A 0.5kg mass stretches a spring by 5cm. What is the spring constant?

F = mg = 0.5×9.8 = 4.9N; k = F/x = 4.9/0.05 = 98 N/m.

How much elastic PE is stored when a spring (k=50 N/m) is stretched 10cm?

E = ½kx² = ½ × 50 × (0.1)² = 0.25 J.

Pro 🔒~15 min

Hooke's Law

Discover the relationship between spring force and extension

How it works

Hooke's Law states that the restoring force of a spring is proportional to its extension (F = kx), where k is the spring constant in N/m. This applies only within the elastic limit — beyond it the spring deforms permanently. Elastic potential energy stored is ½kx². Springs in series have lower effective k (more stretchy); springs in parallel have higher effective k (stiffer).

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Step-by-step

Drag masses onto the spring hook.
The spring stretches; measure extension with the ruler.
Plot force vs. extension to find the spring constant from the slope.
Add masses beyond the elastic limit to observe deformation.

Key formulas

$F = kx$ Hooke's Law
$E_{elastic} = \frac{1}{2}kx^2$ Elastic potential energy
$k_{series} = \frac{k_1 k_2}{k_1 + k_2}$ Springs in series

Frequently asked questions

A 0.5kg mass stretches a spring by 5cm. What is the spring constant?: F = mg = 0.5×9.8 = 4.9N; k = F/x = 4.9/0.05 = 98 N/m.
How much elastic PE is stored when a spring (k=50 N/m) is stretched 10cm?: E = ½kx² = ½ × 50 × (0.1)² = 0.25 J.
Two identical springs (k=100 N/m) in series: what is the effective k?: K_series = k/2 = 50 N/m.