Two springs A and B are identical except that A is stiffer than B i-e- k A - k B- If the two springs are stretched by the same force- thenA- More work is done on B- i-e- WB-WAB- More work is done on A- i-e- WA-WBC- Work done on A and B are equalD- Work done depends on the way in which they are stretched

Given springs are stretched by the same force. We know that F = k x and given nbsp;k_A gt; k_B ∴ F = k_A x_A = k_B x_B......(1) Since W_A = 12k_A x_A^2 ; ...

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Standard XII

Physics

Work Done: Spring

Two springs A...

Question

Two springs A and B are identical except that A is stiffer than B i.e., $k_{A} > k_{B}$ . If the two springs are stretched by the same force, then

More work is done on B, i.e.,

W_{B} > W_{A}

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More work is done on A, i.e.,

W_{A} > W_{B}

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Work done on A and B are equal

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Work done depends on the way in which they are stretched

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Solution

The correct option is A More work is done on B, i.e., $W_{B} > W_{A}$
Given springs are stretched by the same force.
We know that $F = k x$ and given $k_{A} > k_{B}$
$∴ F = k_{A} x_{A} = k_{B} x_{B}$ ......(1)
Since $W_{A} = \frac{1}{2} k_{A} x_{A}^{2}; W_{B} = \frac{1}{2} k_{B} x_{B}^{2}$
$\frac{W_{A}}{W_{B}} = \frac{k_{A} x_{A}^{2}}{k_{B} x_{B}^{2}}$
$\Rightarrow \frac{W_{A}}{W_{B}} = \frac{k_{A}}{k_{B}} {(\frac{k_{B}}{k_{A}})}^{2}$
$(from eqn (1), \frac{x_{A}}{x_{B}} = \frac{k_{B}}{k_{A}})$
$∴ \frac{W_{A}}{W_{B}} = \frac{k_{B}}{k_{A}}$
Because $k_{A} > k_{B}, W_{B} > W_{A}$

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Two springs A and B are identical except that A is stiffer than B i.e., kA>kB. If the two springs are stretched by the same force, then

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Two springs A and B are identical except that A is stiffer than B i.e., $k_{A} > k_{B}$ . If the two springs are stretched by the same force, then