A body weighing 980N is moving with constant speed of 5 m/s on a horizontal surface. The coefficient of sliding friction between the body and the surface is 0.25. What horizontal pull is necessary to maintain the motion?
2 answers
forceFriction=.25*980
M*g = 980N. = Wt. of body = Normal(Fn).
Fk = u*Fn = 0.25 * 980 = 245 N. = Force of kinetic friction.
F-Fk = M*a.
F-245 = M*0 = 0,
F = 245 N.
Fk = u*Fn = 0.25 * 980 = 245 N. = Force of kinetic friction.
F-Fk = M*a.
F-245 = M*0 = 0,
F = 245 N.
1 answer