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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

📚 IB Physics SL Hub / Topical Revision
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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
  • Equation F = kΔx, where F is the force applied, Δx the extension or compression, and k the spring constant.
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
  • Restoring force The spring pulls back with an equal force in the opposite direction, which is why it is called a restoring force.
Limit
  • 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
  • Combining Doubling the extension quadruples the stored energy, because the energy depends on the square of the extension.
Mass on a spring
  • Mass on a spring A hanging mass reaches equilibrium when kΔx = mg, so its extension is mg/k.
Measuring k
  • Measuring k Hang known masses, record the extensions and find the gradient of the force-extension graph.
Limits
  • Limits If the spring is overstretched it deforms permanently and the law fails.
QuantityEquationUnit
ForceF = kΔxN
Spring constantk = F/ΔxN m⁻¹
Elastic energyE = ½kΔx²J

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.

  1. 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)
  2. E = ½kΔx² = 0.5 × 200 × (0.050)² = 0.5 × 200 × 0.0025.
  3. 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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