Between `4 kg` and ground:
` f_(1_t)= f_(k_1) = mu_(1) (2 + 4) g = 6 N`
Between `2kg` and `4kg: f_(1_(2)) = f_(k_(2)) = mu_(2) 2g = 4N`
a. Obviously sliding will occur first between `4 kg` and ground .Because `2 kg` can slide over `4 kg` only after `4 kg` gains some acceleration .
b. `4 kg ` will start sliding on ground if
` F gt f_(1_t) = 6 N or F_(min) = 6N`
c. Maximum acceleration of `2kg` block can be `mu_(2) = 0.2 xx 10 = 2ms^(-2)`
Both blocks can move together up to the combined acceleration of `2ms^(-2)` . To produce the acceleration `2ms^(-2)`
`F - f_(k_1)= (2 + 4) xx 2 rArr F = 18 N`
When `F gt 18 N` the acceleration of `4 kg` will became more than `2 ms^(-2)` but that of `2kg` cannot became more than `2ms^(-2)` , Hence , sliding will occur between `2 kg` and `4 kg` for` F gt 18 N`
i. `F = 3 N` system cannot move because `F lt f_(t_1)` , so acceleration of both blocks is zero. Friction between `4 kg` and ground `= f_(1) = F = 3N` As `2-kg` blocks has no tendency to slide over `4 kg` friction between `2 kg` and `4 kg ` blocks is zero
ii. `F = 12 N lt 18 N, but F lt f_(t_1)` , so both the blocks will move with common acceleration .
`a = (F - f_(k_1))/(2 + 4) = (12 -6)/(6) = 1 ms^(-2)`
Friction between `4 kg` and ground is `f_(k_1) = 6N`
Friction between `2 kg and 4 kg`
` f_(2) = 2a = 2 xx 1 = 2N`
iii. `F = 24N gt 18 N` here slipping will occut at both the contact surface Hence friction between `4 kg` and ground `2 kgand 4 kg` will be `f_(k_1)= 4N`
Acceleration of `4 kg`
`a_(1) = ( 24 - 6 - 4)/(4) = 3.5 ms^(-2)`
Acceleration of `2 kg ` blocks `a_(2) = (f_(k_1))/(2) = (4)/(2) = 2ms^(-2)`
