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 suspeneded 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 relaced from rest?

Answer

`1 kg 0.6s`

Explanation

Since the seurface beneth the block B is smooth , therefore the block B will move to right
Since A remain stationary relative to B therefore acceleration of A and B are identical, let it be a, hence acceleration of block C will also be equal to a but vertically downward
Since A has the tendency to slip to the right over the surface of B hence friction on A will act leftward and on block B rightward
Maximum possible value of mass `m_(A)` of block C will correspond to limiting friction between block B and C hence FBD will be as shown in figure

For vertical equilibrium of A` N_(1) = m_(1)g `
or `N_(1) = 10 N`
For horizontal forces on A
`T - mu N = m_(1) a rArr T = (a + (20)/(3))`
For horizontal forces on B `mu N_(1) = m_(2)a`
`a= (5)/(2) ms^(-2) rArr T = a + (20)/(3) = (25)/(3) `
For force on `C, m_(0)g - T = m_(0).a` or `m_(0) = 1 kg` if mass of blocks C equal to `m_(3) = 2m_(2) g` or `2 kg, A` will slip over the surface at B it mean acceleration of A will be greater than that of B let acceleration of block A be a then acceleration of c will also be equal to a but vertically downward and lot acceleration of block B be b (rightward)
Now FBD will be as shown in figure
For block A
`N_(1) = m_(2)g or N_(1) = 10 N`
`T = mu N_(1) = m_(1) a rArr f = ((a + 20)/(3))`...(i)
For block B `muN_(1) = m_(2)b or b = 5//3 ms^(-2)`
for block C `m_(3)g - T = m_(3)g`...(ii)
from Eqs (i) and (ii)
`T = (100)/(9) N and a = 409 ms^(-2)`
Block A will separate from B other B after through `l = 510 cm` over it .hence considering rightward motion of A retative to B inital velocity `a = 0`
Acceleration `= a - (25)/(9)ms^(-2)`
`s = 50 cm` or `0.50 m t = ?`
Using `v = a + (1)/(2) at^(2) rArr r = 0.6s`