= Solution
{c}
The end displacement relative to the initial unit axial length is
$$
\delta=\ell-1=\alpha\gamma-1.
$$
By <Hooke's law>, the total resisting <force> supplied by the spring is $F_3=-K\delta$. To convert this resultant into the axial <Cauchy stress tensor>, divide by the current end <cross-sectional area>. The transverse total stretches are both $\alpha^{-1/2}$, so that area is
$$
\mathcal A=\alpha^{-1/2}\alpha^{-1/2}=\alpha^{-1}.
$$
Uniform <traction> over that face consequently gives
$$
t_3=\frac{F_3}{\mathcal A}=-K\alpha(\alpha\gamma-1).
$$
Equate this to the expression from part (ii):
$$
\boxed{-K\alpha(\alpha\gamma-1)=\alpha\hat W'(\alpha).}
$$
The extra factor $\alpha$ on the left is essential: the spring law specifies a total <force>, whereas the material <constitutive equation> specifies <force> per current area. The reference end area is one, so the same spring resultant can also be read as a nominal end <traction>.
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