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Potential Energy Formulae

Two similar s...

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

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

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

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W_{P} > W_{Q}; W_{Q} > W_{P}

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

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Solution

The correct option is C $W_{P} > W_{Q}; W_{Q} > W_{P}$
Given, $K_{P} > K_{Q}$
In case (a), the elengation is same
i.e., $x_{1} = x_{2} = x$
So, $W_{P} = \frac{1}{2} K_{P} x^{2}$ and $W_{Q} = \frac{1}{2} K_{Q} x^{2}$
$∴ \frac{W_{P}}{W_{Q}} = \frac{K_{P}}{K_{Q}} > 1 (K_{P} > K_{Q})$
$\Rightarrow W_{P} > W_{Q}$
In case (b), the spring force is same
i.e., $F_{1} = F_{2} = F$
So, $x_{1} = \frac{F}{K_{P}}, x_{2} = \frac{F}{K_{Q}}$
$∴ W_{P} = \frac{1}{2} K_{P} x_{1}^{2} = \frac{1}{2} K_{P} \frac{F^{2}}{K_{P}^{2}} = \frac{1}{2} \frac{F^{2}}{K_{P}}$
and $W_{Q} = \frac{1}{2} K_{Q} x_{2}^{2} = \frac{1}{2} K_{Q} . \frac{F^{2}}{K_{Q}^{2}} = \frac{1}{2} \frac{F^{2}}{K_{Q}}$
$∴ \frac{W_{P}}{W_{Q}} = \frac{K_{Q}}{K_{P}} < 1 \Rightarrow W_{P} < W_{Q}$

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Potential Energy Formulae

Standard XII Physics

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