Question

A 1.3 kg block is attached to a horizontal spring constant k = 19.0 N/m and is free to slide along a frictionless surface. The spring is stretched, and the block is released from rest. The maximum acceleration of the block is observed to be 1.5 m/s2. How far from equilibrium was the spring initially stretched?

Answer 1

To find the initial stretch of the spring, we use Hooke's Law and Newton's second law. The initial stretch is approximately 0.1026 meters.

Step-by-step explanation:

To find the initial stretch of the spring, we can use Hooke's Law, which states that the force exerted by a spring is proportional to its displacement from equilibrium. The formula for Hooke's Law is F = -kx, where F is the force, k is the spring constant, and x is the displacement. Since the block has a maximum acceleration of 1.5 m/s^2, we can determine the force acting on the block using Newton's second law, F = m*a. Plugging in the mass and acceleration, we get F = 1.3 kg * 1.5 m/s^2 = 1.95 N.

Now, we can use Hooke's Law to find the displacement. Rearranging the formula, we have x = -F/k. Plugging in the values, x = -1.95 N / 19.0 N/m = -0.1026 m. However, since displacement is a distance, we take the absolute value, so the initial stretch of the spring is approximately 0.1026 meters.