A block with mass = 5.0 rests on a frictionless table and is attached by a horizontal spring ( = 130N ) to a wall. A second block, of mass = 1.35 , rests on top of the block . 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 will not slip off the first block, we need to consider the forces acting on the blocks.
First, let's calculate the maximum force of static friction that can act between the two blocks. The formula for static friction is:
fs = μs * N
where fs is the force of static friction, μs is the coefficient of static friction, and N is the normal force.
The normal force acting on the second block is equal to its weight, which is given by:
N = m * g
where m is the mass of the second block and g is the acceleration due to gravity.
Next, let's consider the forces acting on the first block. There are two forces acting on it: the force of the horizontal spring and the force of static friction between the first and second blocks.
The force of the spring is given by Hooke's law:
F = -k * x
where F is the force, k is the spring constant, and x is the displacement from the equilibrium position.
Since the force of static friction is the only horizontal force acting on the first block, it must be equal in magnitude and opposite in direction to the force of the spring:
fs = k * x
Setting these two forces equal to each other:
μs * N = k * x
Now, let's substitute the expressions for N and μs:
μs * (m * g) = k * x
Rearranging the equation:
x = (μs * m * g) / k
The maximum possible amplitude of oscillation is when the displacement is equal to the maximum value of x, which is given by:
xmax = A
where A is the maximum amplitude.
Therefore, the maximum possible amplitude of oscillation such that the second block will not slip off is:
A = (μs * m * g) / k
Now we can calculate the value, using the given values in the problem:
μs = 0.40
m = 1.35 kg
g = 9.8 m/s²
k = 130 N/m
A = (0.40 * 1.35 kg * 9.8 m/s²) / 130 N/m
A ≈ 0.0413 m
Therefore, the maximum possible amplitude of oscillation such that the second block will not slip off is approximately 0.0413 meters.