A spring with a constant of 50 N/m is stretched by 0.2 m. Find the restoring force exerted by the spring.
\[ F = kx = 50 \, \text{N/m} \times 0.2 \, \text{m} = 10 \, \text{N} \]
Hooke’s law
Hooke’s Law states that the force exerted by a spring is directly proportional to its extension or compression, as long as the deformation is within the elastic limit.
Mathematically, it is denoted as:
\[ F = -kx \]
- \( F \): Restoring force (N)
- \( k \): Spring constant (N/m)
- \( x \): Displacement from the equilibrium position (m)
- The negative sign indicates that the force opposes the displacement.
Elastic and Plastic Deformation
- Elastic deformation: The object returns to its original shape after the force is removed (follows Hooke’s Law).
- Plastic deformation: The object does not return to its original shape after deformation (exceeds elastic limit).
Energy in a Stretched Spring
The work done to stretch or compress a spring is stored as elastic potential energy:
\[ U = \frac{1}{2}kx^2 \]
Where \( U \) is the elastic potential energy (J).
Key Insight: The energy stored in the spring depends on \( x^2 \), meaning a small increase in \( x \) results in a larger energy increase.
Young’s Modulus
Young’s modulus is a measure of the stiffness of a material. It describes how much a material stretches or compresses under a given stress.
Definition: Young’s modulus \(Y\) is defined as the ratio of stress to strain:
\[ Y = \frac{\text{Stress}}{\text{Strain}} = \frac{F/A}{\Delta L / L} = \frac{F L}{A \Delta L} \]
- \(F\) = Applied force (N)
- \(A\) = Cross-sectional area (m²)
- \(L\) = Original length (m)
- \(\Delta L\) = Extension (m)
Units: Since stress is measured in Pascals (Pa) and strain is dimensionless, Young’s modulus also has units of Pascals (Pa).
High Young’s modulus: The material is stiff (e.g., steel)
Low Young’s modulus: The material is more flexible (e.g., rubber)
- Only applies in the linear (elastic) region of a stress-strain graph.
- Beyond the elastic limit, Hooke’s law and Young’s modulus no longer apply.
- Useful for comparing mechanical properties of different materials.
Applications of Hooke’s Law
- Spring Scales (Weighing machines)
- Shock Absorbers in cars (control motion)
- Bungee Jumping (Elastic motion)
- Atomic Vibrations (Atoms in a molecule vibrate like tiny springs)
A force of 15 N stretches a spring by 0.3 m. Determine the spring constant.
\[ F = kx \Rightarrow k = \frac{F}{x} = \frac{15}{0.3} = 50 \, \text{N/m} \]
A block of 5 kg is dropped onto a vertical spring with a stiffness of 200 N/m. If the block compresses the spring by 10 cm, determine the velocity of the block just before impact.
According to Energy conservation:
\[ \frac{1}{2}kx^2 = \frac{1}{2}mv^2 \Rightarrow v = \sqrt{\frac{kx^2}{m}} = \sqrt{\frac{200 \cdot 0.1^2}{5}} = 0.632 \, \text{m/s} \]
Did you know?
When two identical springs (each of spring constant \( k \)) are connected in:
- (a) Series: Their effective spring constant, \( \frac{1}{k_{\text{eff}}} = \frac{1}{k} + \frac{1}{k} = \frac{2}{k} \Rightarrow k_{\text{eff}} = \frac{k}{2} \)
- (b) Parallel: Their effective spring constant, \( k_{\text{eff}} = k + k = 2k \)
Written by Thenura Dilruk