Find the acceleration of the 500 g block in figure.

Explanation

a. `5a+T-5g=0`
From free body diagram 1, T ltbr. `=5g-5a`………i
`Again (1/2) T-4g-8a=0`
`rarr T-8g-16a=0`

From free ody diagram -2
`T=8g+16a` ……..i
From equations i. and ii. we get
`5g-5a=8g+6a`
`rarr 21a=-3g-a=-9/7`
so, acceleration of 5 kg mass is `9/7` (upward)
and that of 4 kg mass is `2a=(2g)/7` (downward)
b. From free body diagram -3
`4a-T/2=0`

`rarr 8a-T=0`
`rarrT=8a`
Again `, T+5a-5g=0`
From free bodyh diagram -4
`8a+5a-5g=s0`
`rarr 13a-5g=0`
`rarr a= (5g)/13` (down ward)
Acceleration of mass 2 kg is `2a= 10/13 (g) and 5 kg is (5g)/13`
c. `T+ 1a-1g=0`
From free body diagram -5,
`T=1g-1a` .......i
Again, from free body diagram -6,
`T/2 -2g-4a=0` ltbr. `rarr T-4g-8a=0` .........ii
From equation i.
`1g-1a-4g-8a=0`
`rarr a= g/3` (downward),
Acceleration of mass ` 1 kg is g/3 ` (upward)
Acceleration of mass 2 kg is `(2g)/3 ` (downward). .