Question

In the following arrangement, the system is initially at rest. The $5\text{kg}$ block is now released. Assuming the pulleys and string to be massless and smooth, the acceleration of blocks are

• A. $a_A=\frac{g}{7}m/s^2$
• B. $a_B=0m/s^2$
• C. $a_c=\frac{g}{14}m/s^2$
• D. $2a_C=a_A$

Answer 1

Answer: (A), (B), (C), (D)

$2a_C=a_A$

Let the acceleration of the blocks $A$, $B$ and $C$ be $a_A$, $a_b$ and $a_C$ respectively and $a_p$ be the acceleration of the pulley.

So, using the method of intercept we have
${\stackrel{..}{L}}_1+{\stackrel{..}{L}}_2=0$
$\Rightarrow a_A+a_p+a_p-a_B=0$
$\Rightarrow a_p=\frac{a_B-a_A}{2}$...........(i)
Also, ${\stackrel{..}{L}}_3+{\stackrel{..}{L}}_4=0$
$\Rightarrow -a_p+0+0-a_C=0\Rightarrow a_p=-a_C$........(ii)
From eq (i) & (ii) we have
$a_A-a_B=2a_C$...........(iii)

Now, from the FBDs of the blocks we have
$5g-T=5a_A\Rightarrow a_A=(g-\frac{T}{5})$
$2T-8g=8a_C\Rightarrow a_C=(\frac{T}{4}-g)$
$T-10g=10a_B\Rightarrow a_B=(\frac{T}{10}-g)$
On putting above values in the eq (iii) and solving the equation we get
$T=5g$
$\because$ $T=5g<10g$ and the movement of $10kg$ block is restricted downward, its acceleration $a_B=0$
$\therefore$ from eq (iii) we will get $a_A=2a_C$

Again we will write the force equation for the blocks $A$ and $C$
$5g-T=5a_A=10a_C$
$2T-8g=8a_C$
Solving these two equations we get,
$a_C=\frac{g}{14}m/s^2$
Similarly, $a_A=\frac{g}{7}m/s^2$