The spring constant of this spring is 9810 N/m.
Hooke's law states that the force (F) needed to extend a spring is proportional to the distance from the equilibrium position of spring.
F =kx
where k is the constant of proportionality called spring constant.
As, the mass attached with the spring is hung, then the force is equal to the weight (W).
F = W
So, kx =mg
Substituting W = mg into equation (ii) above, we have
F = mg
Substituting F = mg into equation (i) above, we have
mg = kx
Substitute, extension x = 2cm = 0.02m, mass m= 40g = 0.04kg, g is the acceleration due to gravity =9.81m/s²
Substituting these values into equation, we have
k = 20 x 9.81 /0.02
k = 9810 N/m
Therefore, the spring constant is k = 9810 N/m.
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