(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?

Similar Questions

Select the incorrect statement regarding the mechanical energy of a body/system.

$A$ ball of mass $0.4 \ kg$ is thrown vertically upwards by applying a force by hand. If the hand moves $0.5 \ m$ while applying the force and the ball goes up to $5 \ m$ height further,find the magnitude of the force (in $N$) $(g = 10 \ m/s^2)$.

The potential energy of a $1 \ kg$ particle free to move along the $x-$axis is given by: $U(x) = (\frac{x^4}{4} - \frac{x^2}{2}) \ J$. The total mechanical energy of the particle is $2 \ J$. Then,the maximum speed (in $m/s$) is:

$A$ particle is placed at the point $A$ of a frictionless track $ABC$ as shown in the figure. It is gently pushed toward the right. The speed of the particle when it reaches the point $B$ is: (Take $g = 10 \ m/s^2$).

$A$ particle is released from height $S$ above the surface of the earth. At a certain height,its kinetic energy is three times its potential energy. The height from the surface of the earth and the speed of the particle at that instant are respectively:

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