Question

A person wants to determine the spring constant of an exercise stretch cord. He pulls the cord with a force probe that exerts a 70-N force on the cord, causing it to stretch 20cm. Using the Hooke's law find the spring constant of the cord. Assume that the stretch does not exceed the elastic range of the material.

Answer 1

Answer:

350 N/M

Explanation:

• According to Hook's law Force is directly proportional to  extended length in a substance behind elastic region.

Here for a force of 70 N a elongation of 20 cm is observed

⇒

• Force ∝ extension

force = k × extension

k =$\frac{force}{extension}$

extension = 20 cm = 0.2 m.

k = $\frac{70}{0.2}$ = 350 N/M

 

Answer 2

Answer:

350 N/m

Explanation:

If we are assuming the stretch does not exceed the elastic range of the material, then by Hooke's law the spring constant of the cord is simply the ratio between the force 70N acting on the cord to stretch 20cm or 0.2m

k = 70 / 0.2 = 350 N/m

The spring constant is 350 N/m