Repeat the above problem, if instead of `mu`, we are given `mu_(s)` and `mu_(k)`,where `mu_(s) =0.6` and `mu_(k) =0.4`.

Explanation

(a) `R =mg =20N`
`mu_(s) R =0.6xx20=12N`
`mu_(k) R =0.4xx20=8N`
Upto `6_(s)`situation is same but after `6_(s)`, a constant kinetic friction of `8N` will act. At `6_(s)`, friction will suddenly change from `12N(=mu_(s)R)"to" 8N(=mu_(s)R)` and direction of friction is opposite to its motion. Therefore, at `6_(s)` it will start with an inital acceleration.
`a =("decreasein figure")/(mass) =(12-8)/(2) =2m//s^(2)`
For `tle6_(s)`
`f =F=2t` ...(i)
`F_("net") =F-f =0`
`a = (F_("net"))/(m) =0`
For `tge6_(s)`
`F =2t`
`f = muR=8N`
`F_("net") =F-f =2t-8`
`a = (F_("net"))/(m) =(2t-8)/(2) =(t-4)`
At `t =6_(s)`,we can see that, `a_(i) =2m//s^(2)`
Further, `a-t` graph is a straight line of slop ` =1` and intercept ` =-4`. Corresponding `a-t` graph is as shown in figure.