A 40-kg block is dragged along a horizonatl surface by an applied force as shown. The coefficient of kinetic friction is µk = 0.4 and the block moves at constant velocity. What is the magnitude of F ?
2 answers
the friction force is µmg, right?
M*g = 40 * 9.8 = 392 N. = Wt. of block = normal force(Fn).
u*Fn = 0.4 * 392 = 157 N. = Force of kinetic friction.
F - uFn = M*a
F - 157 = 40*0
F =
u*Fn = 0.4 * 392 = 157 N. = Force of kinetic friction.
F - uFn = M*a
F - 157 = 40*0
F =