Question

A 2.3 kg block is pulled across a friction-free horizontal surface, stretching a spring that has a spring constant of 18 N/m. At the moment it is released, the block accelerates at 0.27 m/s2.

What is the net force that the spring exerts on the block?
N
How far was the spring stretched at the moment it was released?

Answer 1

The net force exerted by the spring is 0.621 N. The distance the spring is stretched at the moment it is released is 0.0345 m.

Step-by-step explanation:

The net force that the spring exerts on the block can be found using Newton's second law, which states that force equals mass times acceleration (F = ma). In this case, the mass of the block is 2.3 kg and the acceleration is 0.27 m/s^2. So the net force exerted by the spring can be calculated as: F = (2.3 kg)(0.27 m/s^2) = 0.621 N.

The distance the spring is stretched at the moment it is released can be found using Hooke's Law, which states that the force exerted by a spring is proportional to the distance it is stretched or compressed. The equation for Hooke's Law is F = kx, where F is the force, k is the spring constant, and x is the displacement. Rearranging this equation, we can solve for x: x = F/k. In this case, the force exerted by the spring is 0.621 N and the spring constant is 18 N/m. So the distance the spring is stretched can be calculated as: x = (0.621 N)/(18 N/m) = 0.0345 m.

Answer 2

To find the force on the block, use Newton #1: F = m A

Force = (2.3 kg) x (0.27 m/s²)

Force = (2.3 x 0.27) kg-m/s²

Force = 0.621 Newton

The spring stretches 1 meter when to pull with 18 N of force.

To pull with only 0.621 N of force, it only has to stretch (0.621/18) of a meter. That's 0.0345 meter, or 3.45 cm .