A 0.40-kg block is attached to the end of a horizontal ideal spring and rests on a frictionless surface. The block is pulled so that the spring stretches for 2.0 cm relative to its unstrained length. When the block is released, it moves with an acceleration of 8.0 m/s2. What is the spring constant of the spring

160N/m

Explanation:

According to Hooke's law which states that the extension of an elastic material is directly proportional to the applied force provided that the elastic limit is not exceeded. Mathematically,

F = ke where

F is the applied force

k is the spring constant

e is the extension

From the formula k = F/e

Since the body accelerates when the block is released, F = ma according to Newton's second law of motion.

The spring constant k = ma/e where

m is the mass of the block = 0.4kg

a is the acceleration = 8.0m/s²

e is the extension of the spring = 2.0cm = 0.02m

K = 0.4×8/0.02

K = 3.2/0.02

K = 160N/m

The spring constant of the spring is therefore 160N/m

answer///isobars;