A block of mass m is pushed against a spring | vedclass

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(B) The potential energy stored in the spring when it is compressed by an amount $x$ is given by $U = \frac{1}{2}kx^2$.Here,the natural length is $L_0

Vedclass · Accepted Answer

$\sqrt {\frac{k}{m}} .\frac{{{L_0}}}{2}$

Vedclass · Answer

(B) The potential energy stored in the spring when it is compressed by an amount $x$ is given by $U = \frac{1}{2}kx^2$.Here,the natural length is $L_0$ and it is compressed to $L_0/2$,so the compression $x = L_0 - L_0/2 = L_0/2$.By the principle of conservation of mechanical energy,the potential energy stored in the spring is converted into the kinetic energy of the block when the spring returns to its natural length.$\frac{1}{2}mv^2 = \frac{1}{2}k\left(\frac{L_0}{2}\right)^2$$mv^2 = k\frac{L_0^2}{4}$$v^2 = \frac{k}{m} \cdot \frac{L_0^2}{4}$$v = \sqrt{\frac{k}{m}} \cdot \frac{L_0}{2}$

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