Put the relationship ½mv² − ½mu² = Fs into words.

The work done by the resultant force acting on a body is equal to the change in the body's kinetic energy.

Under what conditions may ½mv² − ½mu² = Fs be applied?

It applies when the resultant force is constant in magnitude and direction and acts along the straight line in which the body moves.

Show that ½mv² − ½mu² = Fs follows from the equations of motion and Newton's second law.

From v² = u² + 2as, multiplying through by ½m gives ½mv² = ½mu² + mas, and since the resultant force is F = ma this becomes ½mv² − ½mu² = Fs.

A 1500 kg car accelerates from 8.0 m s⁻¹ to 20 m s⁻¹ over a distance of 100 m. Calculate the constant resultant force acting on it.

The change in kinetic energy is ½ × 1500 × (400 − 64) = 2.52 × 10⁵ J, so F = 252 000 ÷ 100 = 2.52 × 10³ N, and 2520 N × 100 m gives back 2.52 × 10⁵ J as a check.

CCEA A-Level Physics 1210

Physical Quantities, Scalars and Vectors

Topic overview Physical quantities and SI base units Prefixes and standard form Scalars and vectors Resolving vectors and finding resultants Coplanar forces in equilibrium

Moments and Equilibrium

Topic overview Moment of a force Centre of gravity The principle of moments Determining unknown masses using moments

Linear Motion and Dynamics

Topic overview Displacement, velocity and acceleration The equations of motion for uniform acceleration Measuring the acceleration of free fall with light gates Displacement–time and velocity–time graphs Projectile motion Projectile calculations

Newton's Laws of Motion

Topic overview Newton's first law and inertia Newton's second law and F = ma Applying Newton's laws to simple situations Newton's third law Friction as a force opposing motion

Linear Momentum and Impulse

Topic overview Momentum and its calculation Conservation of linear momentum Collisions in one dimension Elastic and inelastic collisions Impulse and the impulse–momentum equation Impacts and reducing the force of a collision Newton's second law in terms of momentum

Work, Energy and Power

Topic overview Work done and energy transfer Calculating work done, including forces at an angle Gravitational potential energy Kinetic energy Conservation of energy and energy exchanges Work done by a resultant force and change in kinetic energy Power, efficiency and energy use in society

Current, Charge and Potential Difference

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Resistance and Resistivity

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Internal Resistance and Potential Dividers

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Waves

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Superposition, Interference and Diffraction

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Wave-Particle Duality

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Astronomy

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Practical Skills and Data Analysis

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Deformation of Solids

Topic overview Hooke's law and the spring constant Elastic and plastic deformation and the elastic limit Stress, strain and the Young modulus Measuring the Young modulus of a wire Strain energy and force–extension graphs Choosing materials: strength, stiffness and cost

Thermal Physics

Topic overview Boyle's law, Charles's law and the pressure law The ideal gas equation pV = nRT The Avogadro constant, the mole and pV = NkT Internal energy and the distribution of molecular energy The kinetic theory equation pV = ⅓Nm⟨c²⟩Average molecular kinetic energy and temperature Absolute zero and the kelvin scale Specific heat capacity and the electrical method

Uniform Circular Motion

Topic overview Angular velocity, period and frequency Linear speed and angular velocity: v = rω Centripetal acceleration and centripetal force Applying F = mv²/r to motion at constant speed

Simple Harmonic Motion

Topic overview Defining simple harmonic motion and a = −ω²x Displacement, velocity and acceleration: x = A cos ωt The simple pendulum and T = 2π√(l/g)The loaded spiral spring and T = 2π√(m/k)SHM graphs and finding velocity from a gradient Free vibrations, forced vibrations and resonance Damping: light, critical and over-damping

The Nucleus and Radioactive Decay

Topic overview Alpha-particle scattering and the nuclear atom Nuclear radius and nucleon number The density of nuclear matter Alpha, beta and gamma radiation: nature, penetration and range Nuclear equations and changes in nucleon and proton number The random and exponential nature of radioactive decay Activity, the decay constant and A = −λN Exponential decay calculations with A = A₀e^(−λt)Half-life and measuring it

Nuclear Energy, Fission and Fusion

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Gravitational Fields

Topic overview Fields of force and field lines Gravitational field strength Newton's law of universal gravitation The field strength of a point mass Circular motion of planets and satellites Kepler's third law Geostationary satellites

Electric Fields

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Capacitors

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Magnetic Fields

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Charged Particles and Accelerators

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Fundamental Particles

Topic overview Fundamental particles The four fundamental forces Classifying particles Quarks and the structure of hadrons Conservation laws Beta decay in the quark model

Work, Energy and Power

Work done by a resultant force and change in kinetic energy

3 things to know
Unit AS 1

Overview

Recall questions on the equation ½mv² − ½mu² = Fs in words and in symbols, its derivation from suvat and the second law, when it may be applied, what a negative Fs means, and worked problems on an accelerating car, braking, an arrow and a bullet brought to rest.

What you need to know

Answer each one before you reveal it.

Your recallNot started

Key term

State the equation relating the work done by a constant resultant force to the change in kinetic energy, and define every symbol.

Numbers

A 900 kg car travelling at 30 m s⁻¹ is brought to rest by a constant braking force of 6.0 kN. Calculate the braking distance.

Key term

Why must F in ½mv² − ½mu² = Fs be the resultant force rather than just the applied force?

Questions from this subtopic

Each one with the model answer underneath.

1
State the equation relating the work done by a constant resultant force to the change in kinetic energy, and define every symbol.
Model answer
The equation is ½mv² − ½mu² = Fs, where m is the mass in kilograms, u the initial speed and v the final speed in metres per second, F the constant resultant force in newtons and s the distance moved in metres.
2
Put the relationship ½mv² − ½mu² = Fs into words.
Model answer
The work done by the resultant force acting on a body is equal to the change in the body's kinetic energy.
3
Under what conditions may ½mv² − ½mu² = Fs be applied?
Model answer
It applies when the resultant force is constant in magnitude and direction and acts along the straight line in which the body moves.
4
Show that ½mv² − ½mu² = Fs follows from the equations of motion and Newton's second law.
Model answer
From v² = u² + 2as, multiplying through by ½m gives ½mv² = ½mu² + mas, and since the resultant force is F = ma this becomes ½mv² − ½mu² = Fs.
5
A 1500 kg car accelerates from 8.0 m s⁻¹ to 20 m s⁻¹ over a distance of 100 m. Calculate the constant resultant force acting on it.
Model answer
The change in kinetic energy is ½ × 1500 × (400 − 64) = 2.52 × 10⁵ J, so F = 252 000 ÷ 100 = 2.52 × 10³ N, and 2520 N × 100 m gives back 2.52 × 10⁵ J as a check.
6
A 900 kg car travelling at 30 m s⁻¹ is brought to rest by a constant braking force of 6.0 kN. Calculate the braking distance.
Model answer
The kinetic energy to be removed is ½ × 900 × 900 = 4.05 × 10⁵ J, so s = 405 000 ÷ 6000 = 67.5 m, and 6000 N × 67.5 m = 4.05 × 10⁵ J, confirming the result.

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