LESSON 01
What Is Work?
In mechanics, work is done when a force causes a displacement. Work measures the amount of energy transferred by a force.
The work done by a force is the product of the force and the displacement in the direction of the force.
Work is a scalar quantity. It has magnitude but no direction. Its SI unit is the joule, J.
What does one joule mean?
One joule is the work done when a force of 1 N causes a displacement of 1 m in the direction of the force.
Do not automatically use W = Fs. First check whether the force is parallel to the displacement. If it is at an angle, the component of the force in the direction of motion must be used.
LESSON 02
Work Done by a Force at an Angle
Suppose a constant force F acts at an angle θ to the displacement s. Only the component of the force parallel to the displacement does work.
The component of F in the direction of motion is:
Therefore:
Special cases
| Angle | cosθ | Work |
|---|---|---|
| θ = 0° | 1 | W = Fs — maximum positive work |
| θ = 90° | 0 | W = 0 — no work |
| θ = 180° | -1 | W = -Fs — negative work |
Positive work
If the force has a component in the direction of displacement, it transfers energy to the particle. The work is positive.
Negative work
If the force has a component opposite to the displacement, the force removes mechanical energy from the particle. The work is negative.
Zero work
A force perpendicular to the displacement does no work. This is particularly important for the normal reaction on a particle moving along a straight fixed surface.
A force can be large and still do zero work. For example, if a reaction force is perpendicular to the displacement, its work is zero.
LESSON 03
Work Done by Weight
The weight of a particle acts vertically downward with magnitude mg. The work done by weight depends on the change in vertical position.
If the particle moves vertically upward through height h, weight acts opposite to the displacement:
If the particle moves vertically downward through height h:
Connection with gravitational potential energy
When a particle rises, its gravitational potential energy increases. That energy comes from work done against gravity.
Therefore:
Moving upward: gravity does negative work. Moving downward: gravity does positive work.
LESSON 04
Kinetic Energy
Kinetic energy is the energy associated with the motion of a particle. A particle of mass m moving with speed v has kinetic energy:
Kinetic energy is always non-negative because speed is squared.
Important consequences
- Doubling speed makes kinetic energy four times as large.
- Tripling speed makes kinetic energy nine times as large.
- Doubling mass doubles kinetic energy at the same speed.
- A particle at rest has zero kinetic energy.
Change in kinetic energy
If the initial speed is u and final speed is v:
or:
LESSON 05
Gravitational Potential Energy
Gravitational potential energy depends on vertical position relative to a chosen reference level, called a datum.
Here:
- m = mass in kg
- g = acceleration due to gravity in m s⁻²
- h = vertical height above the chosen datum
The datum
The zero level for gravitational potential energy can be chosen conveniently. The important quantity in energy equations is usually the change in height, not the absolute value of GPE.
If a particle moves from height h₁ to height h₂, the change in gravitational potential energy is:
On an inclined plane, h is the vertical height change, NOT the distance travelled along the slope.
LESSON 06
The Work–Energy Principle
The work done by the resultant force on a particle is equal to its change in kinetic energy.
What does resultant work mean?
Every force acting on the particle may do work. The total work is the algebraic sum of the work done by all forces.
For example, if a particle is pulled by a force P while resistance R acts against its motion:
This is often much faster than resolving forces and using Newton's second law.
If the question gives forces, distances, speeds, heights or resistance and asks for an unknown speed/distance/force, consider whether an energy equation can connect the initial and final states.
LESSON 07
Conservation of Mechanical Energy
Mechanical energy consists of kinetic and gravitational potential energy:
If there is no energy transfer to resistance or another external system:
The mass can often be cancelled:
Physical meaning
As a particle moves upward, gravitational potential energy increases. If no energy is lost, this increase comes from kinetic energy, so the particle slows down.
As a particle moves downward, gravitational potential energy decreases. That energy becomes kinetic energy, so the particle speeds up.
LESSON 08
Energy with Resistance
Resistance is a non-conservative force. It transfers mechanical energy into other forms, usually thermal/internal energy.
If a constant resistance R acts over distance s:
For a particle moving from state 1 to state 2:
Alternative work equation
You can also use work-energy:
For a particle pulled by force P against resistance R:
Do not put Rs on the left as positive if resistance opposes the motion. Resistance removes mechanical energy.
LESSON 09
Energy on an Inclined Plane
Inclined-plane problems are especially suitable for energy methods because the vertical height change can be obtained geometrically.
If a particle moves a distance s along an incline at angle θ:
Therefore the change in gravitational potential energy is:
Smooth incline
If the plane is smooth, there is no frictional energy loss. If the particle moves upward:
Rough incline
If resistance R acts along the plane:
The normal reaction does no work when it is perpendicular to the particle's displacement along the plane.
LESSON 10
Stopping Distance
Suppose a particle is moving with speed u and is brought to rest by a constant resistance R.
Initial kinetic energy is:
At the stopping point, final kinetic energy is zero. The resistance has removed all the initial kinetic energy:
What does this tell us?
Stopping distance is proportional to the square of the initial speed when mass and resistance are constant.
So doubling the initial speed requires four times the stopping distance under the same constant resistance.
LESSON 11
Work Done from a Force–Displacement Graph
When force varies with displacement, W = Fs cannot generally be used with one constant force value.
The work done is the area under the force–displacement graph.
Common shapes
Area = base × height
Area = ½ × base × height
Area = ½(a + b)h
Use the appropriate area method or integration if required.
Areas below the displacement axis represent negative work.
Do not simply add all areas if some parts of the graph lie below the axis. Work is signed.
LESSON 12
Power
Power measures the rate at which work is done or energy is transferred.
The SI unit of power is the watt:
Power when a force acts at constant velocity
Since:
This applies when the force and velocity are in the same direction.
General form
where θ is the angle between the force and velocity.
Average versus instantaneous power
| Quantity | Formula |
|---|---|
| Average power | Pavg = total work / total time |
| Instantaneous power | P = Fv cosθ |
LESSON 13
Efficiency and Energy Transfer
Efficiency compares useful output energy with total input energy.
As a percentage:
The same relationship can be written using power:
Energy cannot disappear
If a machine is less than 100% efficient, some input energy is transferred into unwanted forms such as thermal energy or sound.
Keep the quantities in the same units before calculating efficiency. For example, do not divide a value in kJ by a value in J without converting first.
LESSON 14
Energy Methods: The Universal Workflow
Many Mechanics 2 problems become much easier if you treat the motion as a transfer of energy between clearly defined initial and final states.
LESSON 15
Master Formula Atlas
Constant force parallel to displacement.
Constant force at angle θ to displacement.
Kinetic energy.
Change in kinetic energy.
Gravitational potential energy relative to datum.
Change in gravitational potential energy.
Work–energy principle.
Energy dissipated by constant resistance R over distance s.
Vertical height on an incline.
Average power.
Power when force and velocity are parallel.
General instantaneous power relationship.
WORKED EXAMPLE 01
Constant Force Parallel to Motion
Question
A constant horizontal force of 18 N moves a particle through 7 m. Find the work done by the force.
Since the force is parallel to the displacement:
W = Fs
W = 18 × 7
W = 126 J
WORKED EXAMPLE 02
Force at an Angle
Question
A force of 30 N acts at 60° to the direction of motion of a particle. The particle moves 8 m. Find the work done by the force.
W = Fs cosθ
W = 30 × 8 × cos60°
W = 240 × 0.5
W = 120 J
WORKED EXAMPLE 03
Negative Work by Resistance
Question
A resistance of 14 N acts on a particle moving 12 m forwards. Find the work done by the resistance.
The resistance acts opposite to the displacement.
W = Fs cos180°
W = 14 × 12 × (-1)
W = −168 J
WORKED EXAMPLE 04
Change in Kinetic Energy
Question
A particle of mass 5 kg accelerates from 4 m/s to 10 m/s. Find the increase in kinetic energy.
ΔKE = ½m(v² − u²)
= ½ × 5 × (10² − 4²)
= 2.5 × (100 − 16)
= 2.5 × 84
ΔKE = 210 J
WORKED EXAMPLE 05
Work–Energy Principle
Question
A 3 kg particle initially moves at 5 m/s. A constant resultant force does 96 J of work on it. Find its final speed.
Work done = change in kinetic energy.
96 = ½ × 3 × (v² − 5²)
96 = 1.5(v² − 25)
64 = v² − 25
v² = 89
v = √89 ≈ 9.43 m/s
WORKED EXAMPLE 06
Vertical Lift
Question
A 4 kg particle is lifted vertically through 6 m at constant speed. Find the work done by the lifting force. Take g = 9.8 m/s².
Constant speed means there is no change in kinetic energy.
The lifting force balances the weight:
F = mg = 4 × 9.8 = 39.2 N
Work done = Fs
= 39.2 × 6
Work = 235.2 J
This is also exactly the increase in gravitational potential energy:
mgh = 4 × 9.8 × 6 = 235.2 J.
WORKED EXAMPLE 07
Speed from a Vertical Drop
Question
A particle is released from rest and falls through 5 m. Assuming no resistance, find its speed just before impact. Take g = 9.8 m/s².
Use conservation of mechanical energy:
Initial KE + initial GPE = final KE + final GPE.
0 + mgh = ½mv² + 0
mgh = ½mv²
gh = ½v²
v² = 2gh
v² = 2 × 9.8 × 5 = 98
v = √98 ≈ 9.90 m/s
WORKED EXAMPLE 08
Particle Moving Up a Smooth Incline
Question
A particle moves up a smooth plane inclined at 30° to the horizontal. Its initial speed is 14 m/s. Find the distance travelled before it comes to rest. Take g = 9.8 m/s².
At the highest point:
v = 0
Vertical height gained:
h = s sin30°
Conservation of energy gives:
½m(14²) = mgh
½(14²) = 9.8s sin30°
98 = 9.8s × 0.5
98 = 4.9s
s = 20 m
WORKED EXAMPLE 09
Rough Incline
Question
A particle of mass 2 kg moves up a rough plane inclined at 30°. The particle has initial speed 12 m/s and travels 5 m before its speed falls to 4 m/s. The resistance is constant. Find the resistance. Take g = 9.8 m/s².
Use:
Initial KE = final KE + gain in GPE + energy lost to resistance.
½(2)(12²) = ½(2)(4²) + (2)(9.8)(5 sin30°) + 5R
144 = 16 + 49 + 5R
144 = 65 + 5R
79 = 5R
R = 15.8 N
WORKED EXAMPLE 10
Stopping Distance
Question
A 1200 kg car travels at 20 m/s. A constant total resistance of 6000 N acts on it. Find its stopping distance.
At rest, final KE = 0.
Initial kinetic energy = work done against resistance.
½mu² = Rs
½ × 1200 × 20² = 6000s
240000 = 6000s
s = 40 m
WORKED EXAMPLE 11
Force–Displacement Graph
Question
A force increases uniformly from 0 N to 20 N while a particle moves through 6 m. Find the work done.
The F–s graph is a triangle.
Work = area under graph.
W = ½ × base × height
W = ½ × 6 × 20
W = 60 J
WORKED EXAMPLE 12
Average Power
Question
A motor transfers 18 000 J of useful energy in 12 s. Find its average useful power.
P = W/t
P = 18000 / 12
P = 1500 W = 1.50 kW
WORKED EXAMPLE 13
Instantaneous Power
Question
A driving force of 2500 N acts on a car travelling at 16 m/s. The force is parallel to the velocity. Find the power supplied.
P = Fv
P = 2500 × 16
P = 40 000 W = 40 kW
WORKED EXAMPLE 14
Power at an Angle
Question
A force of 80 N acts at 60° to the velocity of a particle moving at 5 m/s. Find the instantaneous power supplied by the force.
P = Fv cosθ
P = 80 × 5 × cos60°
P = 400 × 0.5
P = 200 W
WORKED EXAMPLE 15
Efficiency
Question
A machine receives 5000 J of energy and produces 3600 J of useful output energy. Find its efficiency.
Efficiency = useful output / input × 100%
= 3600 / 5000 × 100%
Efficiency = 72%
PRACTICE
Exam-Style Practice
Show answer
W = Fs = 25 × 8.
W = 200 J.
Show answer
W = Fs cos60° = 40 × 5 × 0.5.
W = 100 J.
Show answer
KE = ½ × 6 × 10².
KE = 300 J.
Show answer
ΔGPE = mgh = 2 × 9.8 × 7.
ΔGPE = 137.2 J.
Show answer
ΔKE = ½ × 4 × (9² − 3²).
ΔKE = 144 J.
Show answer
E = Rs = 20 × 15.
E = 300 J.
Show answer
mgh = ½mv².
v² = 2gh = 2 × 9.8 × 4 = 78.4.
v ≈ 8.85 m/s.
Show answer
P = W/t = 45000/15.
P = 3000 W.
EXAM TRAPS
Common Mistakes to Eliminate
| Mistake | Correct approach |
|---|---|
| Using W = Fs for every force. | Use W = Fs cosθ when the force is not parallel to displacement. |
| Using slope distance as h. | Use the vertical height change. |
| Making resistance positive in a work equation. | Resistance opposes motion, so its work is negative. |
| Including normal reaction work unnecessarily. | If reaction is perpendicular to displacement, its work is zero. |
| Using P = Fv when F and v are not parallel. | Use P = Fv cosθ. |
| Confusing energy and power. | Energy is measured in J; power is measured in W. |
| Using initial speed instead of final speed. | Write ½mu² and ½mv² explicitly before substituting. |
| Forgetting resistance energy. | Add Rs as energy dissipated. |
| Using height instead of vertical height difference. | Use Δh. |
| Forgetting that work can be negative. | Consider the direction of the force relative to displacement. |
EXAM METHOD
How to Attack a Work & Energy Question
Do not automatically use Newton's second law just because the question contains forces. If the question connects force/work with displacement, speed or height, an energy method may be significantly shorter.
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