A forward Horizontal force of 12N is used to pull a 240 N crate at a constant velocity across a horizontal floor. The coefficient of friction is?
F=maF=maF=ma
a=0, v=consta=0,\ v=consta=0, v=const
F−Ff=0→Ff=F→μN=F→μ=F/NF-F_f=0\to F_f=F\to \mu N=F\to\mu=F/NF−Ff=0→Ff=F→μN=F→μ=F/N
μ=FN=Fmg=12240=0.05\mu=\frac{F}{N}=\frac{F}{mg}=\frac{12}{240}=0.05μ=NF=mgF=24012=0.05 . Answer
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