If a spring extends by x on loading,then the | vedclass

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(B) The potential energy $U$ stored in a spring extended by a distance $x$ is given by the formula: $U = \frac{1}{2} Kx^2$.According to Hooke's Law,th

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$\frac{T^2}{2K}$

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(B) The potential energy $U$ stored in a spring extended by a distance $x$ is given by the formula: $U = \frac{1}{2} Kx^2$.According to Hooke's Law,the tension $T$ in the spring is related to the extension $x$ by the equation: $T = Kx$.From this,we can express the extension as $x = \frac{T}{K}$.Substituting this value of $x$ into the energy formula:$U = \frac{1}{2} K \left( \frac{T}{K} \right)^2$$U = \frac{1}{2} K \left( \frac{T^2}{K^2} \right)$$U = \frac{T^2}{2K}$.

If a spring extends by $x$ on loading,then the energy stored by the spring is ($T$ is tension in the spring,$K$ is the spring constant).

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