A block with mass = 5.0 rests on a frictionless table and is attached by a horizontal spring ( = 130 ) to a wall. A second block, of mass = 1.35 , rests on top of . The coefficient of static friction between the two blocks is 0.40.
What is the maximum possible amplitude of oscillation such that will not slip off ?
To determine the maximum possible amplitude of oscillation such that the second block does not slip off, we need to consider the frictional force between the two blocks.
The force of static friction (F_f) between the two blocks can be calculated using the coefficient of static friction (μ) and the normal force (N) between the blocks.
The normal force (N) is equal to the weight of the second block (m2) which is given by:
N = m2 * g
where g is the acceleration due to gravity.
The static frictional force (F_f) is given by:
F_f = μ * N
To prevent the second block from slipping off, the static frictional force (F_f) must be greater than the force applied by the spring when the block is at maximum displacement.
The force applied by the spring (F_spring) at maximum displacement is given by:
F_spring = k * maximum displacement
where k is the spring constant.
Therefore, we can set up the following equation:
F_f ≥ F_spring
μ * N ≥ k * maximum displacement
Substituting the expressions for N and F_f:
μ * (m2 * g) ≥ k * maximum displacement
Simplifying:
μ * m2 * g ≥ k * maximum displacement
Substituting the given values:
(0.40) * (1.35) * g ≥ (130) * maximum displacement
Since the acceleration due to gravity (g) is approximately 9.8 m/s^2:
(0.40) * (1.35) * 9.8 ≥ (130) * maximum displacement
Simplifying further:
Maximum displacement ≥ [(0.40) * (1.35) * 9.8] / (130)
Calculating the value:
Maximum displacement ≥ 0.009 m
Therefore, the maximum possible amplitude of oscillation such that the second block does not slip off is approximately 0.009 meters.
To find the maximum amplitude of oscillation such that the second block will not slip off, we need to consider the forces acting on the system and the conditions for equilibrium.
Let's start by analyzing the forces acting on the second block, which rests on top of the first block.
1. Weight (mg2): The second block has a mass of 1.35 kg, so the weight acting on it is given by mg2, where g is the acceleration due to gravity (9.8 m/s^2 in the SI system).
2. Normal force (N2): The normal force between the two blocks counteracts the weight and is equal in magnitude but opposite in direction.
3. Friction force (fs): The force of static friction between the two blocks must be greater than or equal to the force causing it to slip.
Using the coefficient of static friction (μs = 0.40), we can calculate the maximum static friction force between the two blocks as fs_max = μs N2.
To prevent slipping, the maximum static friction force must be equal to or greater than the force causing the slipping, which is the product of the acceleration (a) and the mass of the second block (m2). Rearranging the equation, we have:
fs_max ≥ m2 * a
Next, we need to relate the acceleration (a) to the maximum amplitude of oscillation (A) using the equation of motion for simple harmonic motion (SHM). In SHM, the maximum acceleration is given by a = ω^2 * A, where ω is the angular frequency.
The angular frequency is related to the spring constant (k) and the mass of the two blocks (m). For a mass-spring system, ω = sqrt(k / m).
Substituting this expression for acceleration into the inequality, we have:
fs_max ≥ m2 * ω^2 * A
Substituting the expressions for the spring constant (k = 130 N/m) and the total mass (m = m1 + m2), we can further simplify the inequality:
μs * N2 ≥ m2 * sqrt(k / m) * A
We can now calculate the values of N2 and fs_max using the equations below:
N2 = m2 * g
fs_max = μs * N2
Finally, we can determine the maximum amplitude of oscillation (A_max) by rearranging the inequality:
A_max ≤ (fs_max * m) / (m2 * sqrt(k))
Substituting the given values into the equation, we have:
A_max ≤ (μs * m2 * g * m) / (m2 * sqrt(k))
Now, we can calculate the maximum amplitude of oscillation using the given values of m1 = 5.0 kg, m2 = 1.35 kg, k = 130 N/m, g = 9.8 m/s^2, and μs = 0.40.