In fig `m_(1)` and `m_(2)` are two blocks of masses `m_(1)=1.5kg` and `m_(2)=3kg` attached by a massless rod parallel to the incline on which both slide, travel down along the plane with `m_(1)` trailing `m_(2)`. The angle of the inclined plane is `theta=45^(@)`. The coefficient of kinetic friction between `m_(1)` and the incline is `mu=0.3` and that between `m_(2)` and the incline is `mu_(2)=0.15`. Find (a) the tension in the rod linking `m_(1)` and `m_(2)` and (b) the common acceleration of the two masses.

Explanation

`m_(1)=1.5kg`, `m_(2)=3kg`, `theta=45^(@)`
The common acceleration of the system
`=("Net force")/("Total mass")`
`a=((m_(1)+m_(2))gsintheta-(m_(1)mu_(1)+m_(2)mu_(2))gcostheta)/(m_(1)+m_(2))`
`=(4.5xx9.8xxsin45-(1.5xx0.3+3xx0.15)xx9.8xxcos45)/(4.5)`
`=8.3m//s^(2)`
Let us consider the motion of `m_(2)`. Let T be the tension in the rod linking `m_(1)` and `m_(2)`.
`m_(2)gsintheta-(T+mu_(2)m_(2)gcostheta)=m_(2)a`
`T=m_(2)gsintheta-mu_(2)m_(2)gcostheta-m_(2)a)`
`=m_(2)g[sintheta-mu_(2)costheta-(a)/(g)]`
`=3xx9.8[sin45-0.15cos45-(8.3)/(9.8)]=0.036N`