Energy methods offer a powerful alternative to Newton's Second Law, especially when you need to find speeds after a height change or over a displacement with friction. This topic links directly to every dynamics question in Cambridge A-Level 9709 Mechanics 1.
W = Fd cosθ | KE = ½mv² | GPE = mgh | P = Fv | Wnet = ΔKE
Learning Objectives
Calculate work done by a force: W = Fd cosθ, identifying the angle correctly
Find work done by gravity, tension, friction and applied forces
Calculate kinetic energy KE = ½mv² and gravitational potential energy GPE = mgh
Apply conservation of mechanical energy on smooth surfaces and slopes
Use the work-energy theorem: W_net = ΔKE = ½mv² − ½mu²
Account for energy lost to friction: KE₁ + GPE₁ = KE₂ + GPE₂ + Fd (friction)
Calculate power as P = W/t and P = Fv (instantaneous power)
Determine the maximum speed of a vehicle using P = Fv and equilibrium of forces
Use energy methods as an efficient alternative to SUVAT for speed problems
Convert correctly between J, kJ, W and kW
Work Done
W = Fd cosθ, positive and negative work, work by gravity and friction
KE & GPE
½mv², mgh, units, change in energy quantities
Conservation
Mechanical energy constant on smooth surfaces; friction removes energy
Work-Energy Theorem
Net work = change in KE; powerful problem-solving tool
Power
P = W/t = Fv; maximum speed; engine power problems
Worked Examples
8 fully worked with mark-scheme steps
Common Mistakes
8 errors that regularly cost marks in exams
Exercises + Practice
80+ self-marking questions across all topics
Learn 1 — Work Done by a Force
Definition of Work Done
Work is done when a force causes (or acts in the direction of) a displacement. The unit is the Joule (J), where 1 J = 1 N m.
W = Fd cosθ
Where:
• F = magnitude of the force (N)
• d = displacement of the object (m)
• θ = angle between the force vector and the displacement vector
Special cases:
• θ = 0° (force parallel to displacement): W = Fd (maximum work)
• θ = 90° (force perpendicular to displacement): W = 0 (no work done)
• θ = 180° (force opposes displacement): W = −Fd (negative work done)
Work Done by Gravity
Gravity acts vertically downward. When an object moves through a vertical height h:
Work done by gravity = mgh (when object descends height h)
• Object descends (displacement has downward component): gravity does positive work = mgh.
• Object ascends (displacement has upward component): gravity does negative work = −mgh.
• Horizontal motion: gravity does zero work (perpendicular to displacement).
Example: A 3 kg ball falls 5 m. Work done by gravity = 3 × 10 × 5 = 150 J.
Work Done Against Friction
Friction always opposes motion, so it always does negative work on the moving object.
Work done against friction = Fd (always positive value representing energy removed)
• Friction force F = μN acts opposite to displacement.
• Work done BY friction = −Fd (negative, as friction opposes motion).
• Work done AGAINST friction = +Fd (energy removed from the mechanical system).
• This energy is converted to heat — it is lost from the mechanical system permanently.
Example: A block slides 4 m with friction force 12 N. Work done against friction = 12 × 4 = 48 J.
Work Done by an Applied Force at an Angle
If a force P is applied at angle α above the horizontal to move a particle horizontally by distance d:
W = Pd cosα
Example: Force 20 N at 40° above horizontal, displacement 6 m horizontally.
W = 20 × 6 × cos40° = 120 × 0.766 = 91.9 J
Work Done by String Tension
A string under tension T does work when it pulls a particle through a displacement d along its direction.
W = Td (when tension and displacement are in same direction).
In pulley systems: the tension does positive work on one particle (pulling it up) and negative work on the other (as that particle moves in the opposite direction to the tension). The net work done by an internal tension in a system is zero if the string is inextensible.
Positive and Negative Work — Summary
Force
Situation
Work Done
Applied force P
In direction of motion
+Pd cosθ
Gravity mg
Object descends h
+mgh
Gravity mg
Object ascends h
−mgh
Friction F
Always opposes motion
−Fd (work by friction) / +Fd (against friction)
Normal reaction N
Perpendicular to surface
0 (always)
Tension T
Pulling particle along string
+Td
The work done by a force only depends on the component of force in the direction of displacement. Normal reaction and centripetal forces always do zero work because they are always perpendicular to displacement.
Learn 2 — Kinetic Energy and Gravitational Potential Energy
Kinetic Energy (KE)
The kinetic energy of a moving particle of mass m with speed v:
KE = ½mv²
• Units: Joules (J) — since kg × (m s⁻¹)² = kg m² s⁻² = J
• KE is always ≥ 0 (speed is never negative)
• KE depends on speed squared: doubling the speed quadruples the KE
• KE depends on mass linearly: doubling mass doubles KE
Example: A 4 kg particle moving at 6 m s⁻¹.
KE = ½ × 4 × 6² = ½ × 4 × 36 = 72 J
Change in Kinetic Energy
ΔKE = ½mv² − ½mu²
• u = initial speed, v = final speed
• ΔKE > 0: particle speeded up (energy gained)
• ΔKE < 0: particle slowed down (energy lost)
Example: Particle accelerates from 3 m s⁻¹ to 7 m s⁻¹, mass 2 kg.
ΔKE = ½ × 2 × 7² − ½ × 2 × 3² = 49 − 9 = 40 J
Finding Speed from KE
Rearrange KE = ½mv²:
v² = 2 × KE / m
v = √(2 × KE / m)
Example: KE = 200 J, m = 8 kg.
v² = 2 × 200 / 8 = 50 → v = √50 = 7.07 m s⁻¹
Gravitational Potential Energy (GPE)
The gravitational potential energy of a particle of mass m at height h above a chosen reference level:
GPE = mgh
• Units: Joules (J) — since kg × m s⁻² × m = J
• h is measured from a chosen reference level (you decide where h = 0)
• GPE is relative — only changes in GPE matter, not absolute values
• GPE increases when the object moves up, decreases when it moves down
Example: A 5 kg mass raised 3 m above the ground.
GPE = 5 × 10 × 3 = 150 J
Change in GPE
ΔGPE = mgh₂ − mgh₁ = mg(h₂ − h₁) = mgΔh
• Δh = h₂ − h₁ (positive if object moves up, negative if it moves down)
• ΔGPE > 0: object moved up (GPE increased)
• ΔGPE < 0: object moved down (GPE decreased — that energy went elsewhere)
Example: A 6 kg particle descends from height 8 m to height 3 m. g = 10 m s⁻².
ΔGPE = 6 × 10 × (3 − 8) = 60 × (−5) = −300 J (GPE decreased by 300 J)
KE from Falling Height h (Smooth Case)
If a particle falls freely from rest through height h, all GPE converts to KE:
mgh = ½mv² → v = √(2gh)
This derivation uses conservation of energy (next topic). Notice mass cancels — the speed of a freely falling object is independent of its mass.
Example: Ball released from rest at height 5 m. g = 10 m s⁻².
v = √(2 × 10 × 5) = √100 = 10 m s⁻¹
Always choose your reference level clearly (usually the lowest point in the problem). Label it on your diagram. GPE = 0 at the reference level; GPE is positive above it and may be negative below it.
Learn 3 — Conservation of Energy
Principle of Conservation of Mechanical Energy
In the absence of friction and air resistance (smooth/conservative system), the total mechanical energy of a particle is constant throughout its motion.
KE₁ + GPE₁ = KE₂ + GPE₂ (smooth surfaces only)
What this means:
• Total mechanical energy = KE + GPE = constant when no friction.
• Energy converts between KE and GPE but none is lost.
• When particle descends: GPE decreases → KE increases (speeds up).
• When particle ascends: GPE increases → KE decreases (slows down).
Example — Ball rolling down smooth slope:
m = 2 kg, starts from rest at h = 4 m. Find speed at bottom (h = 0). g = 10 m s⁻².
KE₁ + GPE₁ = KE₂ + GPE₂
0 + 2 × 10 × 4 = ½ × 2 × v² + 0
80 = v² → v = √80 = 8.94 m s⁻¹
Method: Using Conservation of Energy
Step 1: Identify two positions: initial (subscript 1) and final (subscript 2). Step 2: Choose a reference level for h (often the lowest point). Step 3: Write KE₁ + GPE₁ = KE₂ + GPE₂. Step 4: Substitute known values. Solve for the unknown (usually v). Note: If the particle starts from rest, KE₁ = 0. If it ends momentarily at rest, KE₂ = 0.
When Friction is Present
Friction converts mechanical energy to heat energy. The mechanical energy at the end is less than at the start.
KE₁ + GPE₁ = KE₂ + GPE₂ + W_friction
Where: W_friction = work done against friction = friction force × distance = Fd
• The total energy equation now has an "energy lost" term on the right.
• W_friction is always positive (always removes energy from the mechanical system).
• This applies to any path — on a slope, on a horizontal surface, or a combination.
Example — Rough slope:
m = 3 kg, starts from rest at h = 5 m, slides down a 10 m long slope. Friction force = 6 N. g = 10 m s⁻².
KE₁ + GPE₁ = KE₂ + GPE₂ + W_f
0 + 3 × 10 × 5 = ½ × 3 × v² + 0 + 6 × 10
150 = 1.5v² + 60
1.5v² = 90 → v² = 60 → v = 7.75 m s⁻¹
Energy Lost to Friction
Energy lost = W_friction = F_friction × d (distance slid)
This is also equal to the loss in mechanical energy:
Energy lost = (KE₁ + GPE₁) − (KE₂ + GPE₂)
Example: Particle has initial ME = 500 J and final ME = 320 J.
Energy lost to friction = 500 − 320 = 180 J
Conservation of Energy vs SUVAT
When energy is quicker: When you need a speed after a height change on a slope or after falling, energy is much faster than SUVAT. You don't need the angle or acceleration — just the height.
When SUVAT is needed: When the question asks for time, or when a force varies with distance (not covered in M1), use SUVAT or F = ma.
Key insight: On a smooth slope, the speed at the bottom depends only on the vertical height — not the slope angle or path taken.
Always check whether the surface is smooth (no friction) or rough (friction). If it says "smooth" or "frictionless", use KE₁ + GPE₁ = KE₂ + GPE₂. If it's rough, you must include W_friction on the right-hand side.
Learn 4 — The Work-Energy Theorem
Statement of the Theorem
The net work done on a particle by all forces equals the change in kinetic energy of that particle.
W_net = ΔKE = ½mv² − ½mu²
Where:
• W_net = algebraic sum of work done by ALL forces on the particle
• u = initial speed, v = final speed
• ΔKE = change in kinetic energy
This theorem holds in all situations — smooth or rough, horizontal or on a slope. It is a more general result than conservation of mechanical energy.
Applying the Work-Energy Theorem
Step 1: List every force acting on the particle. Step 2: Calculate the work done by each force over the displacement d. Step 3: Sum the work done: W_net = W_applied + W_gravity + W_friction + W_tension + … Step 4: Set W_net = ½mv² − ½mu² and solve.
Sign convention: Work done by a force in the direction of motion is positive; against motion is negative.
Horizontal Surface Example (Rough)
m = 5 kg, u = 4 m s⁻¹. Applied force P = 30 N, friction = 10 N, distance d = 6 m.
Work done by P = +30 × 6 = +180 J
Work done by friction = −10 × 6 = −60 J
Work done by N (normal) = 0 J (perpendicular to motion)
Work done by gravity = 0 J (perpendicular to motion on horizontal surface)
On a slope, gravity does work = mgh (h = vertical height change).
The work-energy theorem on a rough slope of length L and height h:
mgh − F_friction × L = ½mv² − ½mu²
This is exactly equivalent to the energy conservation equation with friction:
KE₁ + GPE₁ = KE₂ + GPE₂ + F_friction × L
They are the same equation rearranged. Choose whichever form is more convenient.
Energy Methods vs SUVAT — When to Choose
Task
Better Method
Why
Find speed after descending height h (smooth)
Energy
One equation, no angles needed
Find speed after sliding distance d with friction
Work-Energy Theorem
Direct calculation
Find time of motion
SUVAT (F = ma first)
Energy methods don't give time
Find acceleration
F = ma
Energy gives speed not acceleration
Find distance to stop given initial speed and friction
Work-Energy Theorem
½mu² = Fd → d = mu²/(2F)
Finding Distance to Stop
Particle of mass m, initial speed u, friction force F, decelerates to rest on a horizontal surface.
Work-Energy Theorem: W_friction = ΔKE
−F × d = 0 − ½mu² → F × d = ½mu² d = mu² / (2F) = mu² / (2μmg) = u² / (2μg)
Example: u = 12 m s⁻¹, μ = 0.4, g = 10 m s⁻².
d = 144 / (2 × 0.4 × 10) = 144/8 = 18 m
The work-energy theorem is the most versatile tool in M1. When you see a question involving a force, a displacement, and a speed — think "work-energy theorem" first.
Learn 5 — Power
Definition of Power
Power is the rate at which work is done (or the rate of energy transfer).
P = W / t (average power) | P = Fv (instantaneous power)
• Units: Watts (W), where 1 W = 1 J s⁻¹
• Also common: kilowatts (kW), where 1 kW = 1000 W
• P = W/t gives average power over time t
• P = Fv gives the instantaneous power of a force F when the object moves with speed v (F and v in the same direction)
Example 1: A machine does 6000 J of work in 20 s. Power = 6000/20 = 300 W.
Example 2: An engine exerts a driving force of 2000 N at a speed of 15 m s⁻¹.
Power = 2000 × 15 = 30 000 W = 30 kW.
Deriving P = Fv
P = W / t = (F × d) / t = F × (d/t) = F × v
This assumes F and v are in the same direction (angle between them = 0°).
More generally: P = Fv cosθ, but in M1, the driving force and velocity are along the same direction.
Engine Power Problems
A vehicle engine provides a constant power output P. The driving force varies with speed:
F_drive = P / v
As the vehicle speeds up, v increases, so F_drive decreases.
The net force = F_drive − R (where R is the total resistance).
The acceleration decreases as speed increases.
Example: Car, engine power P = 24 kW = 24 000 W, resistance R = 800 N. Find the acceleration at v = 20 m s⁻¹, m = 1200 kg.
F_drive = 24 000 / 20 = 1200 N
Net force = 1200 − 800 = 400 N
a = 400 / 1200 = 0.333 m s⁻²
Maximum Speed (Terminal Velocity)
A vehicle reaches maximum speed when the driving force exactly equals the resistance force (net force = 0, so acceleration = 0).
At maximum speed: F_drive = Resistance → P / v_max = R → v_max = P / R
Example: Engine power P = 40 kW = 40 000 W. Resistance R = 1600 N. Find maximum speed.
v_max = P / R = 40 000 / 1600 = 25 m s⁻¹
Note: At maximum speed, acceleration = 0. This is the equilibrium condition — NOT that the forces are zero, but that the net force is zero.
Power on a Slope
On a slope, the driving force must overcome both gravity (component down the slope) and road friction/resistance.
For steady speed up a slope of angle θ:
F_drive = mg sinθ + R (net force = 0)
P = F_drive × v = (mg sinθ + R) × v
Example: Car mass 1000 kg, slope 30°, resistance 200 N, speed 10 m s⁻¹ (constant), g = 10 m s⁻².
F_drive = 1000 × 10 × sin30° + 200 = 5000 + 200 = 5200 N
P = 5200 × 10 = 52 000 W = 52 kW
Unit Conversions
Quantity
Unit
Conversion
Work / Energy
Joule (J)
1 kJ = 1000 J
Power
Watt (W)
1 kW = 1000 W
Speed
m s⁻¹
1 km/h = 1/3.6 m s⁻¹ ≈ 0.278 m s⁻¹
Common error: mixing kW and W. If power is given in kW, multiply by 1000 to get W before using P = Fv. Similarly, if energy is given in kJ, multiply by 1000 before calculating time.
Worked Examples
8 fully worked examples covering all key Energy, Work and Power topics. Study each step carefully — mark allocations are indicated.
Example 1 — Work Done by an Applied Force
A force of 50 N is applied at 30° above the horizontal to a 4 kg block. The block moves 8 m along a horizontal surface. Friction force is 15 N. Find the work done by each force and the net work done.
Work done by applied force F: W_F = 50 × 8 × cos30° = 400 × 0.866 = 346.4 JM1 A1
Work done by friction: W_friction = −15 × 8 = −120 J (opposes motion) M1 A1
Work done by gravity and normal reaction: Both are perpendicular to horizontal displacement → W_gravity = W_N = 0 JB1
Net work done: W_net = 346.4 − 120 + 0 + 0 = 226.4 JA1
Example 2 — Kinetic Energy Calculation
A particle of mass 3 kg accelerates from 4 m s⁻¹ to 10 m s⁻¹. Find the change in kinetic energy.
Initial KE: KE₁ = ½ × 3 × 4² = ½ × 3 × 16 = 24 JM1
Final KE: KE₂ = ½ × 3 × 10² = ½ × 3 × 100 = 150 JA1
A particle of mass 5 kg slides up a smooth slope, travelling 6 m along the slope. The slope is inclined at 25° to the horizontal. Find the change in GPE and the speed lost if initial speed is 8 m s⁻¹.
Vertical height gained: h = 6 × sin25° = 6 × 0.4226 = 2.535 m M1
Note: Slope angle is irrelevant — only vertical height matters for smooth slopes. B1
Example 5 — Work-Energy Theorem with Friction
A 6 kg particle slides down a rough slope of length 10 m inclined at 20°. The coefficient of friction is 0.3. Initial speed at top is 2 m s⁻¹. g = 10 m s⁻².
Height lost: h = 10 × sin20° = 10 × 0.342 = 3.42 m B1
Normal reaction and friction: N = mg cos20° = 6 × 10 × 0.940 = 56.4 N F_friction = μN = 0.3 × 56.4 = 16.92 N M1 A1
Solve: 217.2 − 169.2 = 3v² 48 = 3v² → v² = 16 → v = 4 m s⁻¹A1 A1
Example 6 — Maximum Speed of a Car
A car of mass 1500 kg has engine power 45 kW. Road resistance is 1800 N. Find the maximum speed of the car on a horizontal road. g = 10 m s⁻².
At maximum speed, net force = 0: Driving force = Resistance P / v_max = R M1
Convert power to Watts: P = 45 kW = 45 000 W B1
Solve for v_max: v_max = P / R = 45 000 / 1800 = 25 m s⁻¹A1
Example 7 — Power of an Engine at a Given Speed
A lorry of mass 8000 kg travels at 12 m s⁻¹ at constant speed up a slope inclined at sin⁻¹(1/20) to the horizontal. Road resistance is 500 N. Find the engine power. g = 10 m s⁻².
At constant speed, net force = 0: F_drive = mg sinθ + R M1
Power: P = F_drive × v = 4500 × 12 = 54 000 W = 54 kWM1 A1
Example 8 — Particle Sliding Down Rough Slope, Find Speed
A particle of mass 4 kg is released from rest at the top of a rough slope 5 m long, inclined at 35° to the horizontal. The friction force is 8 N. Find the speed at the bottom. g = 10 m s⁻².
Vertical height: h = 5 × sin35° = 5 × 0.5736 = 2.868 m M1
Solve: 114.72 = 2v² + 40 2v² = 74.72 → v² = 37.36 → v = 6.11 m s⁻¹A1 A1
Common Mistakes
These 8 errors appear regularly in Cambridge M1 scripts. Learn to avoid them.
Mistake 1 — Forgetting cosθ in the Work Formula
Wrong: W = Fd (when force is at an angle θ to displacement)
Right: W = Fd cosθ — only the component of force parallel to displacement does work. If F = 40 N at θ = 50° over d = 5 m: W = 40 × 5 × cos50° = 128.6 J, not 200 J.
Mistake 2 — Using the Wrong Reference Level for GPE
Wrong: Measuring h from different points on each side of the equation, or taking h as negative "because it's below".
Right: Choose ONE reference level at the start (e.g., the lowest point in the problem). All heights are measured upward from that level. Only ΔGP = mg(h₂ − h₁) matters — absolute values are irrelevant as long as you're consistent.
Mistake 3 — Wrong Sign for Work Done Against Friction
Wrong: Writing KE₁ + GPE₁ + W_friction = KE₂ + GPE₂ (adding W_friction to the left side)
Right: KE₁ + GPE₁ = KE₂ + GPE₂ + W_friction — Energy lost to friction reduces what's available, so it appears on the right (output) side. Alternatively, W_friction = Fd is always a positive quantity representing energy removed.
Mistake 4 — Maximum Speed Condition Wrongly Stated
Wrong: "Maximum speed occurs when the engine force is maximum" or "when power is greatest".
Right: Maximum speed occurs when acceleration = 0, which means the net force = 0, which means driving force = resistance. At this point: P = F_drive × v_max = R × v_max, so v_max = P/R.
Mistake 5 — Mixing kJ and J (or kW and W)
Wrong: Using P = 30 kW directly in P = Fv to get F = 30/15 = 2 N (instead of 2000 N).
Right: Always convert to SI units first. P = 30 kW = 30 000 W. Then F = 30 000/15 = 2000 N. Similarly, 5 kJ = 5000 J — never substitute kJ directly into formulas that expect J.
Wrong: On a slope of length L at angle θ, writing GPE = mgL (using slope length instead of height).
Right: GPE = mgh where h = L sinθ (the vertical height). GPE depends on vertical height only, not the distance along the slope. Always convert: h = L sinθ.
Mistake 7 — Claiming Normal Reaction Does Work
Wrong: Including N × d as work done by normal reaction in the energy equation.
Right: The normal reaction is always perpendicular to the surface (and hence to the displacement along the surface). Therefore cos90° = 0, and the normal reaction does zero work. Never include it in your work-energy equation.
Mistake 8 — Using Energy Methods When Time is Required
Wrong: Trying to find the time of motion using energy methods (e.g., t = W/P gives time only if P is constant and W is total work — this often leads to errors).
Right: Energy methods efficiently find speeds and distances. For time-based questions, use F = ma to find acceleration, then apply SUVAT (v = u + at or s = ut + ½at²). Energy and SUVAT are complementary tools — use each where it is strongest.
Key Formulas — Energy, Work and Power
Formula
Name
Notes
W = Fd cosθ
Work Done by Force
θ = angle between F and displacement. W = Fd when θ = 0°. W = 0 when θ = 90°.
W = mgh
Work Done by Gravity (descent h)
Positive when object descends. Negative when ascending.
KE = ½mv²
Kinetic Energy
Always ≥ 0. Units: Joules (J).
ΔKE = ½mv² − ½mu²
Change in Kinetic Energy
u = initial speed, v = final speed.
GPE = mgh
Gravitational Potential Energy
h measured from chosen reference level. Units: Joules (J).
ΔGPE = mgΔh
Change in GPE
Δh positive upward. ΔGPE > 0 means object moved up.
KE₁ + GPE₁ = KE₂ + GPE₂
Conservation of Mechanical Energy
Only valid when no friction or air resistance (smooth surfaces).
KE₁ + GPE₁ = KE₂ + GPE₂ + Fd
Energy Conservation with Friction
Fd = friction force × distance slid. Energy removed by friction.
W_net = ΔKE = ½mv² − ½mu²
Work-Energy Theorem
Net work done by ALL forces = change in KE. Universal — works with friction, on slopes, etc.
d = u²/(2μg)
Stopping Distance (friction only)
From work-energy theorem: Fd = ½mu².
P = W / t
Average Power
W = work done (J), t = time (s), P in Watts (W).
P = Fv
Instantaneous Power
F = force in direction of motion (N), v = speed (m s⁻¹).
F_drive = P / v
Driving Force from Power
Rearrangement of P = Fv.
v_max = P / R
Maximum Speed
At maximum speed: F_drive = R (resistance), so P/v_max = R.
v = √(2gh)
Speed after Falling Height h
Particle starts from rest. Smooth (no friction). From conservation of energy.
These formulas cover the entire Energy, Work and Power section of Cambridge A-Level 9709 M1. Memorise them all and practise applying them in the correct contexts.
Proof Bank
Two key derivations that are sometimes asked for in Cambridge examinations, and that deepen your understanding of why the formulas work.
Proof 1 — Deriving the Work-Energy Theorem from F = ma
Given: A constant resultant force F acts on a particle of mass m, causing acceleration a. The particle moves a displacement s, with initial speed u and final speed v.
Step 1: From Newton's Second Law: F = ma Step 2: From SUVAT: v² = u² + 2as → as = (v² − u²)/2 Step 3: Multiply both sides of F = ma by s:
Fs = mas = m × (v² − u²)/2 = ½mv² − ½mu² Step 4: The left side Fs = W (work done by net force). The right side = ΔKE. Result: W_net = ΔKE = ½mv² − ½mu² □
Note: This proof assumes constant force and constant acceleration. The result is valid more generally (varying force) using calculus, but in M1 we deal with constant forces.
Proof 2 — Conservation of Mechanical Energy from the Work-Energy Theorem
Setting: A particle of mass m moves under gravity only (smooth surface, no friction). It starts at height h₁ with speed u and ends at height h₂ with speed v.
Step 1: The only force doing work is gravity (normal reaction is perpendicular → does zero work). Step 2: Work done by gravity = mg(h₁ − h₂) [positive when descending, i.e. h₁ > h₂] Step 3: Apply work-energy theorem:
mg(h₁ − h₂) = ½mv² − ½mu² Step 4: Rearrange:
mgh₁ − mgh₂ = ½mv² − ½mu²
mgh₁ + ½mu² = ½mv² + mgh₂
GPE₁ + KE₁ = KE₂ + GPE₂ Result: Total mechanical energy is conserved. □
When friction acts, the friction force does negative work = −Fd. Then: mg(h₁ − h₂) − Fd = ½mv² − ½mu², which rearranges to KE₁ + GPE₁ = KE₂ + GPE₂ + Fd. Energy is not conserved — Fd of energy is converted to heat.
Proof 3 — Deriving P = Fv from P = W/t
Step 1: Power is defined as rate of work done: P = W/t Step 2: Work done by constant force F over displacement d: W = Fd Step 3: Substitute: P = Fd/t Step 4: Speed v = d/t (for constant speed), so d/t = v. Result: P = Fv □
This holds exactly when force and velocity are in the same direction. More generally, P = F · v = Fv cosθ where θ is the angle between force and velocity.
Energy Bar Chart Visualiser
Enter the mass, height and speed of a particle to see its kinetic energy (KE) and gravitational potential energy (GPE) displayed as a bar chart. The total mechanical energy is also shown.
KE
0 J
GPE
0 J
Total ME
0 J
Kinetic Energy (KE)
Gravitational PE (GPE)
Enter values and click Draw to see the energy breakdown.
Energy Conservation Slider — Falling Object
Drag the height slider to see how KE and GPE trade off as a particle falls from height H. All energy is conserved (smooth).
10.0 m
KE: 0 J
GPE: 300 J
Total ME: 300 J
Speed: 0 m/s
Exercise 1 — Work Done Calculations (10 questions)
Use W = Fd cosθ, W = mgh (gravity), W = Fd (friction). Give answers in J to 2 decimal places where needed. g = 10 m/s².
Exercise 2 — Kinetic Energy and GPE (10 questions)
Use KE = ½mv² and GPE = mgh. Give answers in J (or m/s for speed). g = 10 m/s².
Exercise 3 — Conservation of Energy, Smooth Surfaces (10 questions)
Use KE₁ + GPE₁ = KE₂ + GPE₂. No friction. g = 10 m/s². Answers as speeds (m/s) to 2dp unless stated.
Exercise 4 — Work-Energy Theorem with Friction (10 questions)
Use KE₁ + GPE₁ = KE₂ + GPE₂ + Fd or W_net = ΔKE. g = 10 m/s². Answers to 2dp where needed.
Exercise 5 — Power and Maximum Speed (10 questions)
Use P = W/t, P = Fv, v_max = P/R. Convert kW to W first. g = 10 m/s².
Practice — 30 Mixed Questions
All topics mixed. g = 10 m/s² unless stated. Answers to 2dp where needed. Score 100% for confetti!
Challenge — 15 Harder Questions
Harder multi-step questions. g = 10 m/s². Answers to 2dp. Score 100% for confetti!
Exam Style Questions (8 Questions)
Cambridge A-Level 9709 M1 style questions on Energy, Work and Power. Attempt fully before revealing the mark scheme.
Q1 [4 marks]
A particle of mass 3 kg is pulled along a smooth horizontal surface by a force of 20 N at 40° above the horizontal. The particle moves 5 m. Find the work done by the applied force and the work done by the normal reaction.
Work done by applied force = Fd cosθ = 20 × 5 × cos40° [M1]
= 100 × 0.766 = 76.6 J [A1]
Normal reaction is perpendicular to displacement (horizontal surface). [M1]
Work done by normal reaction = 0 J [A1]
Q2 [5 marks]
A particle of mass 4 kg slides from rest down a smooth slope of length 8 m inclined at 30° to the horizontal. Find the speed of the particle at the bottom of the slope. (g = 10 m s⁻²)
Vertical height: h = 8 × sin30° = 8 × 0.5 = 4 m [B1]
Conservation of energy (smooth surface): [M1]
0 + mgh = ½mv² + 0 → 4 × 10 × 4 = ½ × 4 × v² [M1]
160 = 2v² → v² = 80 → v = √80 = 8.94 m s⁻¹ [A1 A1]
Q3 [6 marks]
A particle of mass 5 kg slides down a rough slope of length 10 m inclined at 25° to the horizontal. The coefficient of friction between the particle and the slope is 0.2. The particle starts from rest. Find the speed of the particle at the foot of the slope. (g = 10 m s⁻²)
h = 10 × sin25° = 4.226 m [B1]
N = mg cos25° = 5 × 10 × 0.906 = 45.3 N [M1]
Friction = μN = 0.2 × 45.3 = 9.06 N. Work done against friction = 9.06 × 10 = 90.6 J [M1 A1]
Energy equation: mgh = ½mv² + W_friction [M1]
5 × 10 × 4.226 = ½ × 5 × v² + 90.6
211.3 − 90.6 = 2.5v² → 120.7 = 2.5v² → v² = 48.28 → v = 6.95 m s⁻¹ [A1]
Q4 [5 marks]
A particle of mass 2 kg moving at 8 m s⁻¹ is brought to rest by friction over a distance of 12 m on a horizontal surface. Find the friction force and the coefficient of friction. (g = 10 m s⁻²)
Work-energy theorem: W_net = ΔKE [M1]
−F × 12 = 0 − ½ × 2 × 64 = −64 [M1]
F × 12 = 64 → F = 64/12 = 5.33 N [A1]
N = mg = 2 × 10 = 20 N. μ = F/N = 5.33/20 = 0.267 [M1 A1]
Q5 [5 marks]
A car engine has a power output of 36 kW. The car has mass 1200 kg and moves on a horizontal road against a constant resistance of 900 N. Find the maximum speed and the acceleration when the speed is 20 m s⁻¹. (g = 10 m s⁻²)
Maximum speed: v_max = P/R = 36000/900 = 40 m s⁻¹ [M1 A1]
At v = 20 m s⁻¹: F_drive = P/v = 36000/20 = 1800 N [M1]
Net force = 1800 − 900 = 900 N
a = 900/1200 = 0.75 m s⁻² [M1 A1]
Q6 [6 marks]
A particle of mass 3 kg is projected up a rough slope inclined at 30° to the horizontal with initial speed 10 m s⁻¹. The friction force is 5 N. Find the distance travelled up the slope before the particle comes to rest, and the speed when it returns to the bottom. (g = 10 m s⁻²)
Going up: W_net = ΔKE [M1]
−mg sinθ × d − F × d = 0 − ½mu²
−(3×10×0.5 + 5) × d = −½ × 3 × 100
−(15+5)d = −150 → 20d = 150 → d = 7.5 m [A1] Coming back down: h = 7.5 × sin30° = 3.75 m [B1]
Energy equation down: mgh = ½mv² + F × d (friction now acts up slope) [M1]
3×10×3.75 = ½×3×v² + 5×7.5
112.5 = 1.5v² + 37.5 → 1.5v² = 75 → v² = 50 → v = 7.07 m s⁻¹ [M1 A1]
Q7 [6 marks]
A lorry of mass 12 000 kg travels at constant speed of 15 m s⁻¹ along a road inclined at sin⁻¹(1/30) to the horizontal. The total resistance to motion is 800 N. Find the engine power and the maximum speed on a horizontal road with the same resistance and engine power. (g = 10 m s⁻²)
At constant speed: F_drive = mg sinθ + R [M1]
F_drive = 12000 × 10 × (1/30) + 800 = 4000 + 800 = 4800 N [M1 A1]
P = F_drive × v = 4800 × 15 = 72 000 W = 72 kW [A1]
Maximum speed on horizontal: v_max = P/R = 72000/800 = 90 m s⁻¹ [M1 A1]
Q8 [7 marks]
A particle of mass 6 kg is projected horizontally at 8 m s⁻¹ from the top of a rough slope 12 m long inclined at 35° to the horizontal. The coefficient of friction is 0.25. Find the speed at the foot of the slope. (g = 10 m s⁻²)
The particle slides down the slope. Initial speed component along slope: u_slope = 8 × cos35° = 6.553 m s⁻¹ [B1]
Height lost: h = 12 × sin35° = 6.883 m [B1]
N = mg cos35° = 6×10×0.819 = 49.14 N [M1]
F_friction = 0.25 × 49.14 = 12.29 N. W_friction = 12.29 × 12 = 147.5 J [M1]
Energy equation (using initial KE along slope): [M1]
½ × 6 × 6.553² + 6 × 10 × 6.883 = ½ × 6 × v² + 147.5
128.65 + 412.98 = 3v² + 147.5
394.13 = 3v² → v² = 131.4 → v = 11.46 m s⁻¹ [M1 A1]
Past Paper Questions
5 questions from Cambridge A-Level 9709 Mechanics 1 past papers on Energy, Work and Power. Attempt fully before revealing.
Past Paper Q1 — 9709/41/M/J/20 Q2 [5 marks]
A particle of mass 2 kg is pulled along a rough horizontal surface by a force of 15 N. The particle moves 6 m and its speed increases from 2 m s⁻¹ to 5 m s⁻¹. Find the friction force. (g = 10 m s⁻²)
A particle of mass 3 kg slides from rest down a rough slope of length 8 m inclined at 40° to the horizontal. It reaches the bottom with speed 7 m s⁻¹. Find the work done against friction and the coefficient of friction. (g = 10 m s⁻²)
h = 8 sin40° = 5.143 m. GPE lost = mgh = 3 × 10 × 5.143 = 154.3 J [M1 A1]
KE gained = ½ × 3 × 49 = 73.5 J [M1]
Work done against friction = 154.3 − 73.5 = 80.8 J [A1]
N = mg cos40° = 3 × 10 × 0.766 = 23.0 N. Friction force F = 80.8/8 = 10.1 N [M1]
μ = F/N = 10.1/23.0 = 0.439 [A1]
Past Paper Q3 — 9709/43/M/J/18 Q4 [6 marks]
A car of mass 1000 kg has engine power 30 kW. The resistance to motion is 600 N. Find (i) the maximum speed of the car on a horizontal road, (ii) the acceleration when v = 10 m s⁻¹ on the horizontal road. (g = 10 m s⁻²)
(i) At max speed: F_drive = R → P/v_max = 600 [M1]
v_max = 30000/600 = 50 m s⁻¹ [A1]
(ii) At v = 10 m s⁻¹: F_drive = P/v = 30000/10 = 3000 N [M1]
Net force = 3000 − 600 = 2400 N [M1]
a = 2400/1000 = 2.4 m s⁻² [A1 A1]
Past Paper Q4 — 9709/41/O/N/17 Q3 [7 marks]
Particles A (4 kg) and B (6 kg) are connected by a light inextensible string over a smooth fixed pulley. A hangs on the smooth left face of a wedge inclined at 30° to the horizontal; B hangs vertically on the right side. The system is released from rest. Using energy methods, find the speed of the particles after A has moved 2 m up the slope. (g = 10 m s⁻²)
A moves 2 m up the slope → rises h_A = 2 sin30° = 1 m. B descends 2 m. [M1]
GPE gained by A = 4 × 10 × 1 = 40 J [M1]
GPE lost by B = 6 × 10 × 2 = 120 J [M1]
Net GPE change = 120 − 40 = 80 J converted to KE [A1]
Total KE = ½(4+6)v² = 5v² [M1]
5v² = 80 → v² = 16 → v = 4 m s⁻¹ [A1 A1]
Past Paper Q5 — 9709/42/M/J/16 Q6 [8 marks]
A van of mass 2500 kg travels along a straight road. The engine power is 60 kW and total resistance is 1500 N. (i) Find the maximum speed on a horizontal road. (ii) Find the maximum speed on a road inclined at sin⁻¹(1/25) to the horizontal. (iii) Find the acceleration on the horizontal road when v = 20 m s⁻¹. (g = 10 m s⁻²)
(i) Horizontal: v_max = P/R = 60000/1500 = 40 m s⁻¹ [M1 A1]
(ii) On slope: total resistance = 1500 + mg sinθ = 1500 + 2500×10×(1/25) = 1500 + 1000 = 2500 N [M1 A1]
v_max = 60000/2500 = 24 m s⁻¹ [A1]
(iii) F_drive = P/v = 60000/20 = 3000 N [M1]
Net force = 3000 − 1500 = 1500 N
a = 1500/2500 = 0.6 m s⁻² [M1 A1]