A person wants to determine the spring constant of an exercise stretch cord. He pulls the cord with a force probe that exerts a 35-N force on the cord, causing it to stretch 20 cm. 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. Express your answer to two significant figures and include the appropriate units.

Answer: 180N/m (to 2 significant figures)

Explanation:

According to hooked law which states that the extension of a 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 in newtons

k is the elastic/spring constant in N/m

e is the extension in meters

Given applied force = 35N

extension = 20cm = 0.2m

Since F = ke,

k = F/e = 35/0.2

k = 175N/m

The spring constant is 175N/m

= 180N/m (to 2significant figures)