A spring was hung from a hook in the ceiling. A number of different weights were attached to the spring to make it stretch, and the total length of the spring was measured each time, shown in the following table.

| Weight (kg) | 2 | 4 | 5 | 8 |
|---|---|---|---|---|
| Length (cm

Question

A spring was hung from a hook in the ceiling. A number of different weights were attached to the spring to make it stretch, and the total length of the spring was measured each time, shown in the following table.

| Weight (kg) | 2 | 4 | 5 | 8 |
|---|---|---|---|---|
| Length (cm) | 3 | 4 | 4.5 | 6 |

(i) Draw a graph showing the results.

(ii) Find the equation relating the length of the spring to the weight on it.

(iii) What is the actual (natural, unstretched) length of the spring?

(iv) If the spring has to stretch to $9$ cm long, how much weight should be on it?

(v) How long will the spring be when $6$ kilograms of weight is on it?

Accepted Answer

> [!TLDR]
>
> Check the table is linear (constant slope between consecutive points), fit the line length-vs-weight, then read off the natural length, the weight for $9$ cm, and the length for $6$ kg.

Let $x=$ weight (kg), $y=$ length (cm), from the table $(2,3),(4,4),(5,4.5),(8,6)$.

**Step 1. Part (i) — the graph.** Plotting weight on the $x$-axis and length on the $y$-axis, the four points $(2,3),(4,4),(5,4.5),(8,6)$ all fall on one rising straight line — a graphical check that the spring stretches linearly with the load (Hooke's law).

**Step 2. Verify linearity.** Slope between consecutive points:
$$\frac{4-3}{4-2}=\frac12,\qquad \frac{4.5-4}{5-4}=\frac12,\qquad \frac{6-4.5}{8-5}=\frac{1.5}{3}=\frac12$$
All equal $0.5$, confirming the data is genuinely linear.

**Step 3. Part (ii) — the equation.** Using point-slope through $(2,3)$ with $m=0.5$:
$$y-3=0.5(x-2) \;\Rightarrow\; y=0.5x-1+3=0.5x+2$$
Multiplying by $2$ and rearranging: $2y=x+4$, i.e.
$$x-2y+4=0$$ …

Data given	Equation
Slope $m$ , $y$ -intercept $b$	$y=mx+b$
Slope $m$ , point $(x_1,y_1)$	$y-y_1=m(x-x_1)$
Two points $(x_1,y_1),(x_2,y_2)$	$\dfrac{y-y_1}{y_2-y_1}=\dfrac{x-x_1}{x_2-x_1}$ , equivalently $\begin{vmatrix}x-x_1 & y-y_1\\ x_2-x_1 & y_2-y_1\end{vmatrix}=0$
$x$ -intercept $a$ , $y$ -intercept $b$ (both $\ne0$ )	$\dfrac{x}{a}+\dfrac{y}{b}=1$
Normal length $p$ , angle $\alpha$ of the normal with the $x$ -axis	$x\cos\alpha+y\sin\alpha=p$
Parametric, through $(x_1,y_1)$ at inclination $\theta$ , parameter $r$ = signed distance from $(x_1,y_1)$	$\dfrac{x-x_1}{\cos\theta}=\dfrac{y-y_1}{\sin\theta}=r$

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