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/
a) .47 * 3.7 * 9.8
b) .47 * 3.7 * (9.8 + 2.9)
c) .47 * 3.7 * (9.8 - 2.9)
Thank you so much for your help.
To determine the kinetic frictional force acting on the box in each situation, we can use the formula:
Frictional force (F_friction) = coefficient of kinetic friction (μ) * normal force (Fn)
The normal force is the force exerted by the floor on the box and is equal to the weight of the box in this case.
(a) When the elevator is stationary:
The normal force is equal to the weight of the box, which is the mass (m) * acceleration due to gravity (g).
Normal force (Fn) = m * g = 3.70 kg * 9.8 m/s^2 = 36.26 N
Frictional force (F_friction) = μ * Fn = 0.470 * 36.26 N = 17.03 N
So, the kinetic frictional force on the box when the elevator is stationary is 17.03 N.
(b) When the elevator is accelerating upward:
In this case, the normal force will be reduced by the acceleration force, as the box appears lighter.
Normal force (Fn) = (m * g) - (m * a) = 3.70 kg * (9.8 m/s^2 - 2.90 m/s^2) = 30.01 N
Frictional force (F_friction) = μ * Fn = 0.470 * 30.01 N = 14.06 N
So, the kinetic frictional force on the box when the elevator is accelerating upward is 14.06 N.
(c) When the elevator is accelerating downward:
In this case, the normal force will be increased by the acceleration force, as the box appears heavier.
Normal force (Fn) = (m * g) + (m * a) = 3.70 kg * (9.8 m/s^2 + 2.90 m/s^2) = 46.01 N
Frictional force (F_friction) = μ * Fn = 0.470 * 46.01 N = 21.61 N
So, the kinetic frictional force on the box when the elevator is accelerating downward is 21.61 N.