Work, Energy and Conservation

Energy is perhaps the single most important unifying concept in physics. Unlike forces and velocities, which come and go, energy is conserved — the total amount in a closed system never changes. It only moves from one form to another. This principle, stated with full generality in the 19th century, is the most powerful tool physicists have for predicting the outcome of processes without having to follow every force and motion.

This lesson covers OCR Module 3.3.1 (Work, energy and power) and builds the link between forces (Newton's second law) and energy. By the end you will be able to solve complex problems by tracking where the energy goes — a method that is often quicker and more insightful than using F = ma.

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Work Done by a Force

The work done by a constant force F on an object that moves through displacement s is:

W = F s cos θ

• F is the magnitude of the force (N)
• s is the magnitude of the displacement (m)
• θ is the angle between the force and the displacement vectors

SI unit: joule (J) = N m. One joule is the work done when a force of 1 N moves its point of application by 1 m in the direction of the force.

Interpretation of the Angle

• θ = 0° (force along motion): W = Fs (maximum, positive). Energy is transferred to the object.
• θ = 90° (force perpendicular to motion): W = 0. Example: centripetal force on a satellite in a circular orbit does no work.
• θ = 180° (force opposite to motion): W = −Fs (negative). Energy is taken from the object, e.g. friction on a sliding block.

Exam Tip: A force does no work if it is perpendicular to the motion. This is why a tennis ball swung on a string at constant speed has constant kinetic energy — the tension is always perpendicular to the velocity.

Worked Example — Sledge

A sledge is pulled 12 m along horizontal snow by a rope inclined at 30° above the horizontal. The tension in the rope is 50 N. Calculate the work done by the tension.

W = F s cos θ = 50 × 12 × cos 30° = 50 × 12 × 0.866 = 520 J

Only the horizontal component of the tension (50 cos 30° = 43.3 N) does work on the sledge; the vertical component is perpendicular to the motion and does none.

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Kinetic Energy

The kinetic energy (KE) of a body of mass m moving at speed v is:

E_k = ½ m v²

Derivation: consider a body accelerated from rest by a constant force F through distance s. Work done on it is W = Fs. Using v² = u² + 2as with u = 0, we get s = v²/(2a). Substituting:

W = F × v²/(2a) = (F/a) × v²/2 = m × v²/2 = ½ m v²

This is the work–energy theorem: the work done by the resultant force on a body equals its change in kinetic energy.

W_net = ΔE_k

Worked Example — Braking Car

A 1500 kg car travelling at 25 m s⁻¹ brakes to rest. Find (a) its initial KE and (b) the work done by the brakes.

(a) E_k = ½ × 1500 × 25² = ½ × 1500 × 625 = 469 000 J = 469 kJ

(b) Work done by brakes = −ΔE_k = −(0 − 469 kJ) = −469 kJ. The negative sign indicates the brakes transfer energy away from the car (it ends up as heat in the pads and discs, plus some noise).

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Gravitational Potential Energy (GPE)

Near the Earth's surface, the gravitational potential energy of a mass m raised by height h above a reference level is:

E_p = m g h

Derivation: lifting mass m at constant velocity requires an upward force of mg. Over a height h the work done against gravity is mgh, which is stored as gravitational PE.

The choice of reference level is arbitrary — only changes in PE are physically meaningful. You are free to pick "zero PE" at the ground, the top of a table, or sea level, whatever makes the problem simplest.

Exam Tip: E_p = mgh is only valid near the Earth's surface where g is approximately constant. For problems involving significant changes in height (such as satellites), you must use the full formula E_p = −GMm/r, introduced in the gravitational fields topic.

Worked Example — Lifted Bag

A 5.0 kg bag is lifted 1.5 m onto a shelf. How much gravitational PE does it gain?

E_p = mgh = 5.0 × 9.81 × 1.5 = 73.6 J