1. Recall from Our Spring Lab that the Spring Constant (k) was the slope of the graph of Fs vs. x!
Stronger Spring!
The Spring constant or “Stiffness Factor” provides an indication of spring strength. Bigger (steeper) k, stronger spring!
Weaker Spring!

2. We Derived Hooke’s Law…
Where:
Fs = Force stretching or compressing the spring OR the restoring force built up within the spring.
x = the maximum displacement from the equilibrium position (unstretched).

3. Hooke’s Law
The amount of Force stretching or compressing a spring is directly proportional to the displacement (change in length) of the spring from its equilibrium position.
The restoring force that builds up internally in the spring equals the applied force.

4. Elastic Potential Energy (PEs)!
When we did work to stretch the spring, Elastic PE was stored in the spring as a result. An equation for this relationship can be derived from the area of our graph:
Slope = spring constant Fs
Area = x
A = ½ bh
= ½ Fsx
= ½ kx(x) = ½ kx2
PEs = W (done on the spring) !

5. Elastic Potential Energy (PEs)!
PEs is energy stored in an elastic material, such as a spring, due to an applied force causing a displacement/deformation of the material (work).
Equations:
PEs = Potential Energy of a compressed or stretched spring
Hooke’s Law!
Fs = force compressing or stretching the spring
K = the spring constant in N/m
X = the displacement of the spring from the equilibrium (rest) position
Add to reference tables:
WS = ½ Fsx = PES 5

6. PES = ½ kx2 ? - But we don’t know k! WS = PES = ½ Fsx
It’s easiest to use the work equation from the lab:
WS = PES = ½ Fsx
= ½ (10N)(.2m) = 1.0 J

7. PES = ½ kx2 - or we can find k! k = FS = 10 N x .2m PES = ½ kx2
Calculate k with Hooke’s law:
Thus, PEs can be found:
k = FS = 10 N
x m
PES = ½ kx2
= ½ (50N/m)(.2m)2
= 1.0 N·m = 1.0 J
= 50N/m