A 3.50-kg box is sliding across the horizontal floor of an elevator. The coefficient of kinetic friction between the box and the floor is 0.330. Determine the kinetic frictional force that acts on the box when the elevator is

(a) stationary,

(b) accelerating upward with an acceleration whose magnitude is 2.10 m/s2, and

(c) accelerating downward with an acceleration whose magnitude is 2.10 m/s2.

a) fnormal= mg

b) fnormal= mg+ ma
c) fnormal= mg-ma

(a) mg is the normal force between the floor and the box, and μmg is the frictional force between them.

(b) m(g+2.1) is the normal force between the box and the floor. So the frictional force is ...
(c) m(g-2.1) is the normal force. The frictional force is ....

To determine the kinetic frictional force acting on the box in each scenario, we will use the equation:

\(f_{friction} = \mu_k \cdot m \cdot g\)

where \(f_{friction}\) is the frictional force, \(\mu_k\) is the coefficient of kinetic friction, \(m\) is the mass of the box, and \(g\) is the acceleration due to gravity (approximately 9.8 m/s^2).

(a) When the elevator is stationary, there is no acceleration. Therefore, the kinetic frictional force can be calculated as:

\(f_{friction} = \mu_k \cdot m \cdot g\)

\(f_{friction} = 0.330 \cdot 3.50 \cdot 9.8\) (substituting the given values)

(b) When the elevator is accelerating upward with an acceleration of 2.10 m/s^2, we need to consider the net force acting on the box. The equation for net force is:

\(f_{net} = m \cdot a\)

Rearranging the equation to solve for the frictional force, we get:

\(f_{friction} = m \cdot a - m \cdot g\) (simplifying)

Now substituting the given values:

\(f_{friction} = 3.50 \cdot 2.10 - 3.50 \cdot 9.8\)

(c) When the elevator is accelerating downward with an acceleration of 2.10 m/s^2, we use the same approach as in part (b). The net force equation becomes:

\(f_{net} = m \cdot a\)

And solving for the frictional force:

\(f_{friction} = m \cdot g - m \cdot a\)

Now substituting the given values:

\(f_{friction} = 3.50 \cdot 9.8 - 3.50 \cdot 2.10\)