A 8.70-kg box is sliding across the horizontal floor of an elevator. The coefficient of kinetic friction between the box and the floor is 0.400. Determine the kinetic frictional force that acts on the box when the elevator is
(a) stationary,
N
(b) accelerating upward with an acceleration whose magnitude is 1.00 m/s2, and
N
(c) accelerating downward with an acceleration whose magnitude is 1.00 m/s2.
N
a) There is no kinetic friction if the box is stationary (not sliding)
b) and c) Calculate the force applied to the floor. It will be greater than the weight when accelrating upand less than the weight when acclerating down. Multiply that force by 0.400 for the answers.
To determine the kinetic frictional force acting on the box in each situation, we can use the equation:
Frictional force = coefficient of kinetic friction * normal force
where the normal force is equal to the weight of the box.
(a) When the elevator is stationary, the box is not accelerating, so there is no net force acting on it. Therefore, the kinetic frictional force must be equal in magnitude and opposite in direction to the applied force. Since the box is not moving vertically, the normal force is equal to the weight, which is:
Normal force = mass * acceleration due to gravity
Normal force = 8.70 kg * 9.8 m/s^2
Normal force = 85.26 N
Therefore, the kinetic frictional force when the elevator is stationary is:
Frictional force = 0.400 * 85.26 N
Frictional force = 34.1 N
(b) When the elevator is accelerating upward with an acceleration of 1.00 m/s^2, we need to calculate the net force acting on the box. The net force is given by Newton's second law:
Net force = mass * acceleration
The weight of the box acts downward with a magnitude of:
Weight = mass * acceleration due to gravity
Weight = 8.70 kg * 9.8 m/s^2
Weight = 85.26 N
The net force can be obtained by considering the forces acting on the box:
Net force = applied force - frictional force - weight
Since the box is moving vertically, the normal force is reduced by the upward acceleration:
Normal force = mass * (acceleration due to gravity + acceleration)
Normal force = 8.70 kg * (9.8 m/s^2 + 1.00 m/s^2)
Normal force = 100.98 N
Now we can calculate the net force:
Net force = applied force - frictional force - weight
Net force = mass * acceleration
mass * acceleration = applied force - frictional force - weight
Since the box is moving upward, the frictional force acts in the downward direction, opposing the motion. Therefore:
Frictional force = applied force - mass * (acceleration + acceleration due to gravity)
Frictional force = applied force - 8.70 kg * (1.00 m/s^2 + 9.8 m/s^2)
Frictional force = applied force - 8.70 kg * 10.8 m/s^2
Frictional force = applied force - 93.96 N
The kinetic frictional force can be found by substituting the given coefficient of kinetic friction:
Frictional force = coefficient of kinetic friction * normal force
Frictional force = 0.400 * 100.98 N
Frictional force = 40.39 N
(c) When the elevator is accelerating downward with an acceleration of 1.00 m/s^2, we can follow a similar approach. The only difference is that the net force and frictional force will have opposite directions compared to case (b). Therefore, the net force can be obtained as:
Net force = applied force + frictional force - weight
Similarly, the normal force is increased by the downward acceleration:
Normal force = mass * (acceleration due to gravity - acceleration)
Normal force = 8.70 kg * (9.8 m/s^2 - 1.00 m/s^2)
Normal force = 78.66 N
Now we can calculate the net force:
Net force = applied force + frictional force - weight
Net force = mass * acceleration
mass * acceleration = applied force + frictional force - weight
Since the box is moving downward, the frictional force acts in the upward direction, opposing the motion. Therefore:
Frictional force = -applied force - mass * (acceleration due to gravity - acceleration)
Frictional force = -applied force - 8.70 kg * (9.8 m/s^2 - 1.00 m/s^2)
Frictional force = -applied force - 69.39 N
The kinetic frictional force can be found by substituting the given coefficient of kinetic friction:
Frictional force = coefficient of kinetic friction * normal force
Frictional force = 0.400 * 78.66 N
Frictional force = 31.46 N
In summary:
(a) Kinetic frictional force when the elevator is stationary is 34.1 N.
(b) Kinetic frictional force when the elevator is accelerating upward with an acceleration of 1.00 m/s^2 is 40.39 N.
(c) Kinetic frictional force when the elevator is accelerating downward with an acceleration of 1.00 m/s^2 is 31.46 N.