posted by Sejul on .
A 3.70kg box is sliding across the horizontal floor of an elevator.
The coefficient of kinetic friction between the box and the floor is 0.470.
Determine the kinetic frictional force that acts on the box when
(a) The elevator is stationary.
(b) The elevator is accelerating upward with an acceleration whose magnitude is 2.90 m/s2.
(c) The elevator is accelerating downward with an acceleration whose magnitude is 2.90 m/s2
Wb = 3.70 kg * 9.8 N./kg = 36.3 N. =
Weight of box.
Fb = 36.3 N. @ 0 Deg. = Force of box.
Fp = 36.3*sin(0) = 0. = Force Parallel
Fv = 36.3*cos(0) = 36.3 N. = Force perpendicular to floor.
a. Fk = u*Fv = 0.470 * 36.3 = 17.1 N.=
Force of kinetidc friction.