Question

Two blocks A and B of equal masses m are suspended with ideal pulley and string arrangement as
shown. The acceleration of mass B is​

Answer 1

Answer:

Option (D) $\frac{2g}{5}$

Explanation:

As shown in figure, for second pulley

$T=2T'$

if the displacement of block A and B are $x_A$ and $x_B$ respectively then from work theorem

$T\times x_A=T'\times x_B$

$\implies 2T'\times x_A=T'\times x_B$

$\implies 2x_A=x_B$

differentiating the above twice we get

$2\frac{d^2x_A}{dt^2}=\frac{d^2x_B}{dt^2}$

$2a_A=a_B$

where $a_A$ and $a_B$ are accelerations of block A and B respectively

For block A

$mg-T=ma_A$   .........(1)

For block B

$T'-mg=ma_B$

$\implies \frac{T}{2}-mg=ma_B$ ..........(2)

From (1) and (2)

$mg-2(ma_B+mg)=ma_A$

$\implies mg-2(ma_B+mg)=m\frac{a_B}{2}$

$\implies mg-2ma_B-2mg)=\frac{ma_B}{2}$

$\implies mg-2ma_B-2mg)=\frac{ma_B}{2}$

$\implies -mg=\frac{ma_B}{2}+2ma_B$

$\implies a_B=-\frac{5g}{2}$

Negative sign means the direction of the acceleration is opposite to the direction assumed

Hope this helps.