Two bars of masses `m_(1)` and `m_(2)` connected by a non`-`deformed light spring rest on a horizontal plane. The coefficient of friction between the bars and the surface is equal to `mu`. What minimum constant force has to be appled in the horizontal direction to the bar of mass `m_(1)` in order to shift the other bar?

Options

A. `mum_(1)g+(mum_(2)g)/(2)`
B. `mum_(2)g+(mum_(1)g)/(2)`
C. `mum_(1)g+mum_(2)g`
D. `(1)/(2)(mum_(1)g+mum_(2)g)`

Answer

D

Explanation

`V=(ds)/(dt)=sqrt(9+4s)`
`underset(0)overset(s)int(ds)/(sqrt(9+4s))=underset(0)overset(t)intdt`
`(2sqrt(9+4s))/(4)|_(0)^(s)=t`
`(sqrt(9+4s)-3)=2t`
`(9 + 4s)=(2t + 3)^(2)`
`9+4s=4t^(2)+12t+9`
`s=t^(2)+3t`
`v=(ds)/(dt)=2t + 3`
`(dv)/(dt)=2`
`P_("inst")=F xxv=2xx2xx(2t +3)`
`P_("avg")=(DeltaKE)/("time")=((1)/(2)(2)(2t + 3)^(2))/(t), P_("avg")=(3)/(4)P_("inst")implies ((2t+3)^(2))/(t)=(3)/(4)[4(2t + 3)]`
`(2t + 3)^(2)=3t(2t + 3)implies (2t + 3) = 3t implies t =3sec`