A 5.10 kg box is sliding across the horizontal floor of an elevator. The coefficient of kinetic friction between the box and the floor is 0.450. Determine the kinetic frictional force that acts on the box for each of the following cases.
(a) The elevator is stationary.
(b) The elevator is accelerating upward with an acceleration whose magnitude is 3.00 m/s2.
(c) The elevator is accelerating downward with an acceleration whose magnitude is 3.00 m/s2.
To determine the kinetic frictional force on the box in each case, we need to use the formula for kinetic friction:
\(f_{\text{kinetic}} = \mu_k \cdot N\)
where \(f_{\text{kinetic}}\) is the kinetic frictional force, \(\mu_k\) is the coefficient of kinetic friction, and \(N\) is the normal force.
(a) The elevator is stationary:
In this case, the box is not accelerating, so the normal force is equal to the weight of the box. The normal force is given by:
\(N = m \cdot g\)
Plugging in the values:
\(N = 5.10 \, \text{kg} \cdot 9.8 \, \text{m/s}^2 = 49.98 \, \text{N}\)
Now we can calculate the kinetic frictional force:
\(f_{\text{kinetic}} = \mu_k \cdot N = 0.450 \cdot 49.98 = 22.49 \, \text{N}\)
Therefore, the kinetic frictional force on the box when the elevator is stationary is 22.49 N.
(b) The elevator is accelerating upward with an acceleration of 3.00 m/s^2:
In this case, the normal force will be different because the box is experiencing a net upward force. To find the normal force, we need to consider the forces acting on the box.
The weight of the box is given by:
\(W = m \cdot g\)
\(W = 5.10 \, \text{kg} \cdot 9.8 \, \text{m/s}^2 = 49.98 \, \text{N}\)
The net force acting on the box is:
\(F_{\text{net}} = m \cdot a\)
\(F_{\text{net}} = 5.10 \, \text{kg} \cdot 3.00 \, \text{m/s}^2 = 15.30 \, \text{N}\)
Since the box is sliding and not moving up or down, the normal force can be calculated as:
\(N = W + F_{\text{net}}\)
\(N = 49.98 \, \text{N} + 15.30 \, \text{N} = 65.28 \, \text{N}\)
Now we can calculate the kinetic frictional force:
\(f_{\text{kinetic}} = \mu_k \cdot N = 0.450 \cdot 65.28 = 29.376 \, \text{N}\)
Therefore, the kinetic frictional force on the box when the elevator is accelerating upward with an acceleration of 3.00 m/s^2 is 29.376 N.
(c) The elevator is accelerating downward with an acceleration of 3.00 m/s^2:
In this case, the normal force will be different because the box is experiencing a net downward force. Similar to the previous case, the normal force can be calculated as:
\(N = W - F_{\text{net}}\)
\(N = 49.98 \, \text{N} - 15.30 \, \text{N} = 34.68 \, \text{N}\)
Now we can calculate the kinetic frictional force:
\(f_{\text{kinetic}} = \mu_k \cdot N = 0.450 \cdot 34.68 = 15.606 \, \text{N}\)
Therefore, the kinetic frictional force on the box when the elevator is accelerating downward with an acceleration of 3.00 m/s^2 is 15.606 N.
To determine the kinetic frictional force that acts on the box in each case, we need to use the formula for kinetic friction:
frictional force = coefficient of kinetic friction * normal force
where the normal force is the force exerted by the floor on the box perpendicular to the surface.
Now, let's consider each case separately:
(a) The elevator is stationary:
When the elevator is stationary, there is no acceleration. Therefore, there is no relative movement between the box and the floor. In this case, the box experiences static friction, not kinetic friction. The static friction force adjusts itself to prevent the box from moving. So, the frictional force can have any value less than or equal to the maximum static friction force, which is given by:
maximum static friction force = coefficient of static friction * normal force
Since the box is not moving, the static friction force is not relevant to this case.
(b) The elevator is accelerating upward:
When the elevator is accelerating upward, there is a net force acting in the upward direction on the box. We can calculate the net force using Newton's second law:
net force = mass * acceleration
net force = 5.10 kg * 3.00 m/s^2 = 15.30 N
Now, to find the normal force, we need to consider the forces acting on the box. There are two forces acting on the box: the gravitational force (mg) and the normal force. The normal force cancels out the vertical component of the gravitational force to keep the box in contact with the floor. Therefore, the normal force is equal to the weight of the box, which is:
normal force = mg = 5.10 kg * 9.8 m/s^2 = 49.98 N
Finally, we can calculate the frictional force using the formula for kinetic friction:
frictional force = coefficient of kinetic friction * normal force
frictional force = 0.450 * 49.98 N = 22.49 N
So, the kinetic frictional force that acts on the box when the elevator is accelerating upward is 22.49 N.
(c) The elevator is accelerating downward:
When the elevator is accelerating downward, there is a net force acting in the downward direction on the box. Again, we can calculate the net force using Newton's second law:
net force = mass * acceleration
net force = 5.10 kg * (-3.00 m/s^2) = -15.30 N
Note that the acceleration is negative because it is in the opposite direction to the positive direction we established.
Similarly, the normal force cancels out the vertical component of the gravitational force in this case. Therefore, the normal force is still equal to the weight of the box, which is:
normal force = mg = 5.10 kg * 9.8 m/s^2 = 49.98 N
Now, we can calculate the frictional force using the formula for kinetic friction:
frictional force = coefficient of kinetic friction * normal force
frictional force = 0.450 * 49.98 N = 22.49 N
So, the kinetic frictional force that acts on the box when the elevator is accelerating downward is 22.49 N.