Question

A box of mass m is pushed horizontally on a rough floor with an initial speed of 2 ms-1 the coeficient of kinetic friction between the friction and box is uk=0.1 calculate the distance the box will move before stooping

Answer 1

v₀=2m/s
μk=0.1

in this case, using the FBD of that box, F₁+fk=F
where fk=kinetic friction, F₁=virtual force , F=force applied on the box

now, using the FBD of Y axis, N=mg, means, fk=μk*N=μk*mg

F₁+fk=F
ma₁+(μk*mg)=ma
a₁+(μk*g)=a
a=a₁+(0.1*10)
a=a₁+1

v²-v₀²=2ad
v₀=0 [suppose that the box was stable at start]

v²/21=d
d=4/2a=2/a=2/(a₁+1)

d=2*(a₁+1)⁻¹

all we need is the value of either the Virtual Force , by using it, you'll get the value of a₁ and you'll get the answer. I prefer you checking it out once if possible.