A force of 23.0 N is required to start a 3.0 kg box moving across a horizontal concrete floor.
(a) What is the coefficient of static friction between the box and the floor?
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(b) If the 23.0 N force continues, the box accelerates at 0.50 m/s2. What is the coefficient of kinetic friction?
To find the coefficient of static friction between the box and the floor, we first need to determine the maximum force of static friction. The maximum force of static friction is given by the equation:
\(f_{\text{static}} = \mu_s \cdot N\)
where \(f_{\text{static}}\) is the maximum force of static friction, \(\mu_s\) is the coefficient of static friction, and \(N\) is the normal force.
In this case, the normal force (\(N\)) is equal to the weight of the box, which is the product of the mass of the box (\(m = 3.0 \, \text{kg}\)) and the acceleration due to gravity (\(g = 9.8 \, \text{m/s}^2\)):
\(N = m \cdot g\)
\(N = 3.0 \, \text{kg} \cdot 9.8 \, \text{m/s}^2\)
Now, we can substitute this value of \(N\) into the equation for the maximum force of static friction:
\(f_{\text{static}} = \mu_s \cdot N\)
\(23.0 \, \text{N} = \mu_s \cdot (3.0 \, \text{kg} \cdot 9.8 \, \text{m/s}^2)\)
Now, we can solve for the coefficient of static friction (\(\mu_s\)):
\(\mu_s = \frac{23.0 \, \text{N}}{3.0 \, \text{kg} \cdot 9.8 \, \text{m/s}^2}\)
\(\mu_s \approx 0.78\)
Therefore, the coefficient of static friction between the box and the floor is approximately 0.78.
To find the coefficient of kinetic friction, we can use the equation for the net force acting on the box:
\(f_{\text{net}} = m \cdot a\)
where \(f_{\text{net}}\) is the net force, \(m\) is the mass of the box, and \(a\) is the acceleration.
In this case, the net force (\(f_{\text{net}}\)) is equal to the applied force (\(23.0 \, \text{N}\)) minus the force of kinetic friction (\(f_{\text{kinetic}}\)). The force of kinetic friction is given by:
\(f_{\text{kinetic}} = \mu_k \cdot N\)
where \(f_{\text{kinetic}}\) is the force of kinetic friction and \(\mu_k\) is the coefficient of kinetic friction. Again, the normal force \(N\) is equal to the weight of the box.
We can rearrange the equation for the net force to solve for the force of kinetic friction:
\(f_{\text{net}} = f_{\text{kinetic}} + m \cdot a\)
\(f_{\text{kinetic}} = f_{\text{net}} - m \cdot a\)
Now, we can substitute in the values:
\(f_{\text{kinetic}} = (23.0 \, \text{N}) - (3.0 \, \text{kg} \cdot 0.50 \, \text{m/s}^2)\)
Finally, we can solve for the coefficient of kinetic friction:
\(\mu_k = \frac{f_{\text{kinetic}}}{N}\)
\(\mu_k = \frac{(23.0 \, \text{N}) - (3.0 \, \text{kg} \cdot 0.50 \, \text{m/s}^2)}{3.0 \, \text{kg} \cdot 9.8 \, \text{m/s}^2}\)
\(\mu_k \approx 0.40\)
Therefore, the coefficient of kinetic friction between the box and the floor is approximately 0.40.