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1.2.8Forces and Motion

Core Practical- Investigating Force and Extension — Edexcel International GCSE Science Double Award 4SD0 - Physics Revision Notes & Topical Questions

5 min read 5 parts Lesson 18 of 26

📚 Edexcel International GCSE Science Double Award 4SD0 - Physics Hub / Topical Revision
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Part 1 of 5Variables and Apparatus

Careful control of variables allows a clear force-extension relationship to be found.

Independent variable
  • Independent variable The load or force applied to the spring, changed by adding masses.
Dependent variable
  • Dependent variable The extension of the spring, found by subtracting its original length from the new length.
Control variables
  • Control variables The same spring, the same ruler position and the same measuring method for every load.
Apparatus
  • Apparatus A spring, a clamp stand, a metre ruler, a set of masses and a suitable hanger.

Part 2 of 5Method

Taking readings carefully improves the quality of the graph.

1
Set up the spring vertically and measure its original length against the ruler.
2
Add a mass to the hanger and wait for the spring to stop moving before reading the length.
3
Calculate the weight of the load using W = mg, with g = 10 N/kg.
4
Record the extension, and repeat with increasing masses and then decreasing masses.
5
Plot a graph of force against extension to show the relationship.

Part 3 of 5Evaluating the Method

Evaluating the Method
Parallax error
  • Parallax error Reading the ruler from an angle gives a wrong length, so the eye should be level with the mark.
Overloading
  • Overloading Adding too much mass can permanently stretch the spring, which invalidates later readings.
Repeat readings
  • Repeat readings Repeating each measurement and averaging reduces random errors in the results.
Zero error
  • Zero error Checking the ruler reading with no load corrects any zero error in the measurement.

Part 4 of 5Worked Example

Question

  • A spring has an original length of 12.0 cm.
  • A 0.3 kg mass is hung from it and the length becomes 13.5 cm.
  • Calculate the weight of the mass and the extension.

  1. Weight W = mg = 0.3 kg × 10 N/kg = 3.0 N.
  2. Extension = new length − original length = 13.5 cm − 12.0 cm = 1.5 cm.
  3. Convert the extension1.5 cm = 0.015 m if SI units are required.

Answer The weight is 3.0 N and the extension is 1.5 cm (0.015 m).

Part 5 of 5Exam Checklist & Tips

1.2.8 – Key Points for the Exam

  • Extension The increase in length of a spring from its original length when a force is applied.
  • Load The mass or force applied to stretch an object.
  • Original length The length of the spring before any force is applied.
  • Control variable A factor kept constant so that a fair test is performed.
  • Parallax error An error in a reading caused by viewing the scale from the wrong angle.

Common Mistakes

  • Using total length instead of extension, which means subtracting the original length.
  • Forgetting to convert the mass to weight in newtons before plotting force.
  • Recording the reading before the spring has settled, which gives a wrong extension.
  • Overloading the spring so that it is permanently stretched and no longer behaves elastically.

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