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Work, Energy & Power in Physics: Formulas, Theorem & Units Guide
Work, Energy, and Power form the core foundation of mechanics and physics. Work measures energy transfer when a force displaces an object. Energy represents the capacity to perform work, while Power measures the rate at which work is executed or energy is transformed.
This 4,000+ word comprehensive exam guide covers **Scientific Definition of Work ($W = F \cdot s \cos\theta$)**, **Positive, Negative & Zero Work**, **Kinetic Energy ($KE = \frac{1}{2} m v^2$)**, **Gravitational & Elastic Potential Energy ($PE = mgh, \frac{1}{2}kx^2$)**, **Work-Energy Theorem ($W_{\text{net}} = \Delta KE$)**, **Law of Conservation of Mechanical Energy**, **Power ($P = W/t = \vec{F} \cdot \vec{v}$)**, **Commercial Unit of Energy ($1\text{ kWh} = 3.6 \times 10^6 \text{ J}$)**, **Horsepower ($1\text{ HP} = 746\text{ W}$)**, and solved numerical problems for SSC CGL, RRB NTPC, and UPSC Prelims.
Table of Contents
- 1. Scientific Concept of Work & Formula ($W = F \cdot s \cos\theta$)
- 2. Three Cases of Work: Positive, Negative & Zero Work
- 3. Kinetic Energy ($KE = \frac{1}{2} m v^2$) & Relation with Momentum ($p$)
- 4. Potential Energy (Gravitational $mgh$ & Spring $\frac{1}{2}kx^2$)
- 5. The Work-Energy Theorem ($W_{\text{net}} = \Delta KE$)
- 6. Law of Conservation of Mechanical Energy
- 7. Power Definition ($P = W/t = \vec{F} \cdot \vec{v}$) & Units (Watt, HP)
- 8. Commercial Unit of Electrical Energy ($1\text{ kWh} = 3.6\text{ MJ}$)
- 9. Solved Numerical Examples for Competitive Exams
- 10. Must Remember Points for Quick Revision
- 11. Frequently Asked Questions (FAQ)
Key Takeaways & Core Highlights
- Work Formula: $W = F \cdot s \cdot \cos\theta$. SI Unit: Joule ($\text{J} = \text{N}\cdot\text{m}$). Scalar quantity.
- $\theta = 0^\circ \rightarrow$ **Positive Work** (Maximum $W = F s$).
- $\theta = 90^\circ \rightarrow$ **Zero Work** (Coolie carrying load on head on flat platform does ZERO work against gravity!).
- $\theta = 180^\circ \rightarrow$ **Negative Work** (Friction force $W = -F s$).
- Kinetic Energy & Momentum Relation: $KE = \frac{1}{2} m v^2 = \frac{p^2}{2 m}$. Doubling momentum quadruples kinetic energy ($4\times$)!
- Work-Energy Theorem: Net work done by all forces equals change in kinetic energy ($W_{\text{net}} = KE_f - KE_i$).
- Power Formula: $P = \frac{W}{t} = \vec{F} \cdot \vec{v}$. SI Unit: Watt ($\text{W} = \text{J/s}$).
- Important Conversions:
- $1\text{ kWh (Kilowatt-hour)} = 1000 \text{ W} \times 3600 \text{ s} = \mathbf{3.6 \times 10^6 \text{ Joules (3.6 MJ)}}$.
- $1\text{ HP (Horsepower)} = \mathbf{746 \text{ Watts}}$.
1. Scientific Concept of Work & Formula ($W = F \cdot s \cos\theta$)
In physics, work is said to be done ONLY when a force applied on an object produces displacement in the direction of the force:
$$W = \vec{F} \cdot \vec{s} = F \cdot s \cdot \cos\theta$$Where $\theta$ is the angle between force vector $\vec{F}$ and displacement vector $\vec{s}$.
2. Three Cases of Work: Positive, Negative & Zero Work
| Type of Work | Angle ($\theta$) Range | Physical Conditions & Real-World Examples |
|---|---|---|
| Positive Work | $0^\circ \le \theta < 90^\circ$ ($\cos\theta > 0$) | Force and displacement in same direction. A horse pulling a cart ($\theta=0^\circ$). |
| Zero Work | $\theta = 90^\circ$ ($\cos 90^\circ = 0$) | Force $\perp$ displacement. A porter holding luggage walking horizontally; Earth orbiting Sun (centripetal force $\perp$ velocity). |
| Negative Work | $90^\circ < \theta \le 180^\circ$ ($\cos\theta < 0$) | Force opposes displacement. Friction stopping a skidding car ($\theta=180^\circ$). |
3. Kinetic Energy ($KE = \frac{1}{2} m v^2$) & Relation with Momentum ($p$)
Energy possessed by an object due to its state of motion:
$$KE = \frac{1}{2} m v^2$$Relationship Between Kinetic Energy and Momentum ($p = mv$):
$$KE = \frac{p^2}{2 m} \implies p = \sqrt{2 m \cdot KE}$$If momentum of a body is increased by $100\%$ ($2p$), its Kinetic Energy increases by $300\%$ ($4 \cdot KE$)!
5. The Work-Energy Theorem ($W_{\text{net}} = \Delta KE$)
$$W_{\text{net}} = \Delta KE = \frac{1}{2} m v^2 - \frac{1}{2} m u^2$$7. Power Definition ($P = W/t = \vec{F} \cdot \vec{v}$) & Units (Watt, HP)
$$P = \frac{W}{t} = \frac{F \cdot s}{t} = F \cdot v$$SI Unit: Watt ($\text{W} = \text{J/s}$). $1\text{ HP (Horsepower)} = 746\text{ Watts}$.
8. Commercial Unit of Electrical Energy ($1\text{ kWh} = 3.6\text{ MJ}$)
$$1\text{ kWh} = 1000\text{ W} \times 3600\text{ s} = 3.6 \times 10^6 \text{ Joules (3.6 MegaJoules)}$$9. Solved Numerical Examples for Competitive Exams
Numerical Problem 1 (Kinetic Energy from Momentum):
Question: A body of mass $4\text{ kg}$ has a momentum of $20\text{ kg}\cdot\text{m/s}$. Calculate its kinetic energy.
Solution:
$$KE = \frac{p^2}{2 m} = \frac{(20)^2}{2 \times 4} = \frac{400}{8} = 50 \text{ Joules}$$Answer: Kinetic energy is $50\text{ Joules}$.
Numerical Problem 2 (Commercial Electricity Units):
Question: An electric heater rated $1500\text{ W}$ operates for $4\text{ hours}$ daily. Calculate the electrical energy consumed in 30 days in commercial units ($\text{kWh}$).
Solution:
$$\text{Energy per day} = 1.5 \text{ kW} \times 4 \text{ h} = 6 \text{ kWh}$$ $$\text{Energy in 30 days} = 6 \times 30 = 180 \text{ kWh (Units)}$$Answer: Energy consumed is $180\text{ units}$ ($180\text{ kWh}$).
10. Must Remember Points for Quick Revision
Exam Revision Cheat Sheet:
- Work Formula: $W = F s \cos\theta$. Zero work at $\theta=90^\circ$.
- Kinetic Energy: $KE = \frac{1}{2}mv^2 = \frac{p^2}{2m}$.
- Potential Energy: $PE = mgh$. Spring $PE = \frac{1}{2}kx^2$.
- Work-Energy Theorem: $W_{\text{net}} = \Delta KE$.
- Power: $P = W/t = F \cdot v$. Unit: Watt ($\text{J/s}$).
- 1 kWh: $3.6 \times 10^6 \text{ Joules}$. 1 HP: $746 \text{ Watts}$.
11. Frequently Asked Questions (FAQ)
What is the Work-Energy Theorem?
The Work-Energy Theorem states that the net work done by all forces acting on a body is equal to the change in its Kinetic Energy: W_net = ΔKE = ½ m v² - ½ m u².
How many Joules are there in 1 kilowatt-hour (1 kWh)?
1 kilowatt-hour (1 kWh or 1 commercial unit of electricity) = 1000 Watts × 3600 seconds = 3.6 × 10⁶ Joules (3.6 MegaJoules).
What is 1 Horsepower (HP) in Watts?
1 Imperial Horsepower (1 HP) = 746 Watts (746 J/s). 1 Metric Horsepower ≈ 735.5 Watts.
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