Question
A 300lb car is moving at constant velocity onn a level concrete road. How much force is needed to keep it moving if the u=0.04
Answers
M = 300Lbs * 0.454kg/Lb = 136.2 kg.
Fk = u*Fn = u*Mg = 0.04 * (136.2*9.8) = 53.4 N. = Force of kinetic friction.
Fap-Fk = M*a.
Fap-53.4 = M*0 = 0, Fap = 53.4 N. = Force applied.
Fk = u*Fn = u*Mg = 0.04 * (136.2*9.8) = 53.4 N. = Force of kinetic friction.
Fap-Fk = M*a.
Fap-53.4 = M*0 = 0, Fap = 53.4 N. = Force applied.
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