(B) Let the equilibrium position be at a distance $x$ from the natural length.At equilibrium,the spring force equals the gravitational force: $mg = kx

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(B) Let the equilibrium position be at a distance $x$ from the natural length.At equilibrium,the spring force equals the gravitational force: $mg = kx$.Given $k = mg/a$,we have $mg = (mg/a)x$,which gives $x = a$.Using the principle of conservation of energy,the loss in gravitational potential energy equals the gain in elastic potential energy plus the kinetic energy at the equilibrium position.$mgx = \frac{1}{2}kx^2 + KE$.Substituting $x = a$ and $k = mg/a$:$mg(a) = \frac{1}{2}(mg/a)(a^2) + KE$.$mga = \frac{1}{2}mga + KE$.$KE = mga - \frac{1}{2}mga = \frac{1}{2}mga$.

Class 11 Physics Work, Energy, Power and Collision Work Energy Theorem and Conservation of Mechanical Energy

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$A$ block of mass $m$ is suspended vertically from an elastic string with a force constant $k = mg/a$. Initially,the string is at its natural length and the block is allowed to fall freely. What will be its kinetic energy when the block passes through the equilibrium position?

A
$mga$
B
$mga/2$
C
$0$
D
$2mga$

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Work, Energy, Power and Collision

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Work Energy Theorem and Conservation of Mechanical Energy

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$A$ block of mass $m$ is suspended vertically from an elastic string with a force constant $k = mg/a$. Initially,the string is at its natural length and the block is allowed to fall freely. What will be its kinetic energy when the block passes through the equilibrium position?

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$A$ chain of length $L$ and mass $m$ is placed upon a smooth surface as shown. The length of the chain on the incline is $b$. Calculate the velocity of the chain when its end reaches point $B$.

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