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

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(A) According to the law of conservation of mechanical energy,the initial potential energy stored in the spring is converted into the kinetic energy o

Vedclass · Accepted Answer

$\frac{{\ell _0}}{2}\sqrt {\frac{k}{m}} $

Vedclass · Answer

(A) According to the law of conservation of mechanical energy,the initial potential energy stored in the spring is converted into the kinetic energy of the block when the spring reaches its natural length.Initial compression of the spring,$x = \frac{{\ell _0}}{2}$.Initial potential energy,$U_i = \frac{1}{2}k x^2 = \frac{1}{2}k \left( \frac{{\ell _0}}{2} \right)^2 = \frac{1}{8}k \ell_0^2$.Final kinetic energy of the block,$K_f = \frac{1}{2}mv^2$.By conservation of energy,$U_i = K_f$.$\frac{1}{8}k \ell_0^2 = \frac{1}{2}mv^2$.$v^2 = \frac{k \ell_0^2}{4m}$.$v = \frac{{\ell _0}}{2}\sqrt {\frac{k}{m}}$.

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