Pulling a 1.92 kg. box at a constant speed requires a force of 14.12 N. What is the coefficient of friction (µ) for this box?
ma=F-F(fr)
a=0 =>
F=F(fr) =μ•N= μ•m•g
μ=F/mg=14.12/1.92•9.8=0.75
a=0 =>
F=F(fr) =μ•N= μ•m•g
μ=F/mg=14.12/1.92•9.8=0.75