Two blocks, with masses M1 = 1.4 kg and M2 = 9.9 kg, and a spring with spring constant k = 228 N/m are arranged on a horizontal, frictionless surface. The coefficient of static friction between the two blocks is 0.59. What is the maximum possible amplitude of the simple harmonic motion if no slippage is to occur between the blocks?
78
To find the maximum possible amplitude of the simple harmonic motion without slippage, we need to determine the maximum force of static friction that can be exerted between the two blocks.
First, let's calculate the maximum force of static friction using the coefficient of static friction and the normal force. The normal force on the upper block (M1) is equal to its weight, which is given by the formula: N1 = M1 * g, where g is the acceleration due to gravity (approximately 9.8 m/s²). Plugging in the values, we have N1 = 1.4 kg * 9.8 m/s² = 13.72 N.
The maximum force of static friction (F_static_max) is equal to the coefficient of static friction (μ) multiplied by the normal force (N1): F_static_max = μ * N1 = 0.59 * 13.72 N = 8.09 N.
Since the upper block is connected to the lower block through the spring, the maximum force of static friction will also be the maximum force exerted by the spring, which can be expressed by Hooke's Law: F_spring_max = k * A, where A is the amplitude of the simple harmonic motion.
Therefore, the maximum possible amplitude (A) is given by A = F_spring_max / k = 8.09 N / 228 N/m = 0.0355 m or approximately 3.55 cm.
Hence, the maximum possible amplitude of the simple harmonic motion without slippage between the blocks is approximately 3.55 cm.