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Relation between Work Done and PE

There are two...

Question

There are two mass less springs $A$ 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:

If they are compressed by same distance

W_{A} = W_{B}

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If they are compressed by same force(upto equilibrium state)

W_{A} = W_{B}

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If they are compressed to same distance

W_{A} > W_{B}

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If they are compressed by same force(upto equilibrium state)

W_{A} > W_{B}

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Solution

The correct option is C If they are compressed to same distance $W_{A} > W_{B}$
According to question
$K_{A} > K - B$
$\Rightarrow W_{A} = + \frac{1}{2} K_{A} x^{2}$ $[∴$ Work is done by spring is $= - \frac{1}{\sum} K x^{2}]$
$W_{B} = + \frac{1}{2} K_{B} x^{2}$ $[$ Here, work is done on spring. Hence force, and displacement is in same direction $]$
$∵ K_{A} > K_{B}$
$W_{A} > W_{B}$
If $F_{A} = F_{B}$
$K_{A} x_{A} = K_{B} x_{B} \Rightarrow x_{A} = (\frac{K_{B}}{K_{A}}) x_{B} \Rightarrow x_{A} < x_{B}$
$W_{A} = + \frac{1}{2} K_{A} x_{A}^{2} = - \frac{1}{2} (K_{A} x_{A}) x_{A} = - \frac{1}{2} F_{A} x_{A}$
$W_{B} = + \frac{1}{2} K_{B} x_{B}^{2} = - \frac{1}{2} (K_{B} x_{B}) x_{B} = - \frac{1}{2} F_{B} x_{B}$
$\Rightarrow W_{B} > W_{A}$

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