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
Quantity
Symbol
SI Unit
Typical \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.
Calculate the weight (force) of the mass:
$$F = mg = (0.5\ \text{kg})(9.81\ \text{m s}^{-2}) = 4.905\ \text{N}$$
Use the definition $k = F/x$:
$$k = \frac{4.905\ \text{N}}{0.04\ \text{m}} = 122.6\ \text{N m}^{-1}$$
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 $L_0$, the stretched length $L$, and the extension $x = L - L_0$.
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
A spring has a constant $k = 250\ \text{N m}^{-1}$. How much force is required to stretch it by $0.06\ \text{m}$?
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.
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.