## College Physics (4th Edition)

The tendon stretches a distance of 1.34 cm The tendon stores $31.4~J$ of elastic potential energy.
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.