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Standard XI

Physics

Problem Solving

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 the figure. 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 $5$
The given situation can be depicted in two conditions as shown
(1) Before cutting:

From the FBD we get
$2 k x = 3 g$
$k x = 15$ .......(i)

(2)After cutting:

From the FBD we get,
$a = \frac{3 g - k x}{3} = \frac{30 - 15}{3} = \frac{15}{3} = 5 {m/s}^{2}$

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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 the figure. If spring A is cut from lower point at t=0 then, find acceleration of block in ms−2 at t=0.

The correct option is A 5The given situation can be depicted in two conditions as shown (1) Before cutting: From the FBD we get 2kx=3g kx=15.......(i) (2)After cutting: From the FBD we get, a=3g−kx3=30−153=153=5 m/s2

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 lower point at $t = 0$ then, find acceleration of block in $m s^{- 2}$ at $t = 0$ .

The correct option is A $5$
The given situation can be depicted in two conditions as shown
(1) Before cutting:

From the FBD we get
$2 k x = 3 g$
$k x = 15$ .......(i)

(2)After cutting:

From the FBD we get,
$a = \frac{3 g - k x}{3} = \frac{30 - 15}{3} = \frac{15}{3} = 5 {m/s}^{2}$