recall and use the formula for the spring constant k = F / x

Published by Patrick Mutisya · 14 days ago

Cambridge A-Level Physics 9702 – Stress and Strain

Stress and Strain

Learning Objective

Recall and use the formula for the spring constant

\$k = \frac{F}{x}\$

where k is the spring constant, F is the applied force and x is the extension of the spring.

Key Concepts

  • Stress (\$\sigma\$): The internal force per unit area that develops within a material when an external force is applied.

    \$\sigma = \frac{F}{A}\$

  • Strain (\$\varepsilon\$): The relative deformation of a material, defined as the change in length divided by the original length.

    \$\varepsilon = \frac{\Delta L}{L_0}\$

  • Hooke’s Law: For elastic deformations, the force required to extend or compress a spring is directly proportional to the displacement.

    \$F = kx\$

  • Spring Constant (\$k\$): A measure of the stiffness of a spring. A larger k means a stiffer spring.

Derivation of the Spring Constant Formula

Starting from Hooke’s Law, \$F = kx\$, we isolate \$k\$ to obtain the expression used in calculations:

\$k = \frac{F}{x}\$

Here:

  • \$F\$ – Applied force (N)

  • \$x\$ – Extension or compression from the equilibrium position (m)

  • \$k\$ – Spring constant (N m\(^{-1}\))

Units

QuantitySymbolSI UnitTypical \cdot alue (Example)
Force\$F\$newton (N)10 N
Extension\$x\$metre (m)0.02 m
Spring constant\$k\$newton per metre (N m\(^{-1}\))500 N m\(^{-1}\)

Worked Example

A vertical spring hangs from a support. When a mass of 0.5 kg is attached, the spring stretches by 0.04 m. Determine the spring constant.

  1. Calculate the weight (force) of the mass:

    \$F = mg = (0.5\ \text{kg})(9.81\ \text{m s}^{-2}) = 4.905\ \text{N}\$

  2. Use the definition \$k = F/x\$:

    \$k = \frac{4.905\ \text{N}}{0.04\ \text{m}} = 122.6\ \text{N m}^{-1}\$

  3. Round to an appropriate number of significant figures:

    \$k \approx 1.23 \times 10^{2}\ \text{N m}^{-1}\$

Suggested diagram: A vertical spring with a mass hanging from it, showing the original length \$L0\$, the stretched length \$L\$, and the extension \$x = L - L0\$.

Common Mistakes to Avoid

  • Confusing the total length of the spring with the extension \$x\$; only the change in length is used in the formula.

  • Using mass (kg) directly in the formula instead of converting to force (N) via \$F = mg\$.

  • Neglecting significant figures; the answer should reflect the precision of the given data.

Practice Questions

  1. A spring has a constant \$k = 250\ \text{N m}^{-1}\$. How much force is required to stretch it by \$0.06\ \text{m}\$?

  2. A force of \$15\ \text{N}\$ compresses a spring by \$0.03\ \text{m}\$. Determine the spring constant and state whether the spring is stiffer or softer than the one in question 1.

  3. A mass of \$2\ \text{kg}\$ is attached to a spring and causes an extension of \$0.08\ \text{m}\$. Find the spring constant and the potential energy stored in the spring. (Use \$U = \frac{1}{2}kx^{2}\$.)

Summary

The spring constant \$k\$ quantifies the stiffness of a spring and is obtained from the ratio of applied force to resulting displacement. Mastery of this relationship enables accurate analysis of elastic systems in A‑Level physics.