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Byju's Answer

Standard VIII

Physics

Normal Force

Block of 3 kg...

Question

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 $m / s^{2}$ at $t = 0$ :

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Solution

The correct option is A 5
Let the spring force due to each spring acting be $F$ .
Initially : $2 F = m g$
$∴$ $2 F = 3 g$ $⟹ F = 1.5 g$
Finally : $m a = m g - F$
$∴$ $3 a = 3 g - 1.5 g$ $⟹ a = 0.5 g = 5$ $m / s^{2}$ $(g = 10 m / s^{2})$

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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 m/s2 at t=0 :

The correct option is A 5Let the spring force due to each spring acting be F.Initially : 2F=mg∴ 2F=3g ⟹F=1.5gFinally : ma=mg−F∴ 3a=3g−1.5g ⟹a=0.5g=5 m/s2 (g=10m/s2)

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 $m / s^{2}$ at $t = 0$ :

The correct option is A 5
Let the spring force due to each spring acting be $F$ .
Initially : $2 F = m g$
$∴$ $2 F = 3 g$ $⟹ F = 1.5 g$
Finally : $m a = m g - F$
$∴$ $3 a = 3 g - 1.5 g$ $⟹ a = 0.5 g = 5$ $m / s^{2}$ $(g = 10 m / s^{2})$