Question

Find the value of acceleration of block $B$ (in ${\text{m/s}}^2$) for the system shown in figure. Take $g=10{\text{m/s}}^2$.

• A. $\frac{15}{2}{\text{m/s}}^2$
• B. $\frac{5}{2}{\text{m/s}}^2$
• C. $8{\text{m/s}}^2$
• D. $\frac{25}{2}{\text{m/s}}^2$

Answer 1

The correct option is A $\frac{15}{2}{\text{m/s}}^2$

Total length of string $\left(l\right)$ will remain constant. Let the length of the string between block $B$ and pulley attached to block $A$ be $x_A$.


$\therefore l=x_B+3x_A...\left(i\right)$
where $l$ represents total string length at the level of the pulley attached to $A$.
Differentiating Eq $\left(i\right)$ twice w.r.t time,
$\Rightarrow 0=\frac{d^2{x}_B}{d{t}^2}+3\frac{d^2{x}_A}{d{t}^2}$
$\Rightarrow 0=-a_B+3a_A$
($a_B$ is $-ve$ since distance $x_B$ is decreasing)
$\Rightarrow a_B=3a_A...\left(ii\right)$

Applying Newton's $2^{nd}$ law for block $B$,
$T=m{a}_B....\left(iii\right)$
For block $A$,
$3mg-3T=3m{a}_A$
or $mg-T=m{a}_A...\left(iv\right)$

From Eq. $\left(ii\right),\left(iii\right),\left(iv\right)$:
$mg=m{a}_A+m{a}_B$
$\Rightarrow mg=4m{a}_A$
$\therefore a_A=\frac{10}{4}=\frac{5}{2}{\text{m/s}}^2$
$\&a_B=\frac{15}{2}{\text{m/s}}^2$