Work, Energy, Power and Efficiency (Definitions, Formulas, Examples)
CSEC Physics: Mechanics Unit
Essential Understanding: Work, energy, power, and efficiency are fundamental concepts in physics that explain how forces cause motion, how energy transforms, and how effectively machines operate. These principles apply to everyday situations from walking up stairs to operating electrical devices.
Learning Objectives
By the end of this article, you should be able to:
โ Define work, energy, power, and efficiency in physics terms
โ State and use the correct SI units for each quantity
โ Apply formulas to solve numerical problems
โ Interpret real-world examples involving energy and power
โ Answer CSEC-style questions confidently
Introduction: Why Work and Energy Matter
Work and energy are fundamental concepts that describe how forces cause motion and how energy transforms from one form to another. These principles explain:
Machines
How engines, motors, and simple machines operate by transferring energy
Electricity
Electrical energy conversion to other forms (light, heat, motion)
Transport
How vehicles convert fuel energy into kinetic energy
Human Activity
Biological energy transfer in muscles during physical work
Work in Physics
Definition of Work
Work is done when a force causes an object to move in the direction of the force.
Where:
\(W\) = Work done (in joules, J)
\(F\) = Force applied (in newtons, N)
\(d\) = Distance moved in direction of force (in meters, m)
\(\theta\) = Angle between force and direction of motion
Conditions for Work
Work is ONLY done when:
- A force is applied to an object
- The object moves
- The movement is in the direction of (or has a component in the direction of) the force
Example: Pushing a wall that doesn't move does NO work.
SI Unit: Joule (J)
1 joule = 1 newton meter (1 J = 1 Nยทm)
Definition: 1 joule is the work done when a force of 1 newton moves an object 1 meter in the direction of the force.
Everyday Example: Lifting an apple (about 1 N) 1 meter does approximately 1 joule of work.
Worked Example
Problem: A student pushes a box with a constant force of 50 N along a horizontal floor for a distance of 8 m. If the force is applied parallel to the floor, calculate the work done.
Step 1: Identify known values
\(F = 50 \text{ N}\)
\(d = 8 \text{ m}\)
\(\theta = 0^\circ\) (force parallel to motion)
Step 2: Apply formula
\(W = F \times d \times \cos\theta\)
\(W = 50 \times 8 \times \cos 0^\circ\)
\(W = 50 \times 8 \times 1\)
Step 3: Calculate and state answer
\(W = 400 \text{ J}\)
The work done is 400 joules.
Energy
Definition of Energy
Energy is the capacity to do work. The SI unit of energy is the joule (J), same as work.
Principle of Conservation of Energy: Energy cannot be created or destroyed, only converted from one form to another.
Kinetic Energy (KE)
Energy possessed by a moving object.
Where:
\(m\) = mass (kg)
\(v\) = velocity (m/s)
Example: A 2 kg ball moving at 3 m/s has:
\(KE = \frac{1}{2} \times 2 \times 3^2 = 9 \text{ J}\)
Gravitational Potential Energy (GPE)
Energy stored due to an object's position in a gravitational field.
Where:
\(m\) = mass (kg)
\(g\) = gravitational field strength (โ 10 N/kg on Earth)
\(h\) = height above reference point (m)
Example: A 5 kg object 4 m above ground has:
\(GPE = 5 \times 10 \times 4 = 200 \text{ J}\)
Energy Conversion Example
Falling Object Energy Transfer
Scenario: An apple of mass 0.1 kg falls from a tree branch 3 m above the ground. Calculate its energy at different points.
At the branch (before falling):
\(h = 3 \text{ m}, v = 0 \text{ m/s}\)
\(GPE = mgh = 0.1 \times 10 \times 3 = 3 \text{ J}\)
\(KE = 0 \text{ J}\)
Total energy = 3 J
Halfway down (h = 1.5 m):
Using conservation of energy:
Total energy still = 3 J
\(GPE = 0.1 \times 10 \times 1.5 = 1.5 \text{ J}\)
Therefore \(KE = 3 - 1.5 = 1.5 \text{ J}\)
Just before hitting ground (h = 0 m):
\(GPE = 0 \text{ J}\)
Therefore \(KE = 3 \text{ J}\)
Using \(KE = \frac{1}{2}mv^2\):
\(3 = \frac{1}{2} \times 0.1 \times v^2\)
\(v^2 = 60\)
\(v โ 7.75 \text{ m/s}\)
Power
Definition of Power
Power is the rate at which work is done or energy is transferred.
Where:
\(P\) = Power (in watts, W)
\(W\) = Work done (in joules, J)
\(t\) = Time taken (in seconds, s)
SI Unit: Watt (W)
1 watt = 1 joule per second (1 W = 1 J/s)
Definition: A device has a power of 1 watt if it converts 1 joule of energy per second.
Everyday Examples:
โข LED bulb: 10 W
โข Laptop: 50 W
โข Human climbing stairs: 200-300 W
Alternative Formula
Power can also be calculated using:
Where:
\(F\) = Force (N)
\(v\) = Velocity (m/s)
Derivation:
\(P = \frac{W}{t} = \frac{F \times d}{t} = F \times \frac{d}{t} = F \times v\)
Worked Example
Problem: A crane lifts a 500 kg steel beam vertically through a height of 12 m in 30 seconds. Calculate the power output of the crane. (Take g = 10 N/kg)
Step 1: Calculate weight (force)
\(F = mg = 500 \times 10 = 5000 \text{ N}\)
Step 2: Calculate work done
\(W = F \times d = 5000 \times 12 = 60000 \text{ J}\)
Step 3: Calculate power
\(P = \frac{W}{t} = \frac{60000}{30} = 2000 \text{ W}\)
Answer: The crane's power output is 2000 W or 2 kW.
Power Comparison Chart
Efficiency
Definition of Efficiency
Efficiency measures how effectively a machine or process converts input energy into useful output energy.
Efficiency is always less than 100% due to energy losses.
Energy Losses
Energy is often "lost" as:
- Heat: Friction in moving parts
- Sound: Vibrations in machines
- Light: Unwanted radiation
- Electrical resistance: In wires and components
Example: Incandescent bulb loses 90% of electrical energy as heat, only 10% as light.
Real-World Efficiencies
Typical efficiency ranges:
- Car engine: 20-30%
- Electric motor: 70-95%
- Solar panel: 15-22%
- LED light: 30-40%
- Human body: 20-25%
No machine is 100% efficient due to inevitable energy losses.
Efficiency Calculation Example
Problem: An electric motor consumes 5000 J of electrical energy to lift a load. If the useful work done in lifting the load is 4000 J, calculate the efficiency of the motor.
Step 1: Identify values
Useful energy output = 4000 J
Total energy input = 5000 J
Step 2: Apply efficiency formula
\(\text{Efficiency} = \frac{4000}{5000} \times 100\%\)
Step 3: Calculate
\(\text{Efficiency} = 0.8 \times 100\% = 80\%\)
Step 4: Energy loss
Energy lost = Total input - Useful output
\(5000 - 4000 = 1000 \text{ J}\)
This is lost mainly as heat and sound.
Efficiency Visualizer
Objective: See how input energy is divided between useful output and losses.
Useful Output: 4000 J
Energy Lost: 1000 J
Sankey Diagrams
Sankey diagrams visually represent energy transfers and losses. The width of each arrow is proportional to the amount of energy.
Common Student Errors in CSEC Physics
Calculation Mistakes
- Using mass (kg) instead of weight (N) in work calculations
- Forgetting to square velocity in kinetic energy formula
- Using cm or km instead of meters in formulas
- Not converting time to seconds in power calculations
- Writing efficiency as decimal (0.8) instead of percentage (80%)
Conceptual Errors
- Thinking work is done when force is applied but object doesn't move
- Confusing power (rate) with energy (total quantity)
- Assuming machines can be 100% efficient
- Not recognizing all forms of energy in conservation problems
- Forgetting that GPE depends on height above a reference point
CSEC Exam Focus
How These Topics Appear in CSEC Physics
Multiple-Choice Questions: Often test definitions, units, and simple calculations.
Structured Questions: May include multi-step problems involving work, energy, power, and efficiency.
Calculations and Explanations: You'll need to show working, state formulas, and explain concepts.
Command Words to Know:
- Define: Give the precise meaning
- Calculate: Use numbers to find an answer
- State: Give a specific value or fact
- Explain: Give reasons or make clear
- Describe: Give a detailed account
CSEC-Style Practice Questions
Test Your Understanding
\(W = F \times d = 40 \times 5 = 200 \text{ J}\)
The work done is 200 joules.
\(KE = \frac{1}{2}mv^2 = \frac{1}{2} \times 800 \times (20)^2 = 400 \times 400 = 160000 \text{ J}\)
The kinetic energy is 160,000 J or 160 kJ.
Weight = \(mg = 60 \times 10 = 600 \text{ N}\)
Work done = \(F \times d = 600 \times 12 = 7200 \text{ J}\)
Power = \(\frac{W}{t} = \frac{7200}{15} = 480 \text{ W}\)
The power developed is 480 watts.
\(\text{Efficiency} = \frac{\text{Useful output}}{\text{Total input}} \times 100\% = \frac{1800}{2000} \times 100\% = 90\%\)
The efficiency is 90%.
Initial GPE = \(mgh = 0.5 \times 10 \times 10 = 50 \text{ J}\)
Just before hitting ground, all GPE converted to KE:
\(KE = 50 \text{ J} = \frac{1}{2}mv^2\)
\(50 = \frac{1}{2} \times 0.5 \times v^2\)
\(100 = 0.5 \times v^2\)
\(v^2 = 200\)
\(v = \sqrt{200} โ 14.14 \text{ m/s}\)
The speed is approximately 14.14 m/s.