What is the value of friction `f` for the following value of applied force `F` ?

a. `1 N` b.`2 N` c.`3 N` d.`4 N` e`20 N`
Assume the coefficient of friction to be `mu_(s) = 0.3 : mu_(k) = 0.25` Mass of the body is `m = 1 kg`. (Assume `g = 10 ms^(-2))`

Explanation

`f_(1) = mu_(s) N = mu_(s) mg = 0.3 xx 1 xx 10 = 3 N`
` f_(K)= mu_(K) N = mu_(K) mg = 0.25 xx 1 xx 10 = 2.5 N`
a. `F = 2 N, F lt f_(l)` so the motion will not and the friction will be static ,so friction `f = F = 1 N`
b.`F = 2 N, F lt f_(l)` so the motion will not and the friction will be static ,so friction `f = F = 2 N`
c.`F = 3 N, F lt f_(l)`. Here the motion is just about to start Hence the friction will be limiting ,so friction `f = f_(1) = 3 N`
d.`F = 4 N, F gt f_(l)` so the motion will start and the friction will be kinetic . Hence friction will be `f = f_(k) = 2.5 N`
e.`F = 20 N, F gt f_(l)` so the motion will start and the friction will be kinetic . Hence friction is `f = f_(k) = 2.5 N`