Question

You exert a 100-N pull on the end of a spring. When you increase the force by 20% to 120 N, the spring's length increases 4.0cm beyond its original stretched position.

What is the spring constant of the spring?

Answer 1

100 = k x
120 = k (x+.04) = k x + .04 k
-----------------------------subtract
-20 = kx - kx - .04 k

4 k = 2000
k = 500 N/m

Answer 2

what about
What is its original displacement?

Answer 3

What is it's original displacement?

100=k x

100=(500 N/m) x
x= .2 M