Question

The tension in a ligament in the human knee is approximately proportional to the extension of the ligament, if the extension is not too large.

a) If a particular ligament has an effective spring constant of 155 N/mm as it is stretched, what is the tension in this ligament when it is stretched by 0.730 cm?

b) If a particular ligament has an effective spring constant of 155 N/mm as it is stretched, what is the elastic energy stored in the ligament when stretched by 0.730 cm?

Answer 1

Given data :

Effective spring constant (k) = 155 N/mm (or) 155 *10^3 N/m

Displacement (x) = 0.730 cm (or) 0.73 * 10^-2 m

Calculations :-

a) Tension of force (F) in the ligament when it is streched by displacement (x) :

(By Hooke,s law ) : F= kx

Substitute the given values ;

we get

F = (155*10^3) *(0.73*10^-2)

F = 1131.5 N

b) Elastic potential energy stored in the ligament (U) when it is streched by displacement (x) :

U = (1/2)*kx^2

Substitute the given values ;

we get

U = 0.5 * (155*10^3)*{(0.73*10^-2)^2}

U = 0.5 * (155*1000)*{5.329*10^-5)

U = 4.1299 Joules