CHAPTER 7. MECHANICAL PROPERTIES OF MATTER 7.2
SOLUTION
Step 1 : Determine what information you have:
We know:
F = 0,6 N
The equilibrium spring length is 20 cm
The expanded spring length is 24 cm
Step 2 : Use Hooke’s Law to find the spring constant
First we need to calculate the displacement of the spring from its
equilibrium length:
x = 24 cm− 20 cm
= 4 cm
= 0,04 m
Now use Hooke’s Law to find the spring constant:
F =−kx
0 ,6 =−k· 0 , 04
k =−15 N.m−^1
Step 3 : The second part of the question asks us to find the spring’s
extension if a 0,5 N load is attached to it. We have:
F = 0,5 N
We know from the firstpart of the question that
k = -15 N.m−^1
So, using Hooke’s Law:
F =−kx
x =−
F
k
=−
0 , 5
− 15
= 0,033 m
= 3,3 cm
