Work, energy and power: further applications, elastic strings and springs
Mechanics (M2): Work, Energy and Power – Further Applications, Elastic Strings and Springs
Welcome to the exciting world of work, energy and power! Think of them as the “fuel” that drives everything from a skateboard to a roller‑coaster. In this lesson we’ll explore how to calculate work, energy and power, and then see how these concepts apply to elastic strings and springs.
1. Work and Energy
Work is done when a force moves an object. The amount of work depends on the force, the distance moved and the direction of the force.
Formula for work: $W = \vec{F}\cdot\vec{s} = Fs\cos\theta$ ⚡
Where: $F$ = force, $s$ = displacement, $\theta$ = angle between force and displacement.
Energy is the capacity to do work. The two main types we’ll use are kinetic energy (KE) and potential energy (PE).
Kinetic Energy: $KE = \tfrac{1}{2}mv^2$ 🏎️
Gravitational Potential Energy: $PE_g = mgh$ ⛰️
Elastic Potential Energy (spring): $PE_s = \tfrac{1}{2}kx^2$ 🔧
- Example 1: A 5 kg box is pushed 10 m with a constant force of 20 N. What is the work done?
- Example 2: A 0.5 kg mass is dropped from a height of 4 m. Calculate its kinetic energy just before impact.
Answers:
- 1️⃣ $W = 20\,\text{N} \times 10\,\text{m} = 200\,\text{J}$
- 2️⃣ $PE_g = 0.5\,\text{kg} \times 9.8\,\text{m/s}^2 \times 4\,\text{m} = 19.6\,\text{J}$ → KE ≈ 19.6 J (ignoring air resistance)
2. Power
Power tells us how fast work is done. Think of it as the speed at which energy is transferred.
Average Power: $P_{\text{avg}} = \dfrac{W}{t}$ ⚡
Instantaneous Power: $P = \vec{F}\cdot\vec{v}$ 🚀
- Calculate the average power if 500 J of work is done in 10 s.
- What is the instantaneous power when a 10 N force moves an object at 3 m/s?
Answers:
- 1️⃣ $P_{\text{avg}} = 500\,\text{J} / 10\,\text{s} = 50\,\text{W}$
- 2️⃣ $P = 10\,\text{N} \times 3\,\text{m/s} = 30\,\text{W}$
3. Elastic Strings and Springs
When you stretch a rubber band or compress a spring, you store energy. This is called elastic potential energy.
Hooke’s Law: $F = kx$ 📐
Elastic Potential Energy: $PE_s = \tfrac{1}{2}kx^2$ 🔧
Where: $k$ = spring constant (N/m), $x$ = displacement from equilibrium (m).
| Parameter | Symbol | Units |
|---|---|---|
| Spring constant | $k$ | N m⁻¹ |
| Displacement | $x$ | m |
| Elastic PE | $PE_s$ | J |
- A spring with $k = 200\,\text{N/m}$ is stretched by 0.05 m. Find the force and the elastic potential energy stored.
- How much work is required to compress a 150 N/m spring by 0.1 m?
Answers:
- 1️⃣ $F = 200\,\text{N/m} \times 0.05\,\text{m} = 10\,\text{N}$; $PE_s = \tfrac{1}{2}\times200\times(0.05)^2 = 0.25\,\text{J}$
- 2️⃣ $W = \tfrac{1}{2}k x^2 = \tfrac{1}{2}\times150\times(0.1)^2 = 0.75\,\text{J}$
Exam Tips & Quick Checks
Tip 1: Always check the units – they help you spot mistakes quickly.
Tip 2: For power questions, identify whether you need average or instantaneous power.
Tip 3: When dealing with springs, remember that the work done is the area under the force–displacement graph – a triangle if the force increases linearly.
Tip 4: In multiple‑choice questions, look for “most likely” values – they often use round numbers.
Good luck! 🎓
Revision
Log in to practice.