Force and extension of a spring

Practical method

Aim

Investigate the relationship between force and extension for a spring, and find its spring constant.

Variables

Independent variable: force applied (weight of masses)
Dependent variable: extension of the spring
Control variables: the same spring; loading in equal steps; measuring to the same point on the spring.

Apparatus

• Spring
• Clamp stand with a G-clamp to the bench
• Metre rule clamped vertically
• Pointer attached to the spring
• 100 g slotted masses (weight about 1.0 N each)
• Set square

Method

1. Hang the spring from the clamp and attach a pointer at its bottom. Record the pointer reading on the rule (the natural length).
2. Add a 1.0 N mass and record the new pointer reading at eye level.
3. Add masses one at a time up to 6.0 N, recording each reading.
4. Calculate extension = loaded length − natural length for each force.
5. Unload in steps to check the spring returns to its original length.
6. Plot force (y-axis) against extension (x-axis); the gradient of the straight section is k.

Sample results

Natural length 12.0 cm.

Analysis

Up to 5.0 N the graph is a straight line through the origin: extension is proportional to force, so the spring obeys Hooke's law. k = \(\dfrac{F}{x}\) = 5.0 ÷ 0.100 = 50 N/m. At 6.0 N the extension (12.6 cm) is more than the 12.0 cm the straight line predicts, so the limit of proportionality has been passed.

Errors and improvements

Risk assessment

Common mistakes

• Plotting length instead of extension.
• Using centimetres in F = kx without converting to metres.
• Assuming the whole graph is straight.

Worked analysis

A force of 4.0 N stretches the spring by 8.0 cm (0.080 m). k = \(\dfrac{F}{x}\) = 4.0 ÷ 0.080 = 50 N/m. Elastic energy stored = ½kx² = 0.5 × 50 × 0.080² = 0.16 J.

Common misconception

Extension is not the length of the spring. It is the increase in length from the unstretched (natural) length.

Variations required by some exam boards

Work done: the work done in stretching the spring equals the area under the force–extension graph. In the straight-line region this is ½Fx, which equals the elastic potential energy stored, ½kx². For example, 5.0 N and 0.100 m give ½ × 5.0 × 0.100 = 0.25 J.