Find the acceleration of the 500 g block in figure (5-E18).
Given: Refer the attachment diagram for data
m1 = 100 g = 0.1 kg , m2 = 500 g = 0.5 kg , m3 = 50 g = 0.05 kg
The free-body diagram for the system is given in attachment.
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,
T1 + 0.05g − 0.05a = a ….(ii)
From the free-body diagram of the 100 g block,
T1 + 0.1a − T + 0.5g = 0 ….(iii)
From equation (ii), T1 = 0.05g + 0.05a …..(iv)
From equation (i), T1 = 0.5g − 0.5a …..(v)
Equation (iii) becomes T1 + 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
⇒ a=
g = 
g= 
g downward
So, the acceleration of the 500 gm block is 
(downward).