Asked by hamza
A 2.9 kg brick has a kinetic coefficient of friction of 0.33. what force must be applied to the brick for it to move at a constant velocity?
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Answered by
Henry
M*g = 2.9 * 9.8 = 28.4 N. = Wt. of the
brick = Normal force, Fn.
Fk = u*Fn = 0.33 * 28.4 = 9.38 n. = Force of kinetic friction.
Fap-Fk = M*a.
Fap-9.38 = M*0 = 0.
Fap = 9.38 N.
brick = Normal force, Fn.
Fk = u*Fn = 0.33 * 28.4 = 9.38 n. = Force of kinetic friction.
Fap-Fk = M*a.
Fap-9.38 = M*0 = 0.
Fap = 9.38 N.
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