(III) A scallop force opens its shell with an elastic material called abductin, whose Young’s modulus is about \({\bf{2}}{\bf{.0 \times 1}}{{\bf{0}}^{\bf{6}}}\;{\bf{N/}}{{\bf{m}}^{\bf{2}}}\). If this piece of abductin is 3.0 mm thick and has a cross-sectional area of \({\bf{0}}{\bf{.50}}\;{\bf{c}}{{\bf{m}}^{\bf{2}}}\), how much potential energy does it store when compressed by 1.0 mm?

The stored potential energy is \(U = \frac{1}{2}F\Delta l\).For this problem, first, find the force from the given values of the Young modulus, compression, areas, and initial length.

The Young’s modulus is \(E = 2.0 \times {10^6}\;{\rm{N/}}{{\rm{m}}^{\rm{2}}}\).

The thickness of the material sample is \({l_ \circ } = 3.0\;{\rm{mm}} = 3.0 \times {10^{ - 3}}\;{\rm{m}}\).

The cross-sectional area of the material sample is \(A = 0.50\;{\rm{c}}{{\rm{m}}^{\rm{2}}} = 0.50 \times {10^{ - 4}}\;{{\rm{m}}^{\rm{2}}}\).

The compression of the sample metrical is \(\Delta l = 1.0\;{\rm{mm}} = 1.0 \times {10^{ - 3}}\;{\rm{m}}\).

Let F be the force on the material.

Now, the force on the material can be written as

\(\begin{array}{c}E = \frac{{\frac{F}{A}}}{{\frac{{\Delta l}}{{{l_ \circ }}}}}\\\frac{F}{A} = E\frac{{\Delta l}}{{{l_ \circ }}}\\F = EA\frac{{\Delta l}}{{{l_ \circ }}}\end{array}\).

Now, the stored potential energy stored in the material due to compression is

\(\begin{array}{c}U = \frac{1}{2}F\Delta l\\ = \frac{1}{2}EA\frac{{\Delta l}}{{{l_ \circ }}}\Delta l\\ = \frac{1}{2}EA\frac{{{{\left( {\Delta l} \right)}^2}}}{{{l_ \circ }}}\end{array}\).

Now, substituting the values in the above equation,

\(\begin{array}{c}U = \frac{1}{2}EA\frac{{{{\left( {\Delta l} \right)}^2}}}{{{l_ \circ }}}\\ = \frac{1}{2} \times \left( {2.0 \times {{10}^6}\;{\rm{N/}}{{\rm{m}}^{\rm{2}}}} \right) \times \left( {0.50 \times {{10}^{ - 4}}\;{{\rm{m}}^{\rm{2}}}} \right) \times \frac{{{{\left( {1.0 \times {{10}^{ - 3}}\;{\rm{m}}} \right)}^2}}}{{3.0 \times {{10}^{ - 3}}\;{\rm{m}}}}\\ = 1.7 \times {10^{ - 2}}\;{\rm{J}}\end{array}\).

Hence, the stored potential energy is \(1.7 \times {10^{ - 2}}\;{\rm{J}}\).