Question

In the following arrangement, the system is initially held at rest by an external agent. At $t=0$ , the $5kg$ block is released. If the acceleration of block $C$ is $\frac{x}{10}m{s}^{-2}$ , then find the value of $x$ . Assume that the pulleys and strings are massless and smooth. [ $g=9.8m{s}^{-2}$ ]

Answer 1

Answer: 7

Block B will not move
$5g-T=5a$ ...(i)
$2T-8g=8\frac{a}{2}$ ...(ii)

$2g=14a$
$a=\frac{g}{7}$
$\therefore \frac{a}{2}=\frac{g}{14}=\frac{9.8}{14}=\frac{7}{10}m{s}^{-2}=0.7$
$\therefore x=7$