First slide

Potential energy

Question

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

Moderate

A

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

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

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

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

Solution

$Here, K_{p} > K_{Q}$
$Case (a) : Elongation (x) in each spring is same.$
$W_{P} = \frac{1}{2} K_{p} x^{2}, W_{Q} = \frac{1}{2} K_{Q} x^{2}$
$∴ W_{P} > W_{Q}$
$Case (b): Force of elongation is same.$
$So, x_{1} = \frac{F}{K_{P}} and x_{2} = \frac{F}{K_{Q}}$
$W_{P} = \frac{1}{2} K_{P} x_{1}^{2} = \frac{1}{2} \frac{F^{2}}{K_{P}}$
$W_{Q} = \frac{1}{2} K_{Q} x_{2}^{2} = \frac{1}{2} \frac{F^{2}}{K_{Q}} ∴ W_{P} < W_{Q}$

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