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.

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 ms -2 at t=0. (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/2b65bfe21ae993d41ecd95a0068bb959-.png" /> k x=15 after cut the spring A. a=(3 g-k x/3)=(15/3)=5 m / s 2

Q. Block of $3 k g$ 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 $t = 0$ .

$k x = 15$
after cut the spring $A$ .
$a = \frac{3 g - k x}{3} = \frac{15}{3} = 5 m / s^{2}$

Q. Block of 3kg 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.

kx=15 after cut the spring A. a=33g−kx​=315​=5m/s2

Q. Block of $3 k g$ 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 $t = 0$ .

$k x = 15$
after cut the spring $A$ .
$a = \frac{3 g - k x}{3} = \frac{15}{3} = 5 m / s^{2}$