Find the acceleration of the 500 g block in the following figure.

#### Solution

Given,

m_{1} = 100 g = 0.1 kg

m_{2} = 500 g = 0.5 kg

m_{3} = 50 g = 0.05 kg

The free-body diagram for the system is shown below:

From the free-body diagram of the 500 g block,

T + 0.5a − 0.5g = 0 .....(i)

From the free-body diagram of the 50 g block,

T_{1} + 0.05g − 0.05a = a ....(ii)

From the free-body diagram of the 100 g block,

T_{1}_{ }+ 0.1a − T + 0.5g = 0 ....(iii)

From equation (ii),

T_{1} = 0.05g + 0.05a .....(iv)

From equation (i),

T_{1} = 0.5g − 0.5a .....(v)

Equation (iii) becomes

T_{1} + 0.1a − T + 0.05g = 0

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

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

0.65a = 0.4 g

\[\Rightarrow a = \frac{0 . 4}{0 . 65}g\]

\[ = \frac{40}{65}g = \frac{8}{13}g \left(\text{ downward }\right)\]

So, the acceleration of the 500 gm block is

\[\frac{8g}{13}\] downward.