Block A weighs `4 N` and block weigh `8 N` The coefficient of kinetic friction is `0.25` for all surface find the force `F` to slide `B` at a constant speed when (a)`A` rest on `B` and moves with it (b) `A` is held at rest and (c )`A and B` are connected by a light cord passing over a smooth putting as shown in fig 7.31 (a - c) restively.

Explanation

When A moves with `B` the force opposing the motion is the only force of friction between `B` and `S` the horizontal and as velocity of system is contant,
`F =f_(1) =muR_(1) =0.2 (4 +8) =3N`
(b) When A is held stationary the friction opposing the motion is between `A` and `B`. So
`F -muR_(1) +muR_(2) =3 + 0.25 (4) =4N`
(c ) In this situation for dyamic equilibrium of `B`
`F =muR_(1) +muR_(2)+T.....(i)`
While for the uniform motion of `A`
`T =muR_(2)......(ii)`
Substituting `T` from eqn (ii) in (i) we
get `F =muR_(1) + 2muR_(2) =3+2 xx 1 =5N` .