Answer
(a) The tension in the ligament is $1.125\times 10^3~N$
(b) The elastic potential energy stored in the ligament is $4.22~J$
Work Step by Step
(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 $4.22~J$
