A horizontal force of 3.2 N is applied on a 3.7kg block, which rests on a horizontal surface. If the co-efficient of friction is 0.6, find the acceleration produced in the block ?

Explanation

`R_1-2g=0`
`rarr R_1=2xx10=20`
2a+0.2R_1-12=0`
`rarr 2a+0.2(20)=12`
`rarr 2=12-4`
`rarr a=4m/s^2`
Again `4a-muR_1=0`
`rarr 4a=muR_1=0.2(20)=4`
`rarr a_1=1m/s^2`

`2 kg block has acceleration `4m/s^2 and that of 4 kg is `1 m/s^2`.
`ii R_1=2g=20`
`ma=muR_1=0`
a=0
`4a+0.2xx2xx10-12=0`
`rarr 4a+4=12`
`rarr 4a=8`
`a=2m/s^2`