Question

In the figure shown all contact surfaces are smooth. Acceleration of B block will be (g = 10 m/s'):Pls answer this question diagram in the attached picture​

Answer 1

Answer:

Option (a) 1 m/s²

Explanation:

Taking the dotted portion as one system, the total force of tension acting on the system = 3T

The tension force acting on block A will be T

The work done by the tension force will be equal

Thus if displacement of block A and B be $x_A$, $x_B$ respectively then

$T\times x_A=3T\times x_B$

$x_A=3x_B$

differentiating the above twice we get

$a_A=3a_B$

Where $a_A$and $a_B$ are accelerations of A and B

let $a_B=a$

then $a_A=3a$

From force equation

For Block B

$mg-3T=ma$

For Block A

$T=3ma$

Thus,

$mg-9ma=ma$

$\implies g=10a$

$\implies a=\frac{g}{10}$ m/s²

$\implies a=\frac{10}{10}=1$ m/s²

Therefore, the acceleration of block B is 1 m/s²

Hope this helps.