Course: Combined Science + Separate Physics | Tier: Foundation + Higher | Linked equations: F = ke, Eₑ = ½ke²
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Aim
To investigate the relationship between the force applied to a spring and its extension.
Method
- Hang a spring from a clamp stand, with a metre ruler clamped vertically beside it. Attach a pointer to the bottom of the spring.
- Record the length of the unstretched spring.
- Hang a 100 g mass (weight about 1 N) on the spring and record the new length.
- Add masses one at a time, recording the length each time.
- Calculate extension = new length − original length. Plot force (weight) against extension.
| Independent variable | Force (weight of masses added) |
| Dependent variable | Extension of the spring |
| Control variables | The same spring; position of the ruler |
Questions
Q1 (F) Name the independent and dependent variables.
Show answer
Independent: the force on the spring (weight of the masses). Dependent: the extension of the spring.
Q2 (F) A spring is 4.0 cm long unstretched and 6.5 cm long with a mass on it. Calculate the extension.
Show answer
Extension = new length − original length = 6.5 − 4.0 = 2.5 cm
Q3 (F) State one hazard and how to reduce the risk.
Show answer
Masses could fall on your feet, or the stand could topple – clamp the stand to the bench and stand up while working. The spring could snap – wear safety goggles.
Q4 (F) How can you make the length readings more accurate?
Show answer
Use a pointer on the spring, keep the ruler vertical, and read with your eye level with the pointer. Wait for the spring to stop bouncing.
Q5 (F/H) The graph of force against extension is a straight line through the origin. What does this show?
Show answer
The extension is directly proportional to the force applied (up to the limit of proportionality).
Q6 (F/H) A force of 4.0 N gives an extension of 8.0 cm. Calculate the spring constant.
Show answer
Convert first: 8.0 cm = 0.080 m
F: F = k e
I: 4.0 = k × 0.080
F: k = 4.0 ÷ 0.080
A: 50 N/m
Q7 (F/H) At large forces the graph starts to curve. What does this mean?
Show answer
The spring has gone past its limit of proportionality – the extension is no longer directly proportional to the force.
Q8 (H) Using the spring from Q6, calculate the elastic potential energy stored when the extension is 8.0 cm.
Show answer
F: Eₑ = ½ k e²
I: Eₑ = 0.5 × 50 × 0.080²
A: 0.16 J
