CIE AS & A Level Physics 5.1 Energy conservation Study Notes- 2025-2027 Syllabus
CIE AS & A Level Physics 4.3 Density and pressure Study Notes – New Syllabus
CIE AS & A Level Physics 5.1 Energy conservation Study Notes at IITian Academy focus on specific topic and type of questions asked in actual exam. Study Notes focus on AS/A Level Physics Study Notes syllabus with Candidates should be able to:
- understand the concept of work, and recall and use work done = force × displacement in the direction of the force
- recall and apply the principle of conservation of energy
- recall and understand that the efficiency of a system is the ratio of useful energy output from the system to the total energy input
- use the concept of efficiency to solve problems
- define power as work done per unit time
- solve problems using P = W / t
- derive P = Fv and use it to solve problems
Concept of Work
Work is done when a force causes an object to move in the direction of that force. It represents the energy transferred when a force moves its point of application through a distance.
\( \mathrm{W = F s \cos\theta} \)
- \( \mathrm{W} \): work done (J)
- \( \mathrm{F} \): applied force (N)
- \( \mathrm{s} \): displacement (m)
- \( \mathrm{\theta} \): angle between the force and the direction of motion
Special Cases:
- If the force acts in the same direction as displacement (\( \mathrm{\theta = 0^\circ} \)): \( \mathrm{W = F s} \)
- If the force acts opposite to displacement (\( \mathrm{\theta = 180^\circ} \)): \( \mathrm{W = -F s} \) (negative work, energy taken from object)
- If the force acts perpendicular to displacement (\( \mathrm{\theta = 90^\circ} \)): \( \mathrm{W = 0} \) (no work done, e.g., circular motion)
Unit: joule (J) \( \mathrm{1\,J = 1\,N·m = 1\,kg·m^2·s^{-2}.} \)
Work–Energy Relationship:
The net work done on an object equals the change in its kinetic energy.
\( \mathrm{W = \Delta E_k} \)
Example
A constant force of \( \mathrm{50\,N} \) moves a box through a distance of \( \mathrm{4.0\,m} \) horizontally. Calculate the work done if the force acts at an angle of \( \mathrm{30^\circ} \) above the horizontal.
▶️ Answer / Explanation
Work done is given by \( \mathrm{W = F s \cos\theta.} \)
\( \mathrm{W = 50 \times 4.0 \times \cos30^\circ = 200 \times 0.866 = 173\,J.} \)
The work done by the force is \( \mathrm{173\,J.} \)
Principle of Conservation of Energy
Statement:
Energy cannot be created or destroyed, but it can be transferred from one form to another or converted from one body to another. The total energy of an isolated system remains constant.
\( \mathrm{E_{total\ initial} = E_{total\ final}} \)
Explanation:
- When work is done, energy is transformed from one form to another (e.g. chemical → kinetic → thermal).
- Energy may be stored (potential), transferred (mechanically, electrically), or dissipated (heat, sound), but the total remains constant.
- In real systems, some energy becomes non-useful due to inefficiencies (e.g. friction, resistance).
Examples of Energy Conversions:
- Falling object: gravitational potential energy → kinetic energy.
- Electric motor: electrical energy → mechanical (kinetic) energy.
- Car engine: chemical energy → kinetic + heat energy.
Example
A ball of mass \( \mathrm{0.2\,kg} \) is dropped from a height of \( \mathrm{10\,m.} \) Ignoring air resistance, find the speed of the ball just before it hits the ground.
▶️ Answer / Explanation
By conservation of energy: \( \mathrm{mgh = \tfrac{1}{2}mv^2.} \)
Cancel \( \mathrm{m} \):
\( \mathrm{v = \sqrt{2gh} = \sqrt{2(9.8)(10)} = 14\,m/s.} \)
The ball hits the ground at \( \mathrm{14\,m/s.} \)
Efficiency of a System
Efficiency is the ratio of useful energy (or power) output to the total energy (or power) input, usually expressed as a percentage.
\( \mathrm{Efficiency = \dfrac{Useful\ Output}{Total\ Input}} \times 100\% \)
Alternate Forms:
- \( \mathrm{Efficiency = \dfrac{Useful\ Power\ Output}{Total\ Power\ Input} \times 100\%.} \)
- \( \mathrm{Efficiency = \dfrac{Useful\ Energy\ Output}{Total\ Energy\ Input} \times 100\%.} \)
Explanation:
- Efficiency measures how effectively a system converts input energy into useful output.
- It is always less than 100% in real systems due to energy losses (mainly as heat, sound, or friction).
- Efficiency is dimensionless but often expressed as a percentage.
Typical Efficiencies:
| Device | Useful Output | Typical Efficiency |
|---|---|---|
| Electric motor | Kinetic energy | 85–95% |
| Car engine | Kinetic energy | 25–35% |
| Light bulb (filament) | Light energy | 5–10% |
Example
An electric motor takes in \( \mathrm{2000\,J} \) of electrical energy and produces \( \mathrm{1500\,J} \) of mechanical (useful) work. Calculate its efficiency.
▶️ Answer / Explanation
Using \( \mathrm{Efficiency = \dfrac{Useful\ Output}{Total\ Input} \times 100\%.} \)
\( \mathrm{Efficiency = \dfrac{1500}{2000} \times 100 = 75\%.} \)
The motor operates at \( \mathrm{75\%} \) efficiency.
Using Efficiency in Problem Solving
Energy-Based Problems:
\( \mathrm{Useful\ Output = Efficiency \times Input.} \)
Power-Based Problems:
\( \mathrm{Useful\ Power = Efficiency \times Input\ Power.} \)
Example
A crane motor of efficiency \( \mathrm{80\%} \) lifts a \( \mathrm{500\,kg} \) load to a height of \( \mathrm{8.0\,m.} \) Calculate the electrical energy supplied to the motor. Take \( \mathrm{g = 9.8\,m/s^2.} \)
▶️ Answer / Explanation
Step 1: Work done in lifting the load (useful output):
\( \mathrm{E_{useful} = mgh = 500 \times 9.8 \times 8.0 = 3.92\times10^4\,J.} \)
Step 2: Total input energy:
\( \mathrm{E_{input} = \dfrac{E_{useful}}{\text{Efficiency}} = \dfrac{3.92\times10^4}{0.80} = 4.9\times10^4\,J.} \)
The motor requires \( \mathrm{4.9\times10^4\,J} \) of electrical energy to lift the load.
Definition of Power
Power is the rate at which work is done or energy is transferred.
\( \mathrm{P = \dfrac{W}{t}} \)
- \( \mathrm{P} \): power (W or joules per second, \( \mathrm{J·s^{-1}} \))
- \( \mathrm{W} \): work done or energy transferred (J)
- \( \mathrm{t} \): time taken (s)
Explanation:
- Power measures how quickly energy is used or work is done.
- It describes the rate of energy conversion or transfer.
- One watt (1 W) equals one joule of work done per second.
Alternative Expression:
Since \( \mathrm{W = F s} \), if the work is done uniformly over time, \( \mathrm{P = \dfrac{F s}{t}} \). As \( \mathrm{\dfrac{s}{t} = v} \), this gives \( \mathrm{P = Fv} \) for constant speed — derived below.
Example
A motor does \( \mathrm{6.0\times10^4\,J} \) of work in \( \mathrm{30\,s.} \) Calculate the power output of the motor.
▶️ Answer / Explanation
Using \( \mathrm{P = \dfrac{W}{t}} \):
\( \mathrm{P = \dfrac{6.0\times10^4}{30} = 2.0\times10^3\,W = 2.0\,kW.} \)
The motor’s power output is \( \mathrm{2.0\,kW.} \)
Solving Problems Using ( \mathrm{P = \dfrac{W}{t}} )
Rearrangements:
- \( \mathrm{W = P t} \) → to find work done or energy transferred.
- \( \mathrm{t = \dfrac{W}{P}} \) → to find time taken for given work and power.
Graphical Meaning:
- In a work–time graph, the slope represents power.
- In a power–time graph, the area under the curve gives total work or energy transferred.
Example
A crane lifts a load of \( \mathrm{600\,kg} \) vertically through a height of \( \mathrm{8.0\,m} \) in \( \mathrm{12\,s.} \) Calculate the average power developed by the crane. Take \( \mathrm{g = 9.8\,m/s^2.} \)
▶️ Answer / Explanation
Step 1: Work done = gain in gravitational potential energy:
\( \mathrm{W = mgh = 600 \times 9.8 \times 8.0 = 4.7\times10^4\,J.} \)
Step 2: Power = \( \mathrm{\dfrac{W}{t}} \):
\( \mathrm{P = \dfrac{4.7\times10^4}{12} = 3.9\times10^3\,W = 3.9\,kW.} \)
The crane develops an average power of \( \mathrm{3.9\,kW.} \)
Example
A car engine provides a constant driving force of \( \mathrm{2500\,N} \) when travelling at \( \mathrm{20\,m/s.} \) Calculate the power developed by the engine.
▶️ Answer / Explanation
Power = \( \mathrm{Fv} \)
\( \mathrm{P = 2500 \times 20 = 5.0\times10^4\,W = 50\,kW.} \)
The engine develops \( \mathrm{50\,kW} \) of power at that speed.
Example
An electric heater rated at \( \mathrm{2.0\,kW} \) transfers \( \mathrm{3.6\times10^6\,J} \) of energy. Calculate how long the heater operates for.
▶️ Answer / Explanation
Using \( \mathrm{t = \dfrac{W}{P}} \):
\( \mathrm{t = \dfrac{3.6\times10^6}{2.0\times10^3} = 1800\,s = 30\,min.} \)
The heater operates for \( \mathrm{30\,minutes.} \)
Derivation of \( \mathrm{P=Fv}\)
To relate power to the applied force and the velocity of the moving object.
Derivation:
Work done by a constant force is given by:
\( \mathrm{W = F s.} \)
Average power is work done per unit time:
\( \mathrm{P = \dfrac{W}{t} = \dfrac{F s}{t}.} \)
Since \( \mathrm{\dfrac{s}{t} = v,} \) the velocity of the object:
\( \mathrm{P = Fv.} \)
Vector Form:
If the force and velocity are not in the same direction:
\( \mathrm{P = \vec{F} \cdot \vec{v} = Fv\cos\theta.} \)
- \( \mathrm{\theta} \): angle between the force and velocity directions.
- If \( \mathrm{\theta = 0^\circ} \), \( \mathrm{P = Fv.} \)
- If \( \mathrm{\theta = 90^\circ} \), \( \mathrm{P = 0} \) (no work done).
Interpretation:
- Power represents the rate of doing mechanical work.
- For constant velocity motion, \( \mathrm{F} \) equals the resistive force (e.g. friction, drag).
- A higher velocity at the same force → greater power requirement.
Example
A car experiences a total resistive force of \( \mathrm{600\,N} \) when travelling at a steady speed of \( \mathrm{25\,m/s.} \) Calculate the power developed by the engine to maintain this speed.
▶️ Answer / Explanation
Using \( \mathrm{P = Fv} \):
\( \mathrm{P = 600 \times 25 = 1.5\times10^4\,W.} \)
The engine must provide \( \mathrm{15\,kW} \) of power to maintain the speed.
Instantaneous Power:
At any instant, \( \mathrm{P = Fv} \) gives the rate of work done if both \( \mathrm{F} \) and \( \mathrm{v} \) vary with time.
Power in Lifting:
When lifting an object vertically at constant speed:
\( \mathrm{P = mgv} \)
where \( \mathrm{v} \) is the upward speed.
Power in Rotational Motion:
In a rotating system:
\( \mathrm{P = \tau \omega} \)
- \( \mathrm{\tau} \): torque (N·m)
- \( \mathrm{\omega} \): angular velocity (rad·s⁻¹)
Example
A crane lifts a \( \mathrm{200\,kg} \) load vertically at a constant speed of \( \mathrm{0.5\,m/s.} \) Find the power output of the motor. Take \( \mathrm{g = 9.8\,m/s^2.} \)
▶️ Answer / Explanation
Force required = weight = \( \mathrm{mg = 200 \times 9.8 = 1960\,N.} \)
Using \( \mathrm{P = Fv} \):
\( \mathrm{P = 1960 \times 0.5 = 980\,W.} \)
The crane motor develops \( \mathrm{980\,W} \) of power (≈ 1.0 kW).