{m}$ . Find its stiffness and the energy stored when it extends by $1.0\,\text{mm}$ .

Hint

The stiffness of a wire is $k = \dfrac{EA}{L}$ . Change the extension to metres before you find the energy.

Next step

Work out $k$ from the first three values, then use $E = \tfrac{1}{2}k(\Delta L)^2$ with $\Delta L = 1.0 \times 10^{-3}\,\text{m}$ .

Full working

\begin{aligned} k &= \frac{EA}{L} = \frac{2.0 \times 10^{11} \times 1.0 \times 10^{-7}}{2.00} = 1.0 \times 10^{4}\,\text{N}\,\text{m}^{-1} \\ E &= \frac{1}{2}k(\Delta L)^2 = \frac{1}{2} \times 1.0 \times 10^{4} \times (1.0 \times 10^{-3})^2 \\ &= 5.0 \times 10^{-3}\,\text{J} \end{aligned}

The stiffness is $1.0 \times 10^{4}\,\text{N}\,\text{m}^{-1}$ and the energy stored is $5.0 \times 10^{-3}\,\text{J}$ .

5. A force-extension graph bends. Explain why the energy stored is not $\tfrac{1}{2}F\Delta L$ , and say how to find it instead.

Hint

Think about the shape of the area under the graph, and about the shape the formula assumes.

Next step

State the shape that $\tfrac{1}{2}F\Delta L$ gives, then say how the real area differs.

Full working

The formula $\tfrac{1}{2}F\Delta L$ is the area of a triangle. A triangle is the right shape only when the graph is a straight line through the origin. If the graph bends, the area under it is not a triangle, so the formula gives the wrong value. Find the true area by counting squares, or by splitting the area into strips and adding the areas of the strips.

6. A spring stores $0.90\,\text{J}$ when it is extended by $0.060\,\text{m}$ . The graph is a straight line through the origin. Find the final force and the spring constant.

Hint

Start from $E = \tfrac{1}{2}F\Delta L$ and rearrange it for the force. Then use the force and the extension to find the spring constant.

Next step

Find $F = \dfrac{2E}{\Delta L}$ , then find $k = \dfrac{F}{\Delta L}$ .

Full working

\begin{aligned} F &= \frac{2E}{\Delta L} = \frac{2 \times 0.90}{0.060} = 30\,\text{N} \\ k &= \frac{F}{\Delta L} = \frac{30}{0.060} = 500\,\text{N}\,\text{m}^{-1} \end{aligned}

The final force is $30\,\text{N}$ and the spring constant is $500\,\text{N}\,\text{m}^{-1}$ .