5.4 Investigating Hooke's law — Practise with help

Warm-up: remember

State Hooke's law.
What is the weight of a 200 g mass? Take $g = 9.8\,\text{N/kg}$ .

Answers and feedback

The extension of a spring is directly proportional to the force applied, provided the limit of proportionality is not exceeded. In symbols, $F = kx$ .
$W = mg = 0.200 \times 9.8 = 1.96\,\text{N}$ , so about 2.0 N.

The second one is the conversion the investigation depends on: a mass in grams must become a weight in newtons before it can be plotted.

Practise with help

1. A student records the pointer at 0.150 m with no masses, and at 0.210 m with a mass hung on. State the extension.

Hint

The extension is the change in position, not the new position.

Next step

$0.210 - 0.150$ .

Full working

$0.210 - 0.150 = 0.060\,\text{m}$ , which is 60 mm.

The most common mistake in this investigation is to write 0.210 m. That is the length, and using it makes every extension too large by the natural length — so the gradient, and the spring constant, both come out wrong.

2. A mass of 150 g is hung on a spring. Calculate the force on the spring.

Hint

Convert grams to kilograms first, then use $W = mg$ .

Next step

$m = 0.150\,\text{kg}$ , so $W = 0.150 \times 9.8$ .

Full working

W = mg = 0.150 \times 9.8 = 1.47\,\text{N}

So about 1.5 N to two significant figures.

Using the mass directly as a force would give 150 "newtons", which is wrong by a factor of about 100 — and it would still plot as a straight line, so the graph alone would not reveal it.

3. In an experiment, a force of 1.0 N gives an extension of 0.025 m, and a force of 5.0 N gives 0.125 m. Calculate the spring constant from the gradient.

Hint

Gradient $= \dfrac{\Delta F}{\Delta x}$ , using the two points.

Next step

$\dfrac{5.0 - 1.0}{0.125 - 0.025}$ .

Full working

k = \frac{5.0 - 1.0}{0.125 - 0.025} = \frac{4.0}{0.100} = 40\,\text{N/m}

Taking the points as far apart as possible limits the effect of a small reading error. Using 0 and 5.0 N with the same line would give the same answer here, since the line passes through the origin.

4. A spring's force-extension graph is a straight line for the first five points and then curves upwards for the last three. State what this tells you, and which points you should use to find the spring constant.

Hint

What does a straight line mean, and what does the curve mean?

Next step

The straight part shows Hooke's law holding; the curve shows it has stopped.

Full working

The straight line shows that, for the first five readings, extension was proportional to force — so the spring obeyed Hooke's law. The curve shows the spring passed its limit of proportionality somewhere between the fifth and sixth points.

Use only the straight part to find the gradient, and therefore $k$ . Including the curved points would tilt the line of best fit and give a spring constant that is too large.

It is worth reporting where the straight part ended: that is a real measurement about the spring.

5. An extension is measured as 0.040 m with a metre stick read to the nearest millimetre. Calculate the percentage uncertainty in the extension.

Hint

The reading uncertainty is $\pm 0.001\,\text{m}$ . Divide it by the reading.

Next step

$\dfrac{0.001}{0.040} \times 100\%$ .

Full working

\frac{0.001}{0.040} \times 100\% = 2.5\%

So the extension is $0.040\,\text{m} \pm 2.5\%$ .

Compare with the 0.050 m extension in the section, which gave 2%. A shorter extension carries a larger percentage uncertainty, which is why the smallest readings are the least reliable.

6. A student plots force against extension but labels the horizontal axis "length of spring". Explain why the graph is misleading and what the student should do.

Hint

What would the line look like if the axis really were the total length?

Next step

With no masses, the length is not zero — so the line would not pass through the origin.

Full working

If the axis were really the total length, the graph would be a straight line that does not pass through the origin: with no load the length is the natural length, not zero. And its gradient would not be the spring constant in N/m of extension.

The student has almost certainly plotted the total length but called it the extension. The fix is to subtract the natural length from every reading, and to label the axis "extension / m".

The tell is a straight line with a non-zero intercept: Hooke's law gives a line through the origin, so an intercept means the axis is offset.

A question from Section 5.3

7. Using the spring constant you found in question 3, calculate the energy stored when the spring is stretched by 0.125 m.

Hint

$E = \tfrac{1}{2}kx^2$ , with $k = 40\,\text{N/m}$ .

Next step

$E = 0.5 \times 40 \times 0.125^2$ .

Full working

E = \tfrac{1}{2}kx^2 = 0.5 \times 40 \times 0.015625 = 0.31\,\text{J}

To two significant figures, $0.31\,\text{J}$ .

Notice that this is the energy at the largest extension you measured, which is where the spring is nearest its limit of proportionality. Past that point $\tfrac{1}{2}kx^2$ would not give the stored energy.