Mechanical Power

Mechanical power describes how quickly mechanical work is done or how quickly mechanical energy is transferred.

A machine that performs the same amount of work in less time has a greater power output.

Mechanical power is especially useful when studying:

• engines and motors
• vehicles
• cranes and elevators
• conveyor systems
• pumps
• bicycles
• athletes
• industrial machinery

Two machines may perform exactly the same amount of work, but the more powerful machine can perform that work more quickly.

Mechanical Work and Power

Recall that work is done when a force causes an object to move through a displacement.

When the force acts in the same direction as the displacement:

W = Fd

where:

• W = work done (J)
• F = force (N)
• d = displacement (m)

Mechanical power describes how quickly this work is done:

P = W/t

where:

• P = power (W)
• W = work done (J)
• t = time (s)

Since:

1 J = 1 N·m

and:

1 W = 1 J/s

we can also think of mechanical power as the rate at which a force transfers energy through motion.

---

The Watt

The SI unit of power is the watt (W).

1 W = 1 J/s

Therefore, a machine producing 500 W of mechanical power transfers:

500 J of mechanical energy every second

A 10 kW motor transfers mechanical energy at a rate of:

10,000 J/s

---

Worked Example: Mechanical Work

A motor performs 12,000 J of work in 6 s.

Calculate its mechanical power.

Use:

P = W/t

P = 12,000/6

P = 2000 W

or:

P = 2.0 kW

The motor transfers mechanical energy at a rate of 2000 J/s.

---

Worked Example: Comparing Machines

Two motors each perform 30,000 J of work.

Motor A takes 15 s.

Motor B takes 5 s.

For Motor A:

P = 30,000/15

P = 2000 W

For Motor B:

P = 30,000/5

P = 6000 W

Motor B is three times as powerful.

Both motors perform the same amount of work, but Motor B performs it three times faster.

---

Connecting Force, Distance, and Power

We know:

P = W/t

and:

W = Fd

Substitute Fd for W:

P = Fd/t

We also know:

v = d/t

Therefore:

P = Fv

This is one of the most useful equations for mechanical power.

---

Mechanical Power Equation

For a constant force acting in the direction of motion:

P = Fv

where:

• P = power (W)
• F = force (N)
• v = velocity or speed in the direction of the force (m/s)

This equation shows an important relationship:

power depends on both force and speed

A machine requires more power if it must:

• produce a greater force at the same speed
• move at a greater speed while producing the same force

---

Why P = Fv Works

Start with:

P = W/t

For a constant force parallel to the displacement:

W = Fd

Therefore:

P = Fd/t

Since:

v = d/t

we obtain:

P = Fv

So:

P = W/t = Fd/t = Fv

This connects work, force, displacement, velocity, and power.

---

Worked Example: Moving a Load

A machine pushes a load with a force of 500 N at a constant speed of 4 m/s.

Calculate the mechanical power.

Use:

P = Fv

P = 500 × 4

P = 2000 W

or:

P = 2.0 kW

---

Worked Example: Conveyor Belt

A conveyor system moves material while exerting a force of 800 N.

The belt moves at:

2.5 m/s

Calculate the mechanical power.

P = Fv

P = 800 × 2.5

P = 2000 W

Therefore:

P = 2.0 kW

Rearranging P = Fv

The equation:

P = Fv

can be rearranged.

To find force:

F = P/v

To find speed:

v = P/F

These forms are especially useful when analyzing vehicles and machines.

---

Worked Example: Finding Force

A motor provides:

12,000 W

of mechanical power while moving an object at:

6 m/s

Calculate the force.

Use:

F = P/v

F = 12,000/6

F = 2000 N

---

Worked Example: Finding Speed

A machine provides 15 kW of mechanical power while exerting a force of 3000 N.

First convert:

15 kW = 15,000 W

Then:

v = P/F

v = 15,000/3000

v = 5 m/s

---

Constant Velocity

The equation P = Fv is particularly useful for objects moving at constant velocity.

At constant velocity:

resultant force = 0

This does not mean the engine or motor produces no force.

Instead, the driving force balances the resistive forces.

For a vehicle:

driving force = total resistive force

These resistive forces may include:

• air resistance
• rolling resistance
• friction
• mechanical resistance

The engine must continuously transfer energy to overcome these forces.

---

Worked Example: Car at Constant Speed

A car travels at a constant speed of:

20 m/s

The total resistive force is:

1200 N

Because the car moves at constant velocity:

driving force = 1200 N

Therefore:

P = Fv

P = 1200 × 20

P = 24,000 W

or:

P = 24 kW

The engine must provide at least 24 kW of useful mechanical power to maintain this speed under these simplified conditions.

Why Power Is Needed at Constant Speed

Students sometimes think:

"If the car is not accelerating, why does the engine need power?"

The answer is that resistive forces are continually doing negative work on the vehicle.

The engine must continually transfer energy to overcome:

• air resistance
• tyre deformation
• friction in moving components

Without engine power, these forces would cause the vehicle to slow down.

So at constant speed:

driving force = resistive force

but energy is still being transferred.

---

Force, Speed, and Power

From:

P = Fv

we can see several important relationships.

Constant Force

If F remains constant:

P ∝ v

Doubling speed doubles power.

Constant Speed

If v remains constant:

P ∝ F

Doubling force doubles power.

Constant Power

If P remains constant:

F = P/v

Therefore:

F ∝ 1/v

As speed increases, the available force decreases if power remains constant.

This relationship is extremely important in machines and transportation.

---

Constant Power Example

A motor provides constant mechanical power of:

20 kW

At 2 m/s:

F = 20,000/2

F = 10,000 N

At 5 m/s:

F = 20,000/5

F = 4000 N

At 10 m/s:

F = 20,000/10

F = 2000 N

At constant power:

higher speed → lower available force

---

Mechanical Power in Vehicles

Vehicle performance provides a useful application of mechanical power.

An engine or motor must transfer energy to:

• accelerate the vehicle
• overcome air resistance
• overcome rolling resistance
• climb hills
• carry or tow loads

Greater power allows mechanical energy to be transferred more quickly.

However, the relationship between speed and power is not simple because resistive forces can change with speed.

---

Air Resistance and Speed

Air resistance generally increases as vehicle speed increases.

At higher speeds:

• air resistance becomes greater
• more driving force is required
• more power is required to maintain speed

Because:

P = Fv

increasing both the required force and the speed can cause the required power to rise rapidly.

This is one reason high-speed travel can require very large power outputs.

---

Worked Example: Comparing Vehicle Speeds

At 10 m/s, suppose a vehicle experiences a total resistance of:

500 N

Required power:

P = 500 × 10

P = 5000 W

At 20 m/s, suppose resistance has increased to:

900 N

Required power:

P = 900 × 20

P = 18,000 W

The speed doubled, but the required power increased by much more than a factor of two because the resistive force also increased.

---

Power and Acceleration

When a vehicle accelerates, some of the engine's mechanical power increases the vehicle's kinetic energy.

Recall:

Eₖ = ½mv²

Average mechanical power used to increase kinetic energy can be calculated using:

P = ΔEₖ/t

If resistive forces are also present, the engine must provide additional power to overcome them.

So in a real accelerating vehicle:

engine power = power increasing KE + power overcoming resistance

---

Worked Example: Accelerating Vehicle

A 1200 kg car accelerates from rest to 20 m/s in 8 s.

Ignore resistance.

Calculate the average mechanical power transferred into kinetic energy.

Final KE:

Eₖ = ½mv²

Eₖ = 0.5 × 1200 × 20²

Eₖ = 240,000 J

Initial KE:

0 J

Therefore:

ΔEₖ = 240,000 J

Average power:

P = ΔE/t

P = 240,000/8

P = 30,000 W

or:

P = 30 kW

---

Adding Resistive Forces

Suppose the previous car also experiences an average resistive force of 500 N while traveling an average distance of 80 m during the acceleration.

Work done against resistance:

W = Fd

W = 500 × 80

W = 40,000 J

Total mechanical energy supplied:

240,000 + 40,000 = 280,000 J

Average mechanical power:

P = 280,000/8

P = 35,000 W

or:

35 kW

This is more realistic because some engine energy must overcome resistance rather than increase the car's kinetic energy.

---

Mechanical Power in Lifting

Machines such as cranes and elevators perform mechanical work against gravity.

The work done in lifting an object is equal to its increase in gravitational potential energy:

W = mgΔh

Therefore:

P = mgΔh/t

Worked Example: Crane

A crane lifts a 2000 kg load vertically through 12 m in 30 s.

Use:

g = 9.8 N/kg

Work done:

W = mgΔh

W = 2000 × 9.8 × 12

W = 235,200 J

Power:

P = W/t

P = 235,200/30

P = 7840 W

or:

P = 7.84 kW

---

Solving the Same Crane Problem Using P = Fv

The lifting force at constant speed is:

F = mg

F = 2000 × 9.8

F = 19,600 N

The vertical speed is:

v = d/t

v = 12/30

v = 0.40 m/s

Now:

P = Fv

P = 19,600 × 0.40

P = 7840 W

We obtain exactly the same answer.

This demonstrates the connection between:

P = W/t

and:

P = Fv

---

Worked Example: Elevator

An elevator has a total mass of 1500 kg and rises at a constant speed of 2 m/s.

Ignore friction.

At constant speed, the upward motor force equals the elevator's weight:

F = mg

F = 1500 × 9.8

F = 14,700 N

Power:

P = Fv

P = 14,700 × 2

P = 29,400 W

or:

P = 29.4 kW

---

Mechanical Power in Cycling

Cyclists produce mechanical power by applying forces to pedals.

Energy is transferred from chemical energy in the cyclist's body into mechanical energy.

Some mechanical power is used to:

• increase kinetic energy
• climb hills
• overcome air resistance
• overcome rolling resistance

Worked Example: Cyclist

A cyclist travels at a constant:

8 m/s

against a total resistive force of:

35 N

Mechanical power required:

P = Fv

P = 35 × 8

P = 280 W

The cyclist must produce approximately 280 W of useful mechanical power to maintain this speed.

---

Cyclist Climbing a Hill

When climbing, the cyclist must also work against gravity.

Suppose the combined mass of cyclist and bicycle is:

80 kg

and the cyclist gains vertical height at:

0.50 m/s

The force associated with the weight is:

F = mg

F = 80 × 9.8

F = 784 N

Power required to gain GPE:

P = Fv

P = 784 × 0.50

P = 392 W

This is the mechanical power required just for the increase in gravitational potential energy.

Additional power would be required to overcome air resistance and rolling resistance.

---

Machines and Mechanical Advantage

Machines can trade force for speed or distance.

For an ideal machine:

input power = output power

Therefore:

Finput × vinput = Foutput × voutput

If a machine increases force, the output generally moves more slowly.

For example, an ideal machine might produce:

twice the force

but move the load at:

half the speed

The power remains the same.

Real machines also dissipate energy, so:

useful output power < input power

---

Worked Example: Ideal Machine

A machine receives an input force of:

200 N

moving at:

4 m/s

Input power:

P = Fv

P = 200 × 4

P = 800 W

The machine produces an output force of:

800 N

For an ideal machine:

800 = 800 × v

Therefore:

v = 1 m/s

The machine multiplies the force by four but reduces the speed by four.

---

Mechanical Power and Efficiency

Real machines are not perfectly efficient.

Efficiency can be calculated using:

efficiency = useful output power / input power

or:

efficiency (%) = useful output power / input power × 100

---

Worked Example: Motor Efficiency

An electric motor receives:

5000 W

of electrical power.

It provides:

4000 W

of mechanical power.

Efficiency:

efficiency = 4000/5000 × 100

efficiency = 80%

The remaining:

1000 W

is transferred into other forms, such as thermal energy and sound.

---

Worked Example: Finding Input Power

A crane must provide:

24 kW

of useful mechanical power.

Its motor system is:

80% efficient

Use:

efficiency = useful output power / input power

0.80 = 24/input power

Therefore:

input power = 24/0.80

input power = 30 kW

The system requires 30 kW of input power to provide 24 kW of useful mechanical power.

---

Transportation and Power

Power strongly affects transportation systems.

A vehicle needs power to:

• overcome resistive forces
• accelerate
• climb slopes
• tow loads
• maintain high speeds

For constant-speed motion:

P = Fv

This means that high-speed transportation can require enormous power even when the vehicle is not accelerating.

Examples include:

• cars
• trucks
• trains
• aircraft
• ships

---

Heavy Vehicles

A truck moving at the same speed as a smaller vehicle may require more power because it can experience greater:

• rolling resistance
• mechanical resistance
• gravitational forces when climbing
• drag, depending on its shape and size

When climbing a slope, a heavy vehicle may require particularly large power because it is increasing its gravitational potential energy.

---

Worked Example: Truck

A truck moves at a constant 25 m/s against a total resistive force of 6000 N.

Calculate the mechanical power required.

P = Fv

P = 6000 × 25

P = 150,000 W

or:

P = 150 kW

---

Worked Example: Train

A train moves at a constant speed of:

30 m/s

The locomotive provides a driving force of:

100,000 N

Calculate its mechanical power.

P = Fv

P = 100,000 × 30

P = 3,000,000 W

or:

P = 3 MW

Large transportation systems can therefore require power outputs measured in megawatts.

---

Force at Different Speeds

Suppose an engine can deliver a constant mechanical power of:

60 kW

At 5 m/s:

F = 60,000/5 = 12,000 N

At 10 m/s:

F = 60,000/10 = 6000 N

At 20 m/s:

F = 60,000/20 = 3000 N

At 30 m/s:

F = 60,000/30 = 2000 N

This illustrates the inverse relationship between force and speed at constant power.

---

Why Vehicles Use Gears

Gears help vehicles and machines trade force for speed.

At low speed, a vehicle may need a large driving force to:

• start moving
• accelerate
• climb steep slopes
• move heavy loads

At higher speeds, less driving force may be available for a given engine power.

Gear systems allow the motor or engine to operate effectively while changing the relationship between:

force and speed

Power, Force, and Velocity Graphs

The equation:

P = Fv

allows several graphical relationships to be predicted.

Constant Force

If force remains constant:

P ∝ v

A graph of power against velocity is a straight line through the origin.

Constant Velocity

If velocity remains constant:

P ∝ F

A graph of power against force is a straight line through the origin.

Constant Power

If power remains constant:

F = P/v

Force decreases as velocity increases.

The force-velocity relationship is therefore an inverse relationship.

---

Direction Matters

The simple equation:

P = Fv

assumes the force and velocity are in the same direction.

More generally, mechanical power depends on the component of force in the direction of motion.

If force and motion are perpendicular, the force does no mechanical work on the object at that instant.

For the problems in this topic, unless stated otherwise, assume the force acts in the direction of motion.

---

Common Mistakes

Confusing Work and Power

Work measures energy transferred.

Power measures how quickly that energy is transferred.

W = Fd

but:

P = W/t

---

Using P = Fv During the Wrong Situation

The simple form:

P = Fv

is used when the force acts in the direction of motion.

For many introductory problems, it is especially useful for constant-speed motion.

---

Thinking Constant Velocity Means No Forces

Constant velocity means:

resultant force = 0

It does not mean:

all forces = 0

A vehicle can have a large driving force balanced by an equally large resistive force.

---

Thinking Constant Velocity Means Zero Power

A car moving at constant velocity against resistance still requires power.

The engine continuously transfers energy to overcome resistive forces.

---

Forgetting Unit Conversions

Remember:

1 kW = 1000 W

1 MW = 1,000,000 W

---

Using Distance Instead of Velocity

In:

P = Fv

v must be measured in:

m/s

not metres.

---

Forgetting That Power Depends on Both Force and Speed

A large force does not necessarily mean large power.

If the object moves very slowly, the power may still be small.

---

Assuming More Power Means More Force

Not necessarily.

From:

F = P/v

the force produced by a given power depends on speed.

---

Ignoring Energy Losses

Real machines dissipate energy through:

• friction
• thermal energy
• sound
• deformation

Therefore:

useful mechanical output power < input power

---

Check Your Understanding

1. Define mechanical power.
2. State the equation relating power, work, and time.
3. State the equation relating power, force, and velocity.
4. Explain how P = Fv can be derived from P = W/t.
5. A machine performs 20,000 J of work in 10 s. Calculate its power.
6. A motor exerts a force of 600 N while moving an object at 5 m/s. Calculate the mechanical power.
7. A vehicle experiences a resistance of 1500 N while traveling at a constant 20 m/s. Calculate the required mechanical power.
8. Explain why the vehicle in Question 7 needs power even though it is not accelerating.
9. A 25 kW motor moves a load at 5 m/s. Calculate the force.
10. A machine provides 40 kW while exerting a force of 8000 N. Calculate its speed.
11. A crane lifts a 1500 kg load through 10 m in 20 s. Calculate the work done.
12. Calculate the mechanical power of the crane in Question 11.
13. Solve Question 12 again using P = Fv.
14. A cyclist travels at 10 m/s against a resistance of 30 N. Calculate the mechanical power required.
15. Explain why a cyclist generally needs more power to travel faster.
16. A truck travels at 20 m/s against a resistance of 5000 N. Calculate its mechanical power.
17. Convert your answer from Question 16 into kilowatts.
18. A train produces a driving force of 80,000 N while traveling at 25 m/s. Calculate its mechanical power in MW.
19. A motor provides a constant 50 kW. Calculate the available force at 5 m/s.
20. Calculate the available force from the same motor at 10 m/s.
21. What happens to available force when speed doubles at constant power?
22. Explain why gears are useful in vehicles.
23. A motor receives 10 kW of electrical power and produces 8 kW of useful mechanical power. Calculate its efficiency.
24. A machine must provide 12 kW of useful mechanical power and is 75% efficient. Calculate its required input power.
25. Explain how force, velocity, and power are related in a transportation system.
26. Why does maintaining a high vehicle speed usually require substantial power?
27. A 1000 kg car accelerates from rest to 20 m/s in 5 s. Ignoring resistance, calculate its average mechanical power.
28. If the car in Question 27 also transfers 50,000 J to the surroundings during the acceleration, calculate the average engine power required.
29. Explain the difference between input power and useful mechanical output power.
30. Give one practical example where high force is more important than high speed and explain how power relates to the situation.

---

Key Terms

Mechanical power — Rate at which mechanical work is done or mechanical energy is transferred.

Power — Rate of energy transfer.

Work done — Energy transferred when a force causes displacement.

Watt (W) — SI unit of power, equal to one joule per second.

Force — Push or pull capable of changing an object's motion.

Displacement — Change in position in a specified direction.

Velocity — Rate of change of displacement, including direction.

Driving force — Force that propels a vehicle or machine.

Resistive force — Force opposing motion, such as friction or air resistance.

Constant velocity — Motion at constant speed and direction, meaning resultant force is zero.

Useful mechanical power — Rate at which energy is transferred into the intended mechanical output.

Input power — Rate at which energy is supplied to a machine.

Efficiency — Fraction of input power converted into useful output power.

Mechanical advantage — Increase in output force produced by a machine compared with its input force.

Air resistance — Resistive force caused by motion through air.

Rolling resistance — Resistive force associated with rolling objects such as tyres or wheels.

Gear — Mechanical device used to alter the relationship between force and speed.

---

Key Takeaways

• Mechanical power describes how quickly mechanical work is done.
• Mechanical power can be calculated using P = W/t.
• When a constant force acts in the direction of motion, W = Fd.
• Combining P = W/t, W = Fd, and v = d/t gives P = Fv.
• Power is measured in watts.
• 1 W = 1 J/s.
• Greater force requires greater power when speed remains constant.
• Greater speed requires greater power when force remains constant.
• At constant power, force decreases as speed increases.
• An object moving at constant velocity can still require substantial mechanical power.
• Constant velocity means the driving and resistive forces balance; it does not mean that no forces act.
• Vehicles need continuous power at constant speed because energy is continually transferred against resistive forces.
• Air resistance generally increases with speed, so high-speed transportation can require very large power outputs.
• Mechanical power can be calculated for cranes, elevators, vehicles, bicycles, trains, and industrial machines.
• Lifting power can be calculated using P = mgΔh/t.
• Power associated with acceleration can be found from the rate of increase of kinetic energy.
• Real machines have lower useful output power than input power because some energy is dissipated.
• Gears allow machines to trade force for speed.
• In an ideal machine, increasing output force generally decreases output speed while power remains constant.
• The equations P = W/t and P = Fv describe the same fundamental idea: the rate at which mechanical energy is transferred.

The central relationships are:

P = W/t

and:

P = Fv

So for machines and vehicles:

more force at the same speed → more power

more speed at the same force → more power

and, for a fixed power output:

more speed → less available force.