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  4. Two springs of spring constants 1000 N m-1 and 2000 N m-1 are stretched with same force. They will have potential energy in the ratio of

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Q. Two springs of spring constants $1000\, N\, m^{-1}$ and $2000\, N\, m^{-1}$ are stretched with same force. They will have potential energy in the ratio of

Work, Energy and Power

Solution:

$U=\frac{1}{2} k x^{2}$
$=\frac{1}{2} k\left(\frac{F}{k}\right)^{2}$
$=\frac{F^{2}}{2 k}$
$U_{1} \times k_{1}=U_{2} \times k_{2}$
or $\frac{U_{1}}{U_{2}}=\frac{k_{2}}{k_{1}}$
$=\frac{2000}{1000}=\frac{2}{1}$