Describe motion of the block if `F` is equal to
(a) `20 N` (b) `40 N` (c ) `60 N`

Explanation

`N_(0) = 100 N`
`(f_(s))_(max) = mu_(s) N = 0.52 xx 100 = 52 N`
`f_(k) = mu_(k) N = 0.50 xx 100 = 50 N`
(a) When `F = 20 N`
`F lt (f_(s))_(max)` , the block is at rest
Friction ` = f = f_(s) = F = 20 N`
(b) When `F = 40 N`
`F lt (f_(s))_(max)` , the block is at rest
` Friction = f = f_(s) = F = 40 N`
(c ) When `F = 60 N`
`F gt (f_(s))_(max)` , the block is in motion
Friction `f = f_(k) = mu_(k) N = 50 N`
`50 - 50 = 10 a`
`a = 1 m//s^(2)`
The block is moving under constant acceleration `1 m//s^(2)`.