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

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Question

(b) $U = \frac{1}{2}K{x^2}$ but $T = Kx$
So energy stored $ = \frac{1}{2}\frac{{{{(Kx)}^2}}}{K} = \frac{1}{2}\frac{{{T^2}}}{K}$

Vedclass · Accepted Answer

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

Vedclass · Answer

(b) $U = \frac{1}{2}K{x^2}$ but $T = Kx$
So energy stored $ = \frac{1}{2}\frac{{{{(Kx)}^2}}}{K} = \frac{1}{2}\frac{{{T^2}}}{K}$

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