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Relation between Work Done and PE

There are two...

Question

There are two mass less spring A and B of spring constant $K_{A}$ and $K_{B}$ respectively and $K_{A}$ > $K_{B}$ If $W_{A}$ and $W_{B}$ be denoted as work done on A and work done on B respectively, then

A

if they are compressed to same distance, $W_{A} > W_{B}$

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B

if they are compressed by same force (upto equilibrium state) $W_{A} < W_{B}$

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C

if they are compressed by same distance, $W_{A} = W_{B}$

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D

if they are compressed by same force (upto equilibrium state) $W_{A} > W_{B}$

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Solution

The correct options are
A
if they are compressed to same distance, $W_{A} > W_{B}$

B
if they are compressed by same force (upto equilibrium state) $W_{A} < W_{B}$

$W_{A} = \frac{1}{2} K_{A} x^{2}; W_{B} = \frac{1}{2} K_{B} x^{2} ∵ K_{A} > K_{B} \Rightarrow W_{A} > W_{B}$
If the springs are compressed by same force.
$F = K_{A} x_{A} = K_{B} x_{B}; x_{A} = \frac{F}{K_{A}}; x_{B} = \frac{F}{K_{B}} \frac{W_{A}}{W_{B}} = \frac{\frac{1}{2} K_{A} \frac{F^{2}}{K_{A}^{2}}}{\frac{1}{2} K_{B} \frac{F^{2}}{K_{B}^{2}}} = \frac{K_{B}}{K_{A}} W_{A} < W_{B}$

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