Find the acceleration of the 500g block in figure.6005

1.	Find the acceleration of the 500g block in figure.6005
Answer» Given,m1 = 100 g = 0.1 kgm2 = 500 g = 0.5 kgm3 = 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,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 becomesT1 + 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 = 00.65a = 0.4 g⇒a=0.40.65g =4065g=813g (downward) So, the acceleration of the 500 gm block is 8g13downward.

Answer»

Given,m1 = 100 g = 0.1 kgm2 = 500 g = 0.5 kgm3 = 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,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 becomesT1 + 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 = 00.65a = 0.4 g⇒a=0.40.65g =4065g=813g (downward)

So, the acceleration of the 500 gm block is 8g13downward.

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