5.2 The spring constant and combinations of springs — Practise with help

Warm-up: remember

What does a large spring constant tell you about a spring?
Calculate the force that stretches a 50 N/m spring by 0.30 m.

Answers and feedback

It is stiff — it takes a large force to stretch it a little.
$F = kx = 50 \times 0.30 = 15\,\text{N}$ .

A smaller spring constant means a softer spring. That is the idea that decides both answers in this section.

Practise with help

1. Two springs of spring constant 100 N/m and 100 N/m are joined in series. Find the spring constant of the combination.

Hint

Series adds reciprocals: $\frac{1}{k} = \frac{1}{k_1} + \frac{1}{k_2}$ .

Next step

$\frac{1}{k} = \frac{1}{100} + \frac{1}{100} = \frac{2}{100}$ .

Full working

$\frac{1}{k} = \frac{2}{100}$ , so $k = 50\,\text{N/m}$ .

Two identical springs in series give half the spring constant of one of them, which makes sense: two springs end to end stretch twice as far for the same force. And 50 is smaller than 100, which the check requires.

2. Two springs of spring constant 100 N/m and 100 N/m are joined in parallel. Find the spring constant of the combination.

Hint

Parallel adds the constants directly.

Next step

$k = 100 + 100$ .

Full working

$k = 200\,\text{N/m}$ .

Two identical springs side by side give twice the spring constant of one, because the load is shared and each spring stretches only half as far. And 200 is larger than 100, which the check requires.

3. Springs of 40 N/m and 60 N/m are joined in series. Find the combined spring constant.

Hint

A common denominator of 40 and 60 is 120.

Next step

$\frac{1}{k} = \frac{3}{120} + \frac{2}{120} = \frac{5}{120}$ .

Full working

$\frac{1}{k} = \frac{5}{120}$ , so $k = \frac{120}{5} = 24\,\text{N/m}$ .

The check: 24 is smaller than 40, the smaller of the two. Correct for a series combination.

A quicker route when there are only two springs is $k = \dfrac{k_1 k_2}{k_1 + k_2} = \dfrac{40 \times 60}{100} = 24\,\text{N/m}$ .

4. Springs of 40 N/m and 60 N/m are joined in parallel. Find the combined spring constant.

Hint

Add them.

Next step

$k = 40 + 60$ .

Full working

$k = 100\,\text{N/m}$ .

The check: 100 is larger than 60, the larger of the two. Correct for a parallel combination.

5. A spring of spring constant 30 N/m is joined in series with one of 20 N/m, and the combination is pulled with a force of 6.0 N. Find the extension of each spring and the total extension.

Hint

In series the same force acts on both. Use $x = \frac{F}{k}$ for each.

Next step

$x_1 = \frac{6.0}{30}$ and $x_2 = \frac{6.0}{20}$ .

Full working

x_1 = \frac{6.0}{30} = 0.20\,\text{m} \qquad x_2 = \frac{6.0}{20} = 0.30\,\text{m}

Total extension $= 0.20 + 0.30 = 0.50\,\text{m}$ .

Now check it with the combined constant: $\frac{1}{k} = \frac{1}{30} + \frac{1}{20} = \frac{5}{60}$ , so $k = 12\,\text{N/m}$ , and $x = \frac{6.0}{12} = 0.50\,\text{m}$ . It agrees, which is a good way to check a series calculation.

6. A student joins two springs in parallel and calculates a combined spring constant of 15 N/m. The springs are 40 N/m and 60 N/m. What has gone wrong, and what is the right answer?

Hint

Which rule gives a smaller answer than either spring, and which gives a larger one?

Next step

15 N/m is smaller than both, so the student used the series rule.

Full working

The student used the series rule. Two springs side by side share the load and each stretches less, so the combination is stiffer than either spring. Its spring constant must be larger than both.

For parallel, add: $k = 40 + 60 = 100\,\text{N/m}$ .

(The series answer would be $\frac{40 \times 60}{40 + 60} = 24\,\text{N/m}$ , which also is not 15 — so there is an arithmetic slip as well as the wrong rule.)

A question from Section 5.1

7. A spring of spring constant 80 N/m is stretched by 0.05 m. Calculate the force and say whether the spring is stiff or soft compared with one of constant 20 N/m.

Hint

Use $F = kx$ , then compare the two constants.

Next step

$F = 80 \times 0.05 = 4.0\,\text{N}$ .

Full working

F = kx = 80 \times 0.05 = 4.0\,\text{N}

The 80 N/m spring is the stiffer of the two: for the same 0.05 m extension the 20 N/m spring would need only $20 \times 0.05 = 1.0\,\text{N}$ .

This comparison is the whole idea of this section, applied to single springs. Joining springs is a way of building a spring with the constant you want.