Q. Block of 1 kg is initially in equilibrium and is hanging by two identical springs A and B as shown. If spring A is cut from lower point at `t=0,` then find acceleration of block in `m/s^2` at `t=0` . A) 5 B) 10 C) 15 D) 0

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The block in the figure that is suspended by the two springs A and B has a mass of 3 kg. The elongation of the springs due to the downward force, equal to 3*g, is equal to x. If the spring constant of the springs is k, 3*g = 2*k*x

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The block in the figure that is suspended by the two springs A and B has a mass of 3 kg. The elongation of the springs due to the downward force, equal to 3*g, is equal to x. If the spring constant of the springs is k, 3*g = 2*k*x

When the spring A is cut, the net force on the block is `3*g - k*x` = `3*g - (3*g)/2` = `(3*g)/2` . The resulting acceleration of the block is `(3*g - (3*g)/2)/3` = `(3*g)/(2*3)` = `g/2` = 5 m/s^2.

At the instant t = 0, the acceleration of the block is 5 m/s^2.

The correct answer is option A.

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