Q. 35

# Find the accelera 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).

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