If the friction coefficient between the blocks are `mu_(s) = 0.52` and `mu_(k) = 0.50` , in how much time , the upper block will lose contact with the lower block.

Explanation

`N_(0) = 20 N`
`(f_(x))_(max) = mu_(s) N = 0.52 xx 20 = 10.4 N`
Since the applied force `(20 N) gt (f_(s))_(max)`
`(= 10.4 N)`, i.e. friction between the blocks
`f_(k) = mu_(k) N_(0) = 0.5 xx 20 = 10 N`
` 2kg` block :
`20 - f_(k) = 2 a_(1)`
`20 - 10 = 2 a_(1) rArr a_(1) = 5 m//s^(2)`
`5 kg` block :
`10 = 5 a_(2) rArr a_(2) = 2 m//s^(2)`
Acceleration of the upper block w.r.t. the lower block
`a_(1//2) = a_(1) - a_(2) = 5 - 2 = 3 m//s^(2)`
`s = 37.5 cm = 0.375 m`
`s = (1)/(2) a_(1//2) t^(2) rArr 0.375 = (1)/(2) xx 3 xx t^(2) = 0.25`
The upper block will lose contact with the lower block after `0.5 s`