The diagram below represents the graph of the force applied in stretching a spiral spring against the corresponding extension produced within its elastic li...

Answer Details

Within the elastic limit a spiral spring obeys Hooke's law, so the force applied is directly proportional to the extension produced. The force-extension graph is therefore a straight line passing through the origin, as shown below.

Reading the notations on the graph, the line passes through the points: extension \(5\times10^{-2}\) m at force \(15\) N, \(10\times10^{-2}\) m at \(30\) N, \(15\times10^{-2}\) m at \(45\) N, and \(20\times10^{-2}\) m at \(60\) N.

(a) Force constant of the spring

The force constant \(k\) is the slope (gradient) of the force-extension graph:

\[ k = \frac{\Delta F}{\Delta e} = \frac{60 - 0}{(20 - 0)\times10^{-2}} = \frac{60}{20\times10^{-2}} = \frac{60}{0.20} = 300\ \text{N m}^{-1} \]

Any point on the line gives the same value, e.g. \(30 / 0.10 = 300\ \text{N m}^{-1}\). Hence the force constant is 300 N m\(^{-1}\).

(b) Work done in stretching from \(10\times10^{-2}\) m to \(20\times10^{-2}\) m

The work done equals the area under the force-extension graph between the two extensions. From the graph, at \(e_1 = 10\times10^{-2}\) m the force is \(F_1 = 30\) N, and at \(e_2 = 20\times10^{-2}\) m the force is \(F_2 = 60\) N. This shaded region is a trapezium:

\[ W = \tfrac{1}{2}(F_1 + F_2)(e_2 - e_1) = \tfrac{1}{2}(30 + 60)\,(20 - 10)\times10^{-2} \]\[ W = \tfrac{1}{2}(90)(10\times10^{-2}) = \tfrac{1}{2}(90)(0.10) = 4.5\ \text{J} \]

Equivalently, using \(W = \tfrac{1}{2}k(e_2^{2} - e_1^{2})\):

\[ W = \tfrac{1}{2}(300)\big[(0.20)^{2} - (0.10)^{2}\big] = 150\,(0.04 - 0.01) = 150 \times 0.03 = 4.5\ \text{J} \]

The work done in stretching the spring is 4.5 J.