The coefficient of friction between the blocks are `mu_(s) = 0.3` and `mu_(k) = 0.2`. (a) Find the maximum value of `F` so that the blocks move together. (b) If force `F` is twice the value found in `(a)`, find the acceleration of each block.

Explanation

(a) When the blocks move together
`F = (M + m)a`
`F` will be maximum , when `a = a_(max) = mu_(s) g`
`F_(max) = mu_(s) (M + m)g = 0.3 ( 5 + 2) xx 10 = 21 N`
(b) When `F = 2 F_(max) = 42 N` , the upper block will slide on the lower block
`N_(0) = 20 N`
`f_(k) = mu_(k) N_(0) = 0.2 xx 20 = 4 N`
`f_(k) = 2a_(1) rArr 4 = 2 a_(1) rArr a_(1) = 2 m//s^(2)`
For `5 kg` block :
`42 - f_(k) = 5 a_(2)`
`42 - 4 = 5a_(2) rArr a_(2) = 7.6 m//s^(2)`
In this case , the upper block will move in the backward direction w.r.t. lower block with acceleration `a_(2) - a_(1) = 5.6 m//s^(2)`.