The `10 kg` block is moving to the left with a speed of `1.2 m//s` at time `t = 0` A force `F` is applied kinetic friction is `mu_(k) = 0.2`. Determine the time `t` at which the block comes to a slip. `(g = 10 m//s^(2))`

Explanation

`mu_(k) mg = 0.2 xx 10 xx 10 = 20 N` For `t le 0.2 s`
Retardation ` a_(1) = (F + mu_(k) mg)/(m)`
`= (20 + 20)/(10) = 4 m//s^(2)`
At the end of `0.2 s`
`v = 1.2 - 4 xx 0.2 = 0.4m//s`
For `t gt 0.2 s`
Retardation `a_(2) = (10 + 20)/(10) = 3 m//s^(2)`
Block will come to rest after time
`t_(0) = (v)/(a_(2)) = (0.4)/(3) = 0.13 s`
`:.` Total time `= 0.2 + 0.13 = 0.33 s`