In fig. la and fig.1 b above, Ida and Wb represent the respective loads on a spring placedtiear a 30 cm rule, when in air and when in water: (a) Identify th...

(a) The force causing the spring to shrink in fig.(b)

When the load is lowered into the water it experiences an upthrust (buoyant force) from the water, acting vertically upwards. This upthrust partly supports the weight of the load, so the tension pulling on the spring is reduced and the spring contracts (shrinks). The force responsible for the shrink is therefore the upthrust of the water on the load.

(b) Work done by the force in causing the shrink

Reading the contraction from the metre rule. The pointer attached to the load marks the position of the lower end of the spring against the fixed rule:

• In air (fig. a) the pointer reads \(20\ \text{cm}\) (the spring is fully stretched by the whole weight \(W_a\)).
• In water (fig. b) the upthrust lifts the load, so the pointer rises to \(18\ \text{cm}\).

The contraction (shrink) of the spring is the difference of the two readings:

\[x = 20\ \text{cm}-18\ \text{cm}=2\ \text{cm}=2\times10^{-2}\ \text{m}.\]

Work done. For a spring obeying Hooke's law, the work done in changing its length by \(x\) against a force constant \(k\) is

\[W=\tfrac{1}{2}kx^{2}.\]

With \(k=2.0\times10^{11}\ \text{N m}^{-1}\) and \(x=2\times10^{-2}\ \text{m}\):

\[W=\tfrac{1}{2}\times(2.0\times10^{11})\times(2\times10^{-2})^{2}\]

\[W=\tfrac{1}{2}\times(2.0\times10^{11})\times(4\times10^{-4})\]

\[W=\mathbf{4.0\times10^{7}\ \text{J}}.\]

The key step is to read \(x\) as the difference of the two rule readings (\(20\ \text{cm}\) in air and \(18\ \text{cm}\) in water), giving \(x = 2\ \text{cm}\), and only then substitute into \(W=\tfrac12 kx^{2}\).