A block of mass 4.0 kg is put on top of a block of mass M = 6.0 kg. To cause the top block to slip on the bottom one, while the bottom one is held fixed, a horizontal force of at least 19 N must be applied to the top block. The assembly of blocks is now placed on a horizontal, frictionless table.
a.) Find the magnitude of the maximum horizontal force F that can be applied to the lower block so that the blocks will move together.
b.) Find the magnitude of the resulting acceleration of the blocks.
To find the answers to these questions, we can apply Newton's laws of motion. Let's go step by step:
a.) To find the maximum horizontal force (F) that can be applied to the lower block so that both blocks move together, we need to consider the static friction between the blocks.
When two objects are in contact, the maximum static friction that can exist between them is given by:
fs_max = μs * N
Where:
- fs_max is the maximum static friction
- μs is the coefficient of static friction
- N is the normal force between the objects (equal to the weight of the top block in this case since it is resting on the bottom block)
The normal force N can be calculated as the weight of the top block:
N = m * g
Where:
- m is the mass of the top block
- g is the acceleration due to gravity (approximately 9.8 m/s^2)
Given that the mass of the top block (m) is 4.0 kg, we can calculate N:
N = 4.0 kg * 9.8 m/s^2 = 39.2 N
Now, we need to find the coefficient of static friction (μs). Since the question states that a force of at least 19 N is required to cause the top block to slip on the bottom one, we can set up the following inequality:
F ≤ fs_max
F ≤ μs * N
Since we want to find the maximum value of F, we need to find the maximum value of μs. Rearranging the inequality, we have:
μs ≥ F / N
μs ≥ 19 N / 39.2 N
μs ≥ 0.4846
Therefore, the minimum coefficient of static friction is approximately 0.4846.
b.) To find the resulting acceleration of the blocks, we can use Newton's second law, which states that the net force on an object is equal to its mass multiplied by its acceleration:
∑F = m_total * a
Where:
- ∑F is the net force
- m_total is the total mass of the system (sum of the masses of both blocks)
- a is the acceleration of the system
In this case, the force applied to the lower block (F) and the friction force between the blocks (fs) are the only forces acting on the system. The friction force fs can be calculated using the coefficient of static friction:
fs = μs * N
So the net force (∑F) can be written as:
∑F = F - fs
Substituting the values, we have:
∑F = F - μs * N
Since the two blocks move together, the acceleration of both blocks will be the same (denoted as a). Rearranging the equation, we have:
(F - μs * N) = m_total * a
Substituting the mass values (m_total = 4.0 kg + 6.0 kg = 10.0 kg) and the known values of μs and N, we can solve for the acceleration (a):
(F - 0.4846 * 39.2 N) = 10.0 kg * a
Simplifying:
F - 19 N = 10.0 kg * a
a = (F - 19 N) / 10.0 kg
Therefore, the magnitude of the resulting acceleration of the blocks can be calculated using this equation.