1.2.8Tools and Inquiry — IB Physics skills
Hooke’s Law — IB Physics SL Revision Notes & Topical Questions
5 min read 5 parts Lesson 26 of 33 Unit quiz
Section 1 of 5Hooke's Law
Within a limit, the force needed to stretch or compress a spring is proportional to the change in its length.
Equation
- F = kΔx, where F is the force applied, Δx the extension or compression, and k the spring constant.
Spring constant
- k is measured in N m⁻¹ and tells how stiff the spring is, since it is the force per metre of extension.
Restoring force
- The spring pulls back with an equal force in the opposite direction, which is why it is called a restoring force.
Limit
- Beyond the limit of proportionality the extension is no longer proportional to the force.
Section 2 of 5Graphs and Energy
The force-extension graph holds more information than just the spring constant.
1
Plot force on the vertical axis against extension on the horizontal axis for the loading of the spring.
2
In the linear region the gradient of the line equals the spring constant k.
3
The area under the graph is the work done in stretching the spring, which is stored as elastic potential energy.
4
For a triangle of base Δx and height F this area is ½FΔx, which with F = kΔx gives E = ½kΔx².
5
If the spring is released this energy is transferred to kinetic energy of the load, or to gravitational potential energy.
Section 3 of 5Using Springs
Using Springs
Combining
- Doubling the extension quadruples the stored energy, because the energy depends on the square of the extension.
Mass on a spring
- A hanging mass reaches equilibrium when kΔx = mg, so its extension is mg/k.
Measuring k
- Hang known masses, record the extensions and find the gradient of the force-extension graph.
Limits
- If the spring is overstretched it deforms permanently and the law fails.
| Quantity | Equation | Unit |
|---|---|---|
| Force | F = kΔx | N |
| Spring constant | k = F/Δx | N m⁻¹ |
| Elastic energy | E = ½kΔx² | J |
Section 4 of 5Worked Example
Question
- A spring has a spring constant of 200 N m⁻¹.
- It is stretched by 0.050 m.
- Calculate the force needed and the elastic potential energy stored.
- F = kΔx = 200 × 0.050 = 10 N.What the symbols mean: F = force (N) · k = spring constant (N m−1) · x or e = extension (m)
- E = ½kΔx² = 0.5 × 200 × (0.050)² = 0.5 × 200 × 0.0025.
- E = 0.25 J.
Answer A force of 10 N is needed and 0.25 J of elastic energy is stored.
Section 5 of 5Exam Checklist & Tips
1.2.8 – Key Points for the Exam
- Hooke's law Extension is directly proportional to the force applied, up to the limit of proportionality.
- Spring constant The force per unit extension of a spring, in N m⁻¹.
- Elastic potential energy The energy stored in a deformed elastic object.
- Limit of proportionality The point beyond which force and extension are not proportional.
Common Mistakes
- Using the length of the spring rather than the extension in the equation.
- Forgetting to square the extension when calculating the stored energy.
- Using ½FΔx with a force that is not the final value of the load.
- Giving the unit of k as N instead of N m⁻¹.
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