Tutopiya: Study Resources for IGCSE, IB, GCSE & A-Levels

A spring stretches by 5 cm under a 20 N load. Find its spring constant.

$e = 0.05\,\text{m}$

$k = F/e = 20/0.05 = 400\,\text{N/m}$

A spring of k = 500 N/m is stretched 8 cm within its elastic limit. Calculate the elastic PE stored.

$e = 0.08\,\text{m}$

$E_e = \tfrac{1}{2} \times 500 \times 0.08^2 = 250 \times 0.0064 = 1.6\,\text{J}$

Within limit of proportionality only. Recall (not on equation sheet).

$E_e = \tfrac{1}{2} k e^2$

On AQA equation sheet.

The extension of an elastic object is directly proportional to the force applied, within the limit of proportionality.

A measure of stiffness; force per unit extension (N/m).

Deformation that fully reverses when the force is removed.

Permanent deformation; the material does not return to its original shape.

Using new length instead of extension.

Extension = new length − natural length. Always subtract.

Forgetting to square e in elastic PE.

Write the formula in full: ½ × k × e × e.

Forces and Elasticity