Block of 3 kg is initially in equilibrium and is hanging by two identical springs A and B as shown i

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4-1.Newton's Laws of Motion

medium

Block of $3\; kg$ is initially in equilibrium and is hanging by two identical springs $A$ and $B$ as shown in figures. If spring $A$ is cut from lower point at $t=0$ then, find acceleration of block in $ms^{^{-2}}$ at $t = 0.$

A$5$

B$10$

C$15$

D$0$

Solution

Let the spring force due to each spring acting be $F$ Initially: $2 F=m g$
$\therefore \quad 2 F=3 g \quad \Longrightarrow F=1.5 g$
Finally: $m a=m g-F$
$\therefore 3 a=3 g-1.5 g$ $\left(g=10 m / s^{2}\right)$
$\Longrightarrow a=0.5 g=5 \mathrm{m} / \mathrm{s}^{2}$

Standard 11

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(AIIMS-2018)

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easy

Block of $3\; kg$ is initially in equilibrium and is hanging by two identical springs $A$ and $B$ as shown in figures. If spring $A$ is cut from lower point at $t=0$ then, find acceleration of block in $ms^{^{-2}}$ at $t = 0.$

Solution

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