Two springs have force constants $k_{1}$ and $k_{2} (k_{1} > k_{2})$ . On which spring is more work done, if (i) they are stretched by the same force and (ii) they are stretched by same distance?

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Solution

Work done is given by: $W = \frac{1}{2} k x^{2}$ where k is spring constant and x is displacement.
If we have two springs :
$W_{1} = \frac{1}{2} k_{1} x_{1}^{2}$
$W_{2} = \frac{1}{2} k_{2} x_{2}^{2}$
Force is given by $F = - k x$
(i)If force is same for both: $k_{1} x_{1} = k_{2} x_{2}$
$\frac{W_{1}}{W_{2}} = \frac{\frac{1}{2} \frac{(k_{1} x_{1})^{2}}{k_{1}}}{\frac{1}{2} \frac{(k_{2} x_{2})^{2}}{k_{2}}}$
$= \frac{1}{2} \frac{(k_{1} x_{1})^{2}}{k_{1}} \times \frac{2 k_{2}}{(k_{2} x_{2})^{2}} = \frac{k_{2}}{k_{1}}$
$W_{2} > W_{1}$
(ii) If stretched by same distance:
$\frac{W_{1}}{W_{2}} = \frac{\frac{1}{2} k_{1} x_{1}^{2}}{\frac{1}{2} k_{2} x_{2}^{2}}$
$= \frac{1}{2} k_{1} x_{1}^{2} \times \frac{2}{k_{2} x_{2}^{2}}$
And $x_{1} = x_{2}$ ,
$\frac{W_{1}}{W_{2}} = \frac{k_{1}}{k_{2}}$
$W_{1} > W_{2}$

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Two springs have force constants k1 and k2(k1>k2). On which spring is more work done, if (i) they are stretched by the same force and (ii) they are stretched by same distance?

Two springs have force constants $k_{1}$ and $k_{2} (k_{1} > k_{2})$ . On which spring is more work done, if (i) they are stretched by the same force and (ii) they are stretched by same distance?