Question

Find the acceleration of the 500 g block in the figure (5−E18).
Figure

Answer 1

The acceleration of the 500 g block is $\frac{8}{13} g$.

Explanation:

Step 1:

Given data

$m_1$ = 100 g = 0.1 kg

$m_2$ = 500 g = 0.5 kg

$m_3$ = 50 g = 0.05 kg

Step 2:

The system free-body diagram is shown below:

From a free-body diagram of 500 g block

T + 0.5 a - 0.5 g = 0                    .....(i)

From 50 g block free-body diagram,

T_1  + 0.05 g - 0.05 a = a               ….(ii)

Step 3:

From the 100 gram block free-body diagram,

T + 0.1 a - T + 0.5 g = 0           ….(iii)

From (ii) equation  

T = 0.05 g + 0.05 a                 .....(iv)

From (i)equation  

T = 0.5 g - 0.5 a                     .....(v)

Equation is (iii)

T+ 0.1 a - T + 0.05 g = 0

Step 4:

From (iv) and (v) equations, we get:

0.05 g + 0.05 a + 0.1 a - 0.5 g + 0.5 a + 0.05 g = 0

0.65 a = 0.4 g

$a=\frac{0.4}{0.65} g$

$=\frac{40}{65} g=\frac{8}{13} g \quad(\text { downward })$

So, the 500 gm block acceleration is down  $\frac{8}{13} g$.