Two springs of spring constants $\ 1500\ {Nm}^{- 1}$ and $3000\ {Nm}^{- 1}$ respectively are stretched with the same force. They will have potential energy into ratio

Question

Dr. Abubacker · Accepted Answer

(b) The work done in pulling the string is stored as potential energy in the spring
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$U = \frac{1}{2}kx^{2}$ ...(i)
Where k is spring constant and $x$is distance through which it is pulled.
Also in SHM
Force$\  \propto \ $displacement
$F = kx$ .......(ii)
Putting $x = \frac{F}{k}$in Eq. (i),we get
U$= \frac{1}{2}k\left( \frac{F}{k} ight)^{2} = \frac{F^{2}}{2k}$
$	herefore\frac{U_{1}}{U_{2}} = \frac{K_{2}}{K_{1}} = \frac{3000}{1500} = \frac{2}{1}$
$	herefore\ U_{1}:U_{2} = 2:1\ $

Q. Two springs of spring constants $\ 1500\ {Nm}^{- 1}$ and $3000\ {Nm}^{- 1}$ respectively are stretched with the same force. They will have potential energy into ratio

Explanation

Reviewed by Dr. Abubacker

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