Block B of mass `m_(B) = 0.5 kg` rests on block A, with mass `m_(A) = 1.5kg` which in turn is on a horizontal tabletop (as shown in figure) .The coefficient of kinetic friction between block A and the tabletop is `mu_(k) = 0.4` and the coefficient of static friction between block A and blockB is `mu_(s) = 0.6` A light string attached block A passes over a frictionless, massless pulley and block C is suspended from the other end of the string. What is the largest mass `m_(c)` (in kg) that block C can have so that block A and B still slide together when the system is replaced from rest?

Answer

5

Explanation

Denote the common magnitude of the maximum acceleration as a. For block A to remain at rest with respect to block `B, a le mu_(s)g`. Let us assume `a=m_(s)g` for mass of C to be largest. The tension in the cord is then
`T=(m_(A)+m_(B))a+mu_(k)g(m_(A)+m_(B))`
`=(m_(A)+m_(B))(a+mu_(k)g)`
This tension is related to the mass `m_(C)` (Largest) by
`T=m_(C)(g-a)`
Solving for `m_(C)` yields
`m_(C)=((m_(A)+m_(B))(mu_(s)+mu_(k)))/(1-mu_(s))`
`=((1.5+0.5)(0.6+0.4))/(1-0.6)=5kg`