CIE AS/A Level Maths-4.5 Energy, work and power- Study Notes- New Syllabus - 2026-2027
CIE AS/A Level Maths-4.5 Energy, work and power- Study Notes- New Syllabus
Ace AS/A Level Maths Exam with CIE AS/A Level Maths-4.5 Energy, work and power- Study Notes
Key Concepts:
- Work Done by a Force
- Gravitational Potential Energy and Kinetic Energy
- Work–Energy Principle and Principle of Conservation of Energy
- Power
- Instantaneous Acceleration on a Hill
Work Done by a Force
Work Done by a Force
The work done by a force is the product of the force applied and the displacement of its point of application in the direction of the force.
Formula:
\( W = Fd \cos \theta \)
- \( W \) = Work done (in joules, J)
- \( F \) = Magnitude of the constant force (in newtons, N)
- \( d \) = Displacement of the point of application (in metres, m)
- \( \theta \) = Angle between the force and the displacement vector
Notes:
- If the force is in the same direction as the displacement (\(\theta = 0^\circ\)), then \( W = Fd \).
- If the force is perpendicular to the displacement (\(\theta = 90^\circ\)), then \( W = 0 \) (no work is done).
- If the force is opposite to the displacement (\(\theta = 180^\circ\)), then \( W = -Fd \) (the force resists the motion).
- Work is a scalar quantity, though it arises from vectors.
Example:
A force of \(20 \, \text{N}\) acts at an angle of \(60^\circ\) to the direction of motion of a particle. The particle is displaced by \(5 \, \text{m}\). Find the work done by the force.
▶️ Answer/Explanation
Using the formula: \( W = Fd \cos \theta \)
Substitute: \( W = 20 \times 5 \times \cos 60^\circ \)
\( W = 100 \times 0.5 = 50 \, \text{J} \)
\(\boxed{50 \, \text{J}}\) of work is done by the force.
Example:
A box is dragged along a horizontal floor by a rope inclined at \(30^\circ\) above the horizontal. The tension in the rope is \(50 \, \text{N}\) and the box is moved \(10 \, \text{m}\) along the floor. Find the work done by the force of tension.
▶️ Answer/Explanation
Here, \( F = 50 \, \text{N}, \, d = 10 \, \text{m}, \, \theta = 30^\circ \).
Work done: \( W = Fd \cos \theta \)
\( W = 50 \times 10 \times \cos 30^\circ \)
\( W = 500 \times \dfrac{\sqrt{3}}{2} \)
\( W = 250\sqrt{3} \, \text{J} \approx 433 \, \text{J} \)
\(\boxed{433 \, \text{J}}\) of work is done by the tension force.
Gravitational Potential Energy and Kinetic Energy
Gravitational Potential Energy (GPE)
The energy possessed by a body due to its position in a gravitational field.
Formula:
\( E_p = mgh \)
- \( E_p \) = gravitational potential energy (J)
- \( m \) = mass of the object (kg)
- \( g \) = acceleration due to gravity (\( \approx 9.8 \, \text{m/s}^2 \) or \(10 \, \text{m/s}^2\) in simple problems)
- \( h \) = height above a reference level (m)
Notes:
- The choice of reference level (\(h=0\)) is arbitrary.
- Only changes in potential energy matter physically.
Kinetic Energy (KE)
The energy possessed by a body due to its motion.
Formula:
\( E_k = \dfrac{1}{2}mv^2 \)
- \( E_k \) = kinetic energy (J)
- \( m \) = mass of the object (kg)
- \( v \) = speed of the object (m/s)
Notes:
- Always positive (since \(v^2 \geq 0\)).
- If velocity doubles, kinetic energy increases by a factor of four.
Example:
A box of mass \(4 \, \text{kg}\) is lifted vertically upward through a height of \(2 \, \text{m}\). Find the increase in gravitational potential energy. Take \( g = 10 \, \text{m/s}^2 \).
▶️ Answer/Explanation
Using \( E_p = mgh \):
\( E_p = 4 \times 10 \times 2 \)
\( E_p = 80 \, \text{J} \)
\(\boxed{80 \, \text{J}}\) is the increase in gravitational potential energy.
Example:
A car of mass \(1000 \, \text{kg}\) is moving at a speed of \(20 \, \text{m/s}\). Calculate its kinetic energy.
▶️ Answer/Explanation
Using \( E_k = \dfrac{1}{2}mv^2 \):
\( E_k = \dfrac{1}{2} \times 1000 \times (20)^2 \)
\( E_k = 500 \times 400 \)
\( E_k = 200000 \, \text{J} \)
\(\boxed{2.0 \times 10^5 \, \text{J}}\) is the kinetic energy.
Work–Energy Principle and Principle of Conservation of Energy
Work–Energy Principle
Statement:
The net work done by external forces on a system equals the change in the total energy of the system (kinetic energy + potential energy).
Mathematical Form:
\( W = \Delta E = \Delta E_k + \Delta E_p \)
- \( W \) = work done by external forces (J)
- \( \Delta E_k \) = change in kinetic energy (J)
- \( \Delta E_p \) = change in potential energy (J)
Principle of Conservation of Energy
Statement:
Energy cannot be created or destroyed; it can only be transformed from one form to another. In an isolated system (no external work done), the total mechanical energy remains constant.
Mathematical Form:
\( E_k + E_p = \text{constant} \)
Notes:
- If no non-conservative forces (e.g. friction, air resistance) act, mechanical energy is conserved.
- If resistive forces are present, some mechanical energy is transformed into thermal energy (not lost, but no longer useful for motion).
- Applies to linear and curved motion (e.g. slides, swings, roller coasters).
Example:
A force of \(50 \, \text{N}\) is applied to push a \(10 \, \text{kg}\) box along a smooth horizontal floor for a distance of \(4 \, \text{m}\). Find the increase in the box’s kinetic energy.
▶️ Answer/Explanation
Work done: \( W = Fd \cos\theta \)
Here, \( \theta = 0^\circ \), so \( \cos\theta = 1 \).
\( W = 50 \times 4 = 200 \, \text{J} \)
By the work–energy principle, \( \Delta E_k = W = 200 \, \text{J} \).
\(\boxed{200 \, \text{J}}\) increase in kinetic energy.
Example:
A child of mass \(30 \, \text{kg}\) slides from rest down a smooth curved slide of vertical height \(2.5 \, \text{m}\). Find the child’s speed at the bottom. Take \( g = 10 \, \text{m/s}^2 \).
▶️ Answer/Explanation
Initial energy = GPE = \( mgh = 30 \times 10 \times 2.5 = 750 \, \text{J} \).
At bottom: all GPE converts to KE.
\( \dfrac{1}{2}mv^2 = 750 \)
\( \dfrac{1}{2} \times 30 \times v^2 = 750 \)
\( 15v^2 = 750 \)
\( v^2 = 50 \implies v = \sqrt{50} \approx 7.07 \, \text{m/s} \)
\(\boxed{7.1 \, \text{m/s}}\) at the bottom of the slide.
Power
Power
Power is the rate at which work is done, or the rate at which energy is transferred or transformed.
Mathematical Form:
\( P = \dfrac{W}{t} \)
\( P = \dfrac{F d}{t} \)
Since velocity is displacement per time:
\( P = F v \quad \text{(if force is in the direction of velocity)} \)
\( P = F v \cos\theta \quad \text{(if force is at an angle } \theta \text{ to velocity)} \)
- \( P \) = power (Watt, \( \text{W} \))
- \( W \) = work done (Joule, \( \text{J} \))
- \( F \) = force (Newton, \( \text{N} \))
- \( d \) = displacement (m)
- \( v \) = velocity (m/s)
- \( \theta \) = angle between force and velocity
Example:
A motor pulls a crate with a constant force of \(200 \, \text{N}\) at a constant velocity of \(4 \, \text{m/s}\). Calculate the power delivered by the motor.
▶️ Answer/Explanation
Power: \( P = Fv \)
\( P = 200 \times 4 = 800 \, \text{W} \)
\(\boxed{800 \, \text{W}}\)
Example:
A force of \(100 \, \text{N}\) pulls a cart at a speed of \(2 \, \text{m/s}\). The force makes an angle of \(60^\circ\) with the direction of motion. Find the power delivered.
▶️ Answer/Explanation
Formula: \( P = F v \cos\theta \)
\( P = 100 \times 2 \times \cos 60^\circ \)
\( P = 200 \times 0.5 = 100 \, \text{W} \)
\(\boxed{100 \, \text{W}}\)
Instantaneous Acceleration on a Hill
Instantaneous Acceleration on a Hill
A car moving on an incline experiences forces along the slope:
- Driving force from the engine: \( F_{\text{drive}} \)
- Resistive forces (friction + air resistance): \( F_{\text{resist}} \)
- Component of weight along slope: \( mg \sin\theta \)
The resultant force along the slope is:
\( F_{\text{net}} = F_{\text{drive}} – F_{\text{resist}} – mg \sin\theta \)
By Newton’s second law:
\( a = \dfrac{F_{\text{net}}}{m} = \dfrac{F_{\text{drive}} – F_{\text{resist}} – mg \sin\theta}{m} \)
Example:
A car of mass \(1200 \, \text{kg}\) is moving uphill on a slope inclined at \(5^\circ\). The engine provides a driving force of \(4000 \, \text{N}\), and resistive forces (friction + air resistance) total \(800 \, \text{N}\). Calculate the instantaneous acceleration of the car.
▶️ Answer/Explanation
Weight component down slope: \( mg \sin\theta = 1200 \times 9.8 \times \sin 5^\circ \)
\( = 11760 \times 0.0872 \approx 1026 \, \text{N} \)
Net force: \( F_{\text{net}} = 4000 – 800 – 1026 = 2174 \, \text{N} \)
Acceleration: \( a = \dfrac{2174}{1200} \approx 1.81 \, \text{m/s}^2 \)
\(\boxed{1.81 \, \text{m/s}^2}\)
Example:
A car of mass \(1000 \, \text{kg}\) moves downhill on a slope of \(10^\circ\). The engine is not working, and resistive forces total \(500 \, \text{N}\). Find its acceleration down the slope.
▶️ Answer/Explanation
Weight component down slope: \( mg \sin\theta = 1000 \times 9.8 \times \sin 10^\circ \)
\( = 9800 \times 0.1736 \approx 1701 \, \text{N} \)
Net force: \( F_{\text{net}} = 1701 – 500 = 1201 \, \text{N} \)
Acceleration: \( a = \dfrac{1201}{1000} = 1.20 \, \text{m/s}^2 \)
\(\boxed{1.20 \, \text{m/s}^2}\)