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SHM in Vertical Spring Block

Two springs h...

Question

Two springs have their force constant as $k_{1}$ and $k_{2} (k_{1} > k_{2})$ . When they are stretched by the same force

no work is done in case of both the springs.

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equal work is done in case of both the springs

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more work is done in case of second spring

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more work is done in case of first spring.

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Solution

The correct option is C more work is done in case of second spring
From Hooke's law
$F \propto x \Rightarrow F = k x$ , where k is spring constant

Since force is same in Stretching for both spring so

$F = k_{1} x_{1} = k_{2} x_{2} \Rightarrow x_{1} < x_{2}$ because $k_{1} > k_{2}$

so work done in case of first spring is $W_{1} = \frac{1}{2} k_{1} x_{1}^{2}$ and work done in case of second spring is

$W_{2} = \frac{1}{2} k_{2} x_{2}^{2} s o \frac{W_{1}}{W_{2}} = \frac{x_{1}}{x_{2}} \Rightarrow W_{1} < W_{2}$

It means that more work is done in case of second spring (work done on spring is equal to stored elastic potential energy of the spring)

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