Hooke's Law Required Practical, Spring Constant & Elastic Potential Energy
To stretch, bend, or compress an object, more than one force must be applied to it (otherwise a single force would simply accelerate the entire object). Hooke's Law states that the extension of an elastic object, such as a steel spring, is directly proportional to the force applied to it (F = ke), provided the limit of proportionality is not exceeded. Examiners regularly test the Hooke's Law required practical, calculating the spring constant (k) from the gradient of a force-extension graph, and equating work done to elastic potential energy stored in a stretched or compressed spring (E_e = ½ke²).
- Hooke's Law Formula (F = ke)
- The extension of an elastic object (such as a spring) is directly proportional to the force applied, provided the limit of proportionality is not exceeded. Formula: Force, F (Newtons, N) = Spring constant, k (N/m) x Extension, e (metres, m). Note: The equation also applies to the compression of a spring, where e is the compression.
- Spring Constant (k) & Stiffness
- The spring constant (k, measured in N/m or N/cm) is a measure of the stiffness of a spring: a stiffer spring has a larger spring constant (requires more force per metre of extension). On a graph with Force on the y-axis and Extension on the x-axis, the gradient of the linear section equals the spring constant (k = ΔF / Δe). (Warning: If Extension is plotted on the y-axis and Force on the x-axis, k = 1 / gradient).
- Limit of Proportionality & Elastic vs Inelastic Deformation
- Up to the limit of proportionality, F ∝ e (a straight-line graph through the origin). Beyond the limit of proportionality (and the elastic limit), the spring undergoes inelastic deformation: it no longer obeys Hooke's Law, the graph curves, and the spring is permanently stretched when the load is removed.
Hooke's Law Required Practical, Spring Constant & Elastic Potential Energy: Wave Equation (v = fλ) & Force Vector (F = ma) Simulator
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F_net = 80 N − 25 N = 55 NAcceleration: a = F_net ÷ m = 55 ÷ 10 = 5.50 m/s²Exam notes: Hooke's Law (F = ke), Spring Constant Definition & Elastic Potential Energy
- Hooke's Law states that the extension (e) of a spring or elastic object is directly proportional to the force (F) applied, provided the LIMIT OF PROPORTIONALITY is not exceeded: F = k × e.
- The spring constant (k = F / e, measured in N/m) is a measure of the stiffness of the spring: a steeper force-extension graph (when F is on the y-axis and e is on the x-axis) means a larger spring constant k and a stiffer spring.
- Critical exam TRAP: Extension e is always (Stretched Length − Original Unstretched Length), never the total stretched length. Always convert extension from cm or mm into metres (m) before calculating k in N/m or Elastic Potential Energy in Joules (J).
- The work done stretching a spring is stored as Elastic Potential Energy: Ee = ½ × k × e² (which also equals the area of the triangle ½ × F × e under the linear region of the force-extension graph).
| Feature / Region | Linear Elastic Region (Obeys Hooke's Law) | Beyond the Limit of Proportionality / Elastic Limit |
|---|---|---|
| Force-Extension Graph Shape | Straight line passing directly through the origin (0, 0) | Graph curves (gradient changes; extension increases faster than force) |
| Mathematical Relationship | F = k × e (Force is directly proportional to extension) | F ≠ k × e (Hooke's Law is no longer obeyed) |
| Behaviour When Force Removed | Elastic deformation: object returns to its exact original length (e = 0) | Inelastic (plastic) deformation: object stays permanently stretched |
| Energy Stored / Work Done | All work done is stored as recoverable Ee = ½ke² (area under line) | Some work done is dissipated as thermal energy rearranging internal bonds |
Key Concepts & Exam Definitions (6)
- Hooke's Law Formula (F = ke)
- The extension of an elastic object (such as a spring) is directly proportional to the force applied, provided the limit of proportionality is not exceeded. Formula: Force, F (Newtons, N) = Spring constant, k (N/m) x Extension, e (metres, m). Note: The equation also applies to the compression of a spring, where e is the compression.
- Spring Constant (k) & Stiffness
- The spring constant (k, measured in N/m or N/cm) is a measure of the stiffness of a spring: a stiffer spring has a larger spring constant (requires more force per metre of extension). On a graph with Force on the y-axis and Extension on the x-axis, the gradient of the linear section equals the spring constant (k = ΔF / Δe). (Warning: If Extension is plotted on the y-axis and Force on the x-axis, k = 1 / gradient).
- Limit of Proportionality & Elastic vs Inelastic Deformation
- Up to the limit of proportionality, F ∝ e (a straight-line graph through the origin). Beyond the limit of proportionality (and the elastic limit), the spring undergoes inelastic deformation: it no longer obeys Hooke's Law, the graph curves, and the spring is permanently stretched when the load is removed.
- Elastic potential energy formula (E_e = ½ke²)
- A force that stretches or compresses a spring does mechanical work on the spring, storing elastic potential energy in the spring. Provided the spring is not inelastically deformed, the work done equals the elastic potential energy stored: Elastic potential energy, E_e (Joules, J) = 0.5 x Spring constant, k (N/m) x (Extension, e)² (m²). This also equals the triangular area under the linear Force-Extension graph (Area = ½ x base x height = ½ x e x F = ½ke²).
- Required Practical: Investigating Force vs Extension of a Spring
- 1. Clamp a spring and a millimetre ruler vertically to a weighted clamp stand (place a heavy weight on the base so it doesn't topple; wear safety goggles in case the spring snaps). 2. Measure the initial unstretched length of the coiled part of the spring using a horizontal pointer/fiducial marker read at eye level (avoids parallax error). 3. Hang a 100 g (0.98 N or 1.0 N) slotted mass on the spring, measure the new length, and calculate extension = new length, original length. 4. Repeat up to 600 g (6 N) and plot Force against Extension.
- Linear vs Non-Linear Deformation (Spring vs Rubber Band)
- A metal coil spring shows linear elastic deformation (a straight line through the origin obeying Hooke's Law) up to its limit of proportionality. By contrast, a rubber band or polythene strip shows non-linear deformation: its Force-Extension graph is a curve from the very beginning, so it does not obey Hooke's Law even though a rubber band still returns to its original length (elastic).
Quick-Fire Flashcards (6)
Click any card to check the answer, then mark it as Known to track your progress.
State Hooke's Law in words and write the equation with standard SI units.
An unstretched spring is 12 cm long. When a force of 6.0 N is applied, its total length becomes 20 cm. Calculate the spring constant k in N/m.
State two safety precautions and two accuracy improvements for the Hooke's Law required practical.
What does the area under a Force-Extension graph represent?
How can you tell from a table of results or a graph that a spring has been stretched beyond its limit of proportionality?
A toy dart gun compresses a spring of spring constant 200 N/m by 5.0 cm. Calculate the elastic potential energy stored in the spring.
Exam-Style Questions & Mark Schemes (4)
Write or think through your answer first, then click Check Mark Scheme to see the exact points examiners award marks for.
- Q1
A car suspension spring has an unstretched length of 0.40 m and a spring constant of 25,000 N/m. When the car is loaded, the spring is compressed to a length of 0.32 m. Calculate: (a) the compression of the spring, (b) the downward force acting on this spring, and (c) the elastic potential energy stored in the compressed spring. [5 marks]
Click “Check Mark Scheme” to compare your answer, 1. (a) Compression e = Original length: Compressed length = 0.40 m - 0.32 m = 0.08 m [1 mark]. 2. (b) Using Hooke's Law, F = k x e = 25,000 N/m x 0.08 m [1 mark]. 3. Force F = 2,000 N [1 mark]. 4. (c) Elastic potential energy: E_e = 0.5 x k x e² = 0.5 x 25,000 x (0.08)² [1 mark]. 5. E_e = 12,500 x 0.0064 = 80 J [1 mark].
- Q2
Describe how a student could carry out an experiment to investigate whether both a steel coil spring and a rubber band obey Hooke's Law, including how the data should be processed and compared. [6 marks]
Click “Check Mark Scheme” to compare your answer, 1. Set up a clamp stand secured to the bench with a G-clamp (or counterweight on the base) and clamp a millimetre ruler vertically alongside the suspended steel spring [1 mark]. 2. Attach a horizontal pointer to the bottom of the unstretched spring and read its initial position against the ruler at eye level (to avoid parallax error) [1 mark]. 3. Hang a 1.0 N (100 g) slotted mass from the spring, record the new pointer reading once stationary, and calculate the extension (new reading, initial reading) [1 mark]. 4. Add 1.0 N masses one at a time up to 6.0 N, recording the extension each time, and also check whilst unloading the masses whether the spring returns to its original length (elastic vs inelastic) [1 mark]. 5. Repeat the exact same procedure replacing the steel spring with the rubber band, then plot a graph of Force (N) against Extension (m) for both objects on the same axes [1 mark]. 6. Conclusion: The steel spring gives a straight line through the origin (up to its limit of proportionality), proving F ∝ e (obeys Hooke's Law), whereas the rubber band produces a non-linear curve, proving it does not obey Hooke's Law [1 mark].
- Q3
A student stretches a catapult elastic (k = 160 N/m) by 0.25 m to launch a small stone of mass 0.020 kg vertically upwards. Assuming 100% of the elastic potential energy is transferred to the stone, calculate: (a) the elastic potential energy stored, (b) the launch speed of the stone, and (c) the maximum vertical height reached by the stone (g = 9.8 N/kg). [5 marks]
Click “Check Mark Scheme” to compare your answer, 1. (a) Elastic potential energy stored: E_e = 0.5 x k x e² = 0.5 x 160 x (0.25)² = 80 x 0.0625 = 5.0 J [1 mark]. 2. (b) At launch, kinetic energy equals elastic potential energy: E_k = ½ x m x v² = 5.0 J -> 0.5 x 0.020 x v² = 5.0 [1 mark]. 3. 0.010 x v² = 5.0 -> v² = 500 -> v = √500 = 22.4 m/s [1 mark]. 4. (c) At maximum height, gravitational potential energy equals initial energy: E_p = m x g x h = 5.0 J [1 mark]. 5. h = 5.0 / (0.020 x 9.8) = 5.0 / 0.196 = 25.5 m [1 mark].
- Q4
Two springs, Spring A and Spring B, are tested. On a graph with Extension (m) on the y-axis and Force (N) on the x-axis, both give straight lines through the origin, but the line for Spring A is steeper than the line for Spring B. Explain which spring has the higher spring constant (stiffness) and why. [3 marks]
Click “Check Mark Scheme” to compare your answer, 1. By Hooke's Law (F = ke), extension is given by e = (1 / k) x F [1 mark]. 2. Because Extension is plotted on the y-axis and Force on the x-axis, the gradient of the graph is Δe / ΔF = 1 / k (the reciprocal of the spring constant) [1 mark]. 3. Since Spring B has a shallower gradient, it extends less for a given force, meaning Spring B is stiffer and has the higher spring constant (k) [1 mark].
Common Exam Questions (4)
Why must you apply at least two forces to stretch, compress, or bend a spring?
If you apply only one force to an object on a frictionless surface, there is a resultant force in that direction so the whole object simply accelerates and moves away without changing shape. To deform (stretch, compress, or bend) an object while keeping it stationary, you must apply at least two balanced forces in opposite directions (e.g., the downward weight of the hanging mass and the upward pulling force of the clamp stand).
What is the most common mistake students make when calculating extension (e) in Hooke's Law questions?
In F = ke and E_e = ½ke², the symbol 'e' stands for extension (how much longer the spring has stretched), not the total stretched length of the spring. Always calculate Extension = Stretched length: Original unstretched length first, and convert centimetres (cm) into metres (m) by dividing by 100 before substituting into the formula.
What is the difference between elastic deformation and inelastic (plastic) deformation?
Elastic deformation occurs when an object returns to its exact original shape and length once the deforming forces are removed. Inelastic (or plastic) deformation occurs when an object is stretched beyond its elastic limit, so it stays permanently stretched and does not return to its original length when the forces are removed.
What does the limit of proportionality look like on a force-extension graph?
While a spring obeys Hooke's Law, the graph of Force (F) against Extension (e) is a straight line passing through the origin (0,0), because F is directly proportional to e. The limit of proportionality is the exact point where the graph stops being a straight line and begins to curve (becoming nonlinear).
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