Question

A 20-newton weight is attached to a spring.
causing it to stretch, as shown in the diagram.
What is the spring constant of this spring?

Answer 1

Answer:

40 N/m

Explanation:

Assuming that the missing diagram is as attached, then we can deduce from Hooke's law that the extension of a spring is directly proportional to the applied force. This is mathematically expressed as

F=kx

Here, F represent the applied force, x denote the extension of the spring while k is the spring constant.

From the attached diagram, extension of the spring x=1-0.5=0.5m

Substituting 20 N for F and 0.5 m for x then

20=0.5k

k=20/0.5=40 N/m