(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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Class 11

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Work, Energy, Power and Collision

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Work Energy Theorem and Conservation of Mechanical Energy

Work equal to $25\,J$ is done on a mass of $2\,kg$ to set it in motion. If the whole of it is used to increase the kinetic energy,then the velocity acquired by the mass is ............ $m/s$.

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$A$ bullet when fired at a target with a velocity of $100\,m/s$ penetrates $1\,m$ into it. If the bullet is fired at a similar target with a thickness of $0.5\,m$,then it will emerge from it with a velocity of:

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$A$ vertical spring with force constant $k$ is fixed on a table. $A$ ball of mass $m$ at a height $h$ above the free upper end of the spring falls vertically on the spring so that the spring is compressed by a distance $d$. The net work done in the process is

MediumAIIMS 2008

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

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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

Work equal to $25\,J$ is done on a mass of $2\,kg$ to set it in motion. If the whole of it is used to increase the kinetic energy,then the velocity acquired by the mass is ............ $m/s$.

$A$ ball is moving to and fro about the lowest point $A$ of a smooth hemispherical bowl. If it is able to rise up to a height of $20 \,cm$ on either side of $A$,its speed at $A$ must be .......... $m/s$ (Take $g = 10 \,m/s^2$,mass of the body = $5 \,g$).

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