Question

Find the acceleration of the 500 g block in figure.

Answer 1

Given, $m_1$ = 500 g = 0.5 kg

$m_2$ = 100 g = 0.1kg

$m_3$ = 50 g = 0.0.05 kg

Consider the following free body diagrams:

for FBD-1, 2 and 3

$0.5g-T_1=0.5a⟶\left(1\right)$

$T_1-T_2=0.1a⟶\left(2\right)$

$T_2-0.05g=\left(0.05\right)a⟶\left(3\right)$

adding above eq. we get

$a=\frac{(0.5g-0.05g)}{0.65}$

$a=6.92\frac{m}{se{c}^2}$

acceleration of 500g body would be downward with $6.92m/se{c}^2$