Elastic Potential Energy
Work stored in a deformed spring — Hooke's Law and the spring constant
If the force you apply to a spring is directly proportional to stretch, how many times would you have to increase your force to get 4 times the stretch?
Elastic P.E.
work stored in a deformed spring
a) General Equation
F = Average Force needed to deform the spring
x = distance deformed (stretch or compress)
Ideal Spring
b) Hooke's Law
Force needed to deform an ideal spring a given amount is directly proportional to its displacement (deformation).
Graphic from Learning-Connections Clipart
Ex) Ideal Spring
| F (Force to Deform) (N) |
x (distance deformed) (m) |
|---|---|
| 2.0 | 0.30 |
| 4.0 | 0.60 |
| 6.0 | 0.90 |
| 8.0 | 1.2 |
Hooke's Law — University of Colorado PhET
c) Spring Constant (k) — Ideal Spring
quantifies the stiffness of a spring
What is the spring constant of the spring that produced the data above?
| F (N) | x (m) |
|---|---|
| 2.0 | 0.30 |
| 4.0 | 0.60 |
| 6.0 | 0.90 |
| 8.0 | 1.2 |
Which is the stiffer spring?
Ex) Plots and k
Pick any point on the line
d) Elastic P.E.
= area of the triangle under the F vs x graph
Try This at Home!
What crushed this can?
S.M.U. Physics
The can implodes because the atmospheric pressure is greater than the pressure inside the can.
When the can was submerged in the water it caused the water vapor inside the can to rapidly cool. When a gas is cooled, its pressure goes down. This drop in pressure causes the pressure inside the can to be lower than the atmospheric pressure.
Note: Use less than 1 cm of water in this experiment (about ½ inch).
