An ideal spring with spring-constant of $8000 N / m$ is hung from the ceiling and a block of mass $4 k g$ is attached to its lower end. If the mass is released with the spring in its natural length, then the spring force in case of maximum elongation is

400 N

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600 N

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800 N

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Solution

The correct option is C 800 N
At maximum elongation situation, all the kinetic energy is converted to spring potential energy.
So, K.E = P.E
$\frac{1}{2} k x^{2} = m g x$
$k x^{2} = 2 m g$
$x = \frac{1}{10} m$
spring force $= k x = 8000 \times \frac{1}{10} = 800 N$

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An ideal spring with spring-constant of 8000 N/m is hung from the ceiling and a block of mass 4 kg is attached to its lower end. If the mass is released with the spring in its natural length, then the spring force in case of maximum elongation is

The correct option is C 800 NAt maximum elongation situation, all the kinetic energy is converted to spring potential energy. So, K.E = P.E 12kx2=mgx kx2=2mg x=110 m spring force =kx=8000×110=800 N

An ideal spring with spring-constant of $8000 N / m$ is hung from the ceiling and a block of mass $4 k g$ is attached to its lower end. If the mass is released with the spring in its natural length, then the spring force in case of maximum elongation is

The correct option is C 800 N
At maximum elongation situation, all the kinetic energy is converted to spring potential energy.
So, K.E = P.E
$\frac{1}{2} k x^{2} = m g x$
$k x^{2} = 2 m g$
$x = \frac{1}{10} m$
spring force $= k x = 8000 \times \frac{1}{10} = 800 N$