Answer
The tendon stretches a distance of 1.34 cm
The tendon stores $31.4~J$ of elastic potential energy.
Work Step by Step
We can find the distance the tendon stretches:
$kx = F$
$x = \frac{F}{k}$
$x = \frac{4.7~kN}{350~kN/m}$
$x = 0.0134~m = 1.34~cm$
The tendon stretches a distance of 1.34 cm
We can find the elastic potential energy stored in the tendon:
$U_s = \frac{1}{2}kx^2$
$U_s = \frac{1}{2}(3.50\times 10^5~N/m)(0.0134~m)^2$
$U_s = 31.4~J$
The tendon stores $31.4~J$ of elastic potential energy.
