When you push a 1.70 kg book resting on a tabletop it takes 2.60 N to start the book sliding. Once it is sliding, however, it takes only 1.50 N to keep the book moving with constant speed. What are the coefficients of static and kinetic friction between the book and the tabletop?
A 747 jetliner lands and begins to slow to a stop as it moves along the runway. The mass is 3.8 105 kg, its speed is 25.9 m/s, and the net braking force is 4.27 105 N.
To find the coefficients of static and kinetic friction between the book and the tabletop, we can use the following equations:
1. The maximum force of static friction is given by:
\(F_{\text{static}} = \mu_s \times N\)
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.
2. The force of kinetic friction is given by:
\(F_{\text{kinetic}} = \mu_k \times N\)
where \(F_{\text{kinetic}}\) is the force of kinetic friction, \(\mu_k\) is the coefficient of kinetic friction, and \(N\) is the normal force.
3. The normal force is the force exerted by the tabletop on the book and is equal to the weight of the book (\(mg\)), where \(m\) is the mass of the book and \(g\) is the acceleration due to gravity (\(9.8 \, \text{m/s}^2\)).
Let's solve for \(\mu_s\) and \(\mu_k\) step by step:
Step 1: Calculate the normal force (N)
The normal force can be calculated as:
\(N = mg\), where \(m = 1.70 \, \text{kg}\) and \(g = 9.8 \, \text{m/s}^2\)
\(N = 1.70 \, \text{kg} \times 9.8 \, \text{m/s}^2\)
\(N = 16.66 \, \text{N}\)
Step 2: Calculate the coefficient of static friction (\(\mu_s\))
To find \(\mu_s\), we need to determine the maximum force of static friction (\(F_{\text{static}}\)).
\(F_{\text{static}} = 2.60 \, \text{N}\)
\(F_{\text{static}} = \mu_s \times N\)
\(2.60 \, \text{N} = \mu_s \times 16.66 \, \text{N}\)
Now we solve for \(\mu_s\).
\(\mu_s = \frac{2.60 \, \text{N}}{16.66 \, \text{N}}\)
\(\mu_s = 0.156\)
Step 3: Calculate the coefficient of kinetic friction (\(\mu_k\))
The force of kinetic friction is \(F_{\text{kinetic}} = 1.50 \, \text{N}\).
\(F_{\text{kinetic}} = \mu_k \times N\)
\(1.50 \, \text{N} = \mu_k \times 16.66 \, \text{N}\)
Now we solve for \(\mu_k\).
\(\mu_k = \frac{1.50 \, \text{N}}{16.66 \, \text{N}}\)
\(\mu_k = 0.090\)
So, the coefficient of static friction (\(\mu_s\)) is 0.156, and the coefficient of kinetic friction (\(\mu_k\)) is 0.090.
To find the coefficients of static and kinetic friction between the book and the tabletop, we can use the given information about the force required to start the book sliding and the force required to keep the book moving at a constant speed.
First, let's calculate the coefficient of static friction:
The force required to start the book sliding is equal to the maximum static friction force, which can be calculated as the product of the coefficient of static friction (μs) and the normal force (N), where the normal force is the weight of the book.
Step 1: Calculate the normal force (N):
The weight of the book is equal to the mass of the book (m) multiplied by the acceleration due to gravity (g).
Given mass (m) = 1.70 kg
Acceleration due to gravity (g) = 9.8 m/s^2
Weight (W) = m * g = 1.70 kg * 9.8 m/s^2 = 16.66 N
Step 2: Calculate the coefficient of static friction (μs):
Using the equation for static friction: Fs = μs * N
We know that the force required to start the book sliding is 2.60 N. So, we can set up the following equation:
2.60 N = μs * 16.66 N
Simplifying the equation:
μs = 2.60 N / 16.66 N ≈ 0.156
Therefore, the coefficient of static friction (μs) between the book and the tabletop is approximately 0.156.
Next, let's calculate the coefficient of kinetic friction:
The force required to keep the book moving at a constant speed is equal to the kinetic friction force, which can be calculated as the product of the coefficient of kinetic friction (μk) and the normal force (N).
Given force required to keep the book moving (Fk) = 1.50 N
Step 3: Calculate the coefficient of kinetic friction (μk):
Using the equation for kinetic friction: Fk = μk * N
We know that Fk = 1.50 N and N = 16.66 N.
Substituting the values into the equation:
1.50 N = μk * 16.66 N
Simplifying the equation:
μk = 1.50 N / 16.66 N ≈ 0.090
Therefore, the coefficient of kinetic friction (μk) between the book and the tabletop is approximately 0.090.