Forces and Energy Transfers

This lesson covers the relationship between forces, work done and energy transfers — as required by the Edexcel GCSE Physics specification (1PH0), Topic 2: Motion and Forces. You need to understand work done, energy transfers in mechanical systems, and power, and be able to solve multi-step problems combining forces, energy and power.

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Work Done

When a force causes an object to move (or to be displaced), work is done on the object. Work done is a measure of energy transferred.

Work done = Force × Distance moved in the direction of the force

W = F × s

Quantity	Symbol	Unit
Work done	W	Joules (J)
Force	F	Newtons (N)
Distance	s	Metres (m)

Key Points

• Work is only done when the force causes movement in the direction of the force (or has a component in that direction).
• If you carry a box horizontally at constant height, you do no work against gravity — because the movement is perpendicular to the gravitational force.
• 1 joule of work is done when a force of 1 newton moves an object 1 metre in the direction of the force.
• Work done = energy transferred. They are the same quantity, measured in joules.

Exam Tip: The distance must be in the direction of the force. If a force acts horizontally and the object moves horizontally, use that distance. If a force acts at an angle, you may need to resolve either the force or the distance into components (Higher tier).

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Worked Examples: Work Done

Worked Example 1

A person pushes a box with a force of 50 N across a floor for 8 m. How much work is done?

W = Fs = 50 × 8 = 400 J

Worked Example 2

A crane lifts a 200 kg load to a height of 15 m. How much work is done against gravity?

Weight = mg = 200 × 9.8 = 1960 N

W = Fs = 1960 × 15 = 29,400 J

Worked Example 3

A car's engine produces a driving force of 3000 N. The car travels 500 m. How much work does the engine do?

W = Fs = 3000 × 500 = 1,500,000 J (1.5 MJ)

Worked Example 4: Finding Distance

How far can a 400 N force move an object if 2000 J of work is done?

s = W / F = 2000 / 400 = 5 m

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Work Done Against Friction

When an object moves against friction, work is done against the frictional force. This work is not stored as useful energy — it is transferred to the thermal energy of the surfaces in contact, causing them to heat up.

Key Points

• Friction always opposes motion.
• Work done against friction → increase in thermal energy (heat).
• This is why your hands warm up when you rub them together, and why brakes get hot.
• Energy is not destroyed — it is transferred from kinetic energy to thermal energy.

Worked Example 5

A box is dragged 12 m across a rough floor. The friction force is 30 N. How much thermal energy is generated?

Work done against friction = Fs = 30 × 12 = 360 J

This 360 J is transferred to thermal energy of the box and floor surfaces.

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Energy Transfers in Mechanical Systems

Gravitational Potential Energy (GPE)

GPE = mass × gravitational field strength × height

GPE = mgh

Quantity	Unit
GPE	Joules (J)
Mass (m)	kg
g	9.8 N/kg
Height (h)	m

GPE increases when an object is raised and decreases when it is lowered.

Kinetic Energy (KE)

KE = ½ × mass × velocity²

KE = ½mv²

Quantity	Unit
KE	Joules (J)
Mass (m)	kg
Velocity (v)	m/s

KE increases when an object speeds up and decreases when it slows down.

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Conservation of Energy in Mechanical Systems

In the absence of friction and air resistance, energy is conserved — it converts between gravitational potential energy and kinetic energy without any loss.

Roller Coaster Example

At the top of a hill:

• Maximum GPE (high up)
• Minimum KE (momentarily slow)

At the bottom of a hill:

• Minimum GPE (low down)
• Maximum KE (fastest speed)

GPE lost = KE gained (if no friction)

mgh = ½mv²

This can be rearranged to find the speed at the bottom:

v = √(2gh)

Worked Example 6: Roller Coaster

A roller coaster car of mass 500 kg drops from a height of 30 m. Assuming no friction, what is its speed at the bottom?

GPE lost = mgh = 500 × 9.8 × 30 = 147,000 J

KE gained = ½mv²

147,000 = ½ × 500 × v²

v² = 147,000 / 250 = 588

v = √588 = 24.2 m/s

Pendulum Example

A pendulum swings back and forth, continuously converting GPE to KE and back:

flowchart LR
    A["Highest point<br/>(left)<br/>Max GPE<br/>KE = 0<br/>Momentarily at rest"] --> B["Lowest point<br/>(centre)<br/>Max KE<br/>Min GPE<br/>Fastest speed"]
    B --> C["Highest point<br/>(right)<br/>Max GPE<br/>KE = 0<br/>Momentarily at rest"]
    C --> B

    style A fill:#e67e22,color:#fff
    style B fill:#2980b9,color:#fff
    style C fill:#e67e22,color:#fff

In reality, air resistance and friction at the pivot mean the pendulum gradually loses height — energy is transferred to thermal energy, and the swing gets smaller over time.

Exam Tip: In energy conservation problems, always state the assumption that friction and air resistance are negligible (if the question says so). If friction IS present, then some energy is transferred to thermal energy: GPE lost = KE gained + thermal energy produced.

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Power

Power is the rate at which energy is transferred or the rate at which work is done.

Power = Work done ÷ Time

P = W / t

Quantity	Symbol	Unit
Power	P	Watts (W)
Work done	W	Joules (J)
Time	t	Seconds (s)

1 watt = 1 joule per second (1 W = 1 J/s)

Alternative Power Equations

Since W = Fs, we can write:

P = Fs / t = F × (s/t) = F × v

P = Fv

This is useful for calculating the power of a vehicle moving at constant velocity against a resistive force.

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Worked Examples: Power

Worked Example 7

A motor lifts a 50 kg load through 4 m in 10 seconds. What is the power?

Work done = mgh = 50 × 9.8 × 4 = 1960 J

Power = W / t = 1960 / 10 = 196 W

Worked Example 8

A car travels at a constant 25 m/s against a total resistive force of 800 N. What power does the engine provide?

P = Fv = 800 × 25 = 20,000 W (20 kW)

Worked Example 9: Multi-Step Problem

A cyclist of total mass (cyclist + bike) 80 kg rides up a hill that is 200 m long and 15 m high. The journey takes 60 seconds. The average friction force is 20 N. Calculate the power output of the cyclist.

Step 1: Work done against gravity = mgh = 80 × 9.8 × 15 = 11,760 J

Step 2: Work done against friction = Fs = 20 × 200 = 4,000 J

Step 3: Total work done = 11,760 + 4,000 = 15,760 J

Step 4: Power = W / t = 15,760 / 60 = 262.7 W

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Multi-Step Problems: Forces, Energy and Power

Worked Example 10

A car of mass 1200 kg is travelling at 30 m/s. The driver brakes and the car comes to a stop over a distance of 90 m.

(a) What was the initial kinetic energy of the car?