5.1 Hooke's law — Practise with help

Warm-up: remember

What is the unit of force?
A spring is 0.20 m long. It is stretched to 0.26 m. What is its extension?

Answers and feedback

The newton, N. A force of 1 N gives a 1 kg mass an acceleration of $1\,\text{m/s}^2$ .
$0.26 - 0.20 = 0.06\,\text{m}$ . The extension is the change in length, not the new length. If you wrote 0.26 m, that is the length — and this is the mistake that costs most marks in this section.

Practise with help

1. A spring has a spring constant of 25 N/m. Calculate the force needed to stretch it by 0.20 m.

Hint

Which form of $F = kx$ do you need? You are given $k$ and $x$ , and asked for $F$ .

Next step

$F = kx = 25 \times 0.20$ .

Full working

F = kx = 25 \times 0.20 = 5.0\,\text{N}

The unit works out too: $\text{N/m} \times \text{m} = \text{N}$ .

2. A force of 18 N stretches a spring by 0.06 m. Calculate its spring constant.

Hint

Rearrange $F = kx$ for $k$ .

Next step

$k = \dfrac{F}{x} = \dfrac{18}{0.06}$ .

Full working

k = \frac{F}{x} = \frac{18}{0.06} = 300\,\text{N/m}

A spring constant of 300 N/m means 300 N would stretch it by a full metre, so this is a stiff spring.

3. A spring of spring constant 500 N/m is pulled with a force of 15 N. Calculate its extension.

Hint

Now you want $x$ , so divide the force by the spring constant.

Next step

$x = \dfrac{15}{500}$ .

Full working

x = \frac{F}{k} = \frac{15}{500} = 0.030\,\text{m}

That is 30 mm. Check the sense of it: a large spring constant means a small extension for the same force, and 30 mm is small.

4. A spring stretches by 0.08 m when a 2.4 N force is applied. Calculate the force needed to stretch it by 0.20 m, assuming it stays within its limit of proportionality.

Hint

Two steps, or one ratio. Either find $k$ first, or use the fact that force is proportional to extension.

Next step

$k = \dfrac{2.4}{0.08} = 30\,\text{N/m}$ . Then $F = 30 \times 0.20$ .

Full working

k = \frac{2.4}{0.08} = 30\,\text{N/m} \qquad F = kx = 30 \times 0.20 = 6.0\,\text{N}

The ratio way is quicker once you trust it: the extension grew from 0.08 m to 0.20 m, which is 2.5 times bigger, so the force is 2.5 times bigger: $2.4 \times 2.5 = 6.0\,\text{N}$ . Same answer.

5. A force-extension graph for a spring is a straight line through the origin passing through the point $(0.05\,\text{m},\ 2.0\,\text{N})$ . State the spring constant and say what the gradient of the line represents.

Hint

$F = kx$ is a straight line through the origin. What is its gradient?

Next step

Gradient $= \dfrac{\text{rise}}{\text{run}} = \dfrac{2.0}{0.05}$ .

Full working

Gradient $= \dfrac{2.0\,\text{N}}{0.05\,\text{m}} = 40\,\text{N/m}$ .

Because $F = kx$ is $F = k \times x$ , the gradient is the spring constant. So $k = 40\,\text{N/m}$ .

That is why the graph is worth drawing: the gradient of the straight part gives $k$ from all the points at once, rather than from one pair that might be misread.

6. A spring obeys Hooke's law up to an extension of 0.15 m. A student stretches an identical spring by 0.40 m. Describe the graph of force against extension, and explain why $F = kx$ cannot be used at 0.40 m.

Hint

The limit of proportionality is at 0.15 m. What happens to the line after that?

Next step

Straight up to 0.15 m, then bending upwards.

Full working

Up to 0.15 m the graph is a straight line through the origin, with gradient equal to the spring constant. Past 0.15 m it curves upwards: each extra newton gives a larger extension than the one before.

At 0.40 m the spring is well past its limit of proportionality, so the extension is no longer proportional to the force. $F = kx$ describes only the straight part, so using it at 0.40 m gives an answer that is too small.

Stretch it far enough and it passes its elastic limit and does not return to its natural length at all.

A question from Chapter 4

7. A box is pushed 3.0 m along a floor by a force of 12 N. Calculate the work done.

Hint

Work is force times distance along the direction of the force.

Next step

$W = Fd = 12 \times 3.0$ .

Full working

W = Fd = 12 \times 3.0 = 36\,\text{J}

This matters here: stretching a spring also does work, and Section 5.3 works out where that energy goes. The difference is that the force changes as the spring stretches, so you cannot simply multiply force by extension.