Two similar springs P and Q have spring constants K\(_P\) and K \(_Q\) , such that K\(_P\) > K \(_Q\) . They are stretched first by the same amount (case a), then by the same force (case b). The work done by the springs \(W_P\) and W\(_Q\) are related as, in case (a) and case (b) respectively
The correct answer is A:\(W_P>W_Q,W_Q>W_P\) Here, \(K_P > K_Q\) Case (a): Elongation (x) in each spring is same. \(W_P=\frac{1}{2} K_P x^2,W_Q=\frac{1}{2}K_Qx^2\) \(\therefore \, \, \, \, \, \, W_P > W_Q\) Case (b) : Force of elongation is same. So, \(x_1=\frac{F}{K_p}\) and \(x_2=\frac{F}{K_Q}\) \(W_P=\frac{1}{2}K_P x_2^1=\frac{1}{2} \frac{F^2}{K_P}\) \(W_Q=\frac{1}{2}K_Q x_2^2=\frac{1}{2} \frac{F^2}{K_Q}\) \(\therefore \, \, \, \, \, \, W_P < W_Q\)
Hooke’s Law states that for small deformities, the stress and strain are proportional to each other. Thus,
Stress ∝ Strain
Stress = k × Strain … where k is the Modulus of Elasticity.
When a limited amount of Force or deformation is involved then concept of Hooke’s Law is only applicable . If we consider the fact, then we can deviate from Hooke's Law. This is because of their extreme Elastic limits.
