Question

a) Determine the stiffness constant of a vertical spring if it stretches by 1.4 cm when an object of mass 3.2 kg is hung from the free end of the spring.

(b) For the same spring in Part (a), how far will the spring stretch if the 3.2 kg object is replaced by an object of mass 2.5 kg?

(c) For the same spring in Part (a), determine how much mass is hung on the free end of the spring if the spring stretches by 2.9 cm from its equilibrium position.

Answer 1

(a) Downward force on the spring, F = weight of the mass = W = m*g

Elongation in the spring = x = 1.4 cm = 0.014 m

Suppose, stiffness = k

For a spring,

F = k*x

=> m*g = k*x

=> k = (m*g) / x = (3.2*9.8) / 0.014 = 2240 N/m (Answer)

(b) 3.2 kg mass is replaced by 2.5 kg mass.

So, in this case, spring stretch, x1 = (m1*g) / k = (2.5*9.8) / 2240 = 0.0109 m = 1.09 cm (Answer)

(c) In this case, spring stretch is given as, x2 = 2.9 cm = 0.029 m

So, the required mass to be hung to have a stretch of x2 = 0.029 m,

m2 = (k*x2) / g = (2240 * 0.029) / 9.8 = 6.63 kg (Answer)