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Relation between Work Done and PE

There are two...

Question

There are two massless springs $A$ and $B$ of spring constants $K_{A}$ and $K_{B}$ ( $K_{A} > K_{B}$ ) respectively. $W_{A}$ and $W_{B}$ be denoted as work done on springs $A$ and $B$ respectively.

A

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

W_{A} < W_{B}

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B

If they are compressed to same distance,

W_{A} = W_{B}

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C

If they are compressed to 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 option is C If they are compressed to same distance, $W_{A} > W_{B}$
If the springs are compressed to same amount :
$W_{A} = \frac{1}{2} K_{A} x^{2}; W_{B} = \frac{1}{2} K_{B} x^{2}$

$∵ K_{A} > K_{B}$ we can say that, $W_{A} > W_{B}$

If the springs are compressed by same force , $F = K_{A} x_{A} = K_{B} x_{B}$

$x_{A} = \frac{F}{K_{A}}$ and $x_{B} = \frac{F}{K_{B}}$

Thus, $\frac{W_{A}}{W_{B}} = \frac{\frac{1}{2} K_{A} . \frac{F^{2}}{K_{A}^{2}}}{\frac{1}{2} K_{B} . \frac{F^{2}}{K_{B}^{2}}} = \frac{K_{B}}{K_{A}}$
Hence, $W_{A} < W_{B}$

Hence, options (a) and (b) are the correct answers.

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