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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

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$W_{p} > W_{Q}; W_{Q} > W_{P}$

$W_{P} = W_{Q}; W_{P} > W_{Q}$

$W_{P} < W_{Q}; W_{Q} < W_{P}$

$W_{P} = W_{Q}; W_{P} = W_{Q}$

answer is A.

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Detailed Solution

Work done by spring $= \frac{1}{2} k x^{2}$

When they have same displacement

Case (a): $w_{P} = \frac{1}{2} k_{P} x^{2} and w_{Q} = \frac{1}{2} k_{Q} x^{2}$

$\Rightarrow w_{P} > w_{Q} as k_{p} > k_{Q}$

Case (b): $x_{p} = \frac{F}{k_{p}}, x_{Q} = \frac{F}{k_{Q}}$

When same force is applied

$\Rightarrow w_{p} = \frac{1}{2} k_{p} {(\frac{F}{k_{p}})}^{2} = \frac{F^{2}}{2 k_{p}}$ .....(1)

$\Rightarrow W_{Q} = \frac{1}{2} k_{Q} {(\frac{F}{k_{Q}})}^{2} = \frac{F^{2}}{2 k_{Q}}$ ....(2)

from (1) and (2) as $k_{Q} < K_{p}$

$w_{Q} > w_{p}$

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