Asked by Melissa
If the coefficient of kinetic friction is 0.25 , how much horizontal force is needed to pull each of the following masses along a rough desk at a constant speed?
A) 25 kg
A) 25 kg
Answers
Answered by
Henry
Wm = mg = 25kg * 9.8m/kg = 245 N. = Wt.
of the mass.
Fm = 245N @ 0 Deg. = Forc of te mas.
Fp = 245*sin(0) = 0 = Force parallel to
desk.
Fv = 245*cos(0) = 245 N. = Force Perpendicular to the desk.
Fn = Fap -Fp - Fk = ma. a = 0.
Fap -0 - 0.25*245 = 25*0 = 0.
Fap - 61.25 = 0.
Fap = 61.25 N. = Force applied = Force
to move mass.
of the mass.
Fm = 245N @ 0 Deg. = Forc of te mas.
Fp = 245*sin(0) = 0 = Force parallel to
desk.
Fv = 245*cos(0) = 245 N. = Force Perpendicular to the desk.
Fn = Fap -Fp - Fk = ma. a = 0.
Fap -0 - 0.25*245 = 25*0 = 0.
Fap - 61.25 = 0.
Fap = 61.25 N. = Force applied = Force
to move mass.
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