A block of 3,kg is initially in equilibrium a | vedclass

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(A) Initially,the block is in equilibrium,so the total upward force from the two identical springs balances the weight of the block.Let $k$ be the spr

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(A) Initially,the block is in equilibrium,so the total upward force from the two identical springs balances the weight of the block.Let $k$ be the spring constant and $x$ be the extension in each spring.$2kx = mg$Given $m = 3\,kg$ and $g = 10\,m/s^2$,we have:$2kx = 3 	imes 10 = 30\,N$$kx = 15\,N$When spring $A$ is cut,the force exerted by spring $A$ becomes zero instantaneously,while the force exerted by spring $B$ remains $kx = 15\,N$ because the spring extension cannot change instantaneously.The net force on the block at $t=0$ is:$F_{net} = mg - kx = 30\,N - 15\,N = 15\,N$ (downwards).Using Newton's second law,$F_{net} = ma$:$15 = 3 	imes a$$a = 5\,m/s^2$.

$A$ block of $3\,kg$ is initially in equilibrium and is hanging by two identical springs $A$ and $B$ as shown in the figure. If spring $A$ is cut from the lower point at $t=0$,then find the acceleration of the block in $m/s^2$ at $t=0$. (Take $g = 10\,m/s^2$)

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