A `2 kg` block is kept over a ground with coefficient of friction `mu = 0.8` as shown in figure. A time varying force `F = 2t (F` in newton and `t` in second) is applied on the block. Plot a graph between acceleration of block versus time `(g = 10 m//s^(2))`

Explanation

`f_(max) = mu N= mu mg`
`= 0.8 xx 2 xx 10 = 16 N`
Block will not move till driving force `F = 2t` becomes `16 N`. This force becomes `16 N` in `8 s`
For `t le 8 s`
`a = 0`
For `t ge 8 s`
`a = (F - f_(max))/(m) = (2t - 16)/(2) rArr a= t - 8`
i.e. `a - t` graph is a straight line of slop `+ 1` and intercept`-8`
Corresponding `a- t` graph is shown bellow