Consider the situation shown in figure. Calculate a. the acceleration of the 1.0 kg blocks, b. the tension in the strilng connecting the 1.0 kg blocks and c. the tension in the stting attached to 0.50 kg.

Explanation

Form the ree bdoy diagrams,
`T+0.5a-0.5g=`……………i
`muR+1a+T_1-T=0` …ii
`muR+1a-T_1=0`
muR+a=T_1`………..iii
From equations ii and iii we have
`muR+a=T-T_1`
`rarr T-T_1=T_1`
`rarr T=2T_1`
Equation ii becoems `muR+a+T_1-2T_1=0`
`rarr muR+a-T_1=0`
`rarr T_1=muR+a`......iv
`Equation i. becomes `2T_1+0.5a-0.5g=0`

`rarr T_1=(0.5g-0.5a)/2
`=0.25g-0.25a`
From equatins iv and v
`0.2g+a=0.25g-0.25a`

`rarr a=0.05/1.25xx10
`=0.4xx10 m/s^2`
`[g=10m/s^2]`
a. Aceleration of 1 kg block each is0.4 `m/s^2`.
b. Tension `T_1=0.2g+a+0.4=2.4N`
c. t=0.5-0.5a`
`=0.5xx10-0.5xx0.4`
`4.8N`