## College Physics (4th Edition)

(a) The tension in the ligament is $1.125\times 10^3~N$ (b) The elastic potential energy stored in the ligament is $4.22~J$
(a) $k = 150~N/mm = 1.50\times 10^5~N/m$ We can find the tension in the ligament: $T = kx = (1.50\times 10^5~N/m)(0.0075~m) = 1.125\times 10^3~N$ The tension in the ligament is $1.125\times 10^3~N$ (b) We can find the elastic potential energy stored in the ligament: $U_s = \frac{1}{2}kx^2$ $U_s = \frac{1}{2}(1.50\times 10^5~N/m)(0.0075~m)^2$ $U_s = 4.22~J$ The elastic potential energy stored in the ligament is $4.22~J$