A 9.30 kg block (m1) and a 11.85 kg block (m2) rest on a horizontal frictionless surface as shown below. The surface between the top and bottom blocks is roughened so that there is no slipping between the two blocks. A 30.7 N force is applied to the bottom block as suggested in the figure. Use the figure for the next 3 questions. ibb.co/ftFKB5C (the figure)
Calculate the acceleration for the two block system.
What is the force of static friction between the top and bottom blocks?
What is the minimum coefficient of static friction necessary to keep the top block from slipping on the bottom block?
To calculate the acceleration for the two-block system, we need to use Newton's second law of motion, which states that the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass. We can apply this law separately to each block and then analyze them together.
1. To determine the acceleration of the system, we consider the force applied to the bottom block (m2). We can use the equation:
F = m2 * a
where F is the force applied to the bottom block and m2 is its mass.
Plugging in the given values, we have:
30.7 N = 11.85 kg * a
Solving for a, we find:
a = 30.7 N / 11.85 kg ≈ 2.59 m/s²
Therefore, the acceleration for the two-block system is approximately 2.59 m/s².
2. To calculate the force of static friction between the top and bottom blocks, we need to consider the static friction acting in the opposite direction to the applied force. Since the surface between the two blocks is roughened, the force of static friction can be determined using the equation:
F_friction = μ * F_normal
where F_friction is the force of static friction, μ is the coefficient of static friction, and F_normal is the normal force between the two blocks.
In this case, the normal force is equal to the weight of the top block (m1 * g), where g is the acceleration due to gravity (approximately 9.8 m/s²).
Plugging in the values, we have:
F_friction = μ * m1 * g
Substituting the given mass (m1) and the acceleration due to gravity (g), we get:
F_friction = μ * 9.30 kg * 9.8 m/s²
From the figure, we know that the bottom block (m2) is being pulled to the right with a force of 30.7 N. Hence, the force of static friction must be equal to this force:
F_friction = 30.7 N
Equating the two expressions for the force of static friction, we can solve for μ:
μ * 9.30 kg * 9.8 m/s² = 30.7 N
μ ≈ 30.7 N / (9.30 kg * 9.8 m/s²)
Therefore, the force of static friction between the top and bottom blocks is approximately 0.35.
3. To find the minimum coefficient of static friction required to prevent the top block from slipping on the bottom block, we need to consider the maximum value of static friction. The maximum static friction is given by:
F_friction_max = μ_max * F_normal
We know that the maximum static friction is the value that prevents slipping, which can be expressed as:
F_friction_max = μ_max * m1 * g
From the previous calculation, we found the force of static friction (F_friction) to be approximately 0.35 N. Equating this to F_friction_max, we can solve for μ_max:
μ_max * 9.30 kg * 9.8 m/s² = 0.35 N
μ_max ≈ 0.35 N / (9.30 kg * 9.8 m/s²)
Therefore, the minimum coefficient of static friction necessary to keep the top block from slipping on the bottom block is approximately 0.004.