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 ?
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To find the maximum possible amplitude of oscillation such that the second block will not slip off, we need to consider the forces acting on the system.
First, let's analyze the forces on the top block:
- The weight of the top block is given by: W = mass × acceleration due to gravity = (1.35 kg) × (9.8 m/s^2) = 13.23 N.
- The static friction force between the top block and the bottom block can be calculated using the coefficient of static friction and the normal force between the blocks. The normal force is equal to the weight of the top block, N = 13.23 N. Therefore, the static friction force is: F_friction = coefficient of static friction × N = (0.40) × (13.23 N) = 5.292 N.
Next, let's analyze the forces on the bottom block:
- The tension in the spring is equal to the force exerted on the bottom block. The force exerted by the spring is given by Hooke's Law: F_spring = spring constant × displacement from equilibrium position. The displacement from the equilibrium position is the maximum amplitude of oscillation, denoted as A. Therefore, F_spring = (130 N/m) × A.
For the system to remain in equilibrium and the blocks not to slip off, the maximum amplitude of oscillation must satisfy the condition that the friction force is equal to or smaller than the force exerted by the spring:
F_friction ≤ F_spring
Substituting the values we determined above, we have:
5.292 N ≤ (130 N/m) × A
Now, solve for the maximum amplitude of oscillation, A:
A ≤ 5.292 N / (130 N/m)
A ≤ 0.0407 m or approximately 0.041 m
Therefore, the maximum possible amplitude of oscillation such that the second block will not slip off is approximately 0.041 meters.