Work and Kinetic Energy
On 14 October 2012, Felix Baumgartner stepped from a capsule at 39,045 m altitude during the Red Bull Stratos project, reaching 376.9 m/s in free fall. His kinetic energy at peak speed was $KE = \frac{1}{2}(80)(376.9)^2 = 5.69\text{ MJ}$ — converted entirely from gravitational potential energy stored at 39 km altitude. This lesson explains exactly how work and kinetic energy are linked by the work-energy theorem.
Practise this lesson
Work and kinetic energy worksheets.
A waiter carries a tray of food horizontally across a restaurant at constant height and constant speed. He exerts an upward force on the tray to stop it from falling. Has he done any work on the tray in the physics sense?
A 20 N force acts at 90° to the direction of motion of an object. The work done is:
Know
- Definition of work: $W = Fs\cos\theta$ (J)
- Definition of kinetic energy: $KE = \frac{1}{2}mv^2$ (J)
- Work-energy theorem: $W_{net} = \Delta KE$
- $\Delta U = mg\Delta h$ for gravitational PE (preview)
Understand
- Why the waiter does zero work on the tray (force perpendicular to displacement)
- Why only the component of force in the direction of motion does work
- Why $W_{net} = \Delta KE$ follows from Newton's Second Law
Can Do
- Calculate work done by a force at any angle
- Calculate kinetic energy from mass and speed
- Apply the work-energy theorem to find final velocity
- Identify when a force does zero work
Core Content
A waiter holds a full tray at shoulder height and walks the full length of a restaurant. He is clearly exerting effort — his arm muscles ache. But the tray never moves up or down; the force holding it up is vertical while the displacement is horizontal. When you check the angle between force and displacement, you find they are 90° apart — so no energy was transferred to the tray at all.
where $F$ is the force magnitude, $s$ is the displacement magnitude, and $\theta$ is the angle between the force and the displacement vectors. The SI unit is the joule (J).
If $\theta = 0°$: $W = Fs$ (maximum work — force and motion in same direction). If $\theta = 90°$: $W = 0$ (no work — force perpendicular to motion). If $\theta = 180°$: $W = -Fs$ (negative work — force opposes motion, e.g. friction).
The waiter's answer: The tray moves horizontally. The waiter's force is vertical (upward). Angle = 90°. $\cos90° = 0$. Work done = 0 J.
A person pushes a 15 kg box with a force of 80 N at 30° below horizontal across a 5 m horizontal floor. Calculate the work done by the push force.
Work: $W = Fs\cos\theta$ (J), where $\theta$ is the angle between the force and displacement; $\theta = 0°$ gives maximum positive work, $\theta = 90°$ gives zero work (e.g. normal force), and $\theta = 180°$ gives negative work (e.g. friction opposing motion).
Pause — copy the highlighted formula and three angle cases into your book before moving on.
True or false: Friction always does negative work on a sliding object.
We just saw that work is the energy transferred by a force through a displacement. That raises a question: where does that transferred energy go — and how does it connect to an object's motion? This card answers it → net work equals the change in kinetic energy: $W_{net} = \Delta KE$.
Kinetic energy is the energy an object has because it is moving. The work-energy theorem links work done to change in kinetic energy — they are two descriptions of the same physical change.
The work-energy theorem can be derived directly from Newton's Second Law. If a constant net force $F_{net}$ acts over displacement $s$, then $W_{net} = F_{net} \cdot s = ma \cdot s$. Using kinematics: $v^2 = u^2 + 2as$, so $as = (v^2 - u^2)/2$. Substituting: $W_{net} = m \cdot (v^2 - u^2)/2 = \Delta(\frac{1}{2}mv^2) = \Delta KE$.
A 1200 kg car starts from rest and a net force of 3600 N acts over 50 m. Find the final speed.
Kinetic energy: $KE = \frac{1}{2}mv^2$ (J); work-energy theorem: $W_{net} = \Delta KE = KE_f - KE_i$; positive $W_{net}$ → object speeds up, negative $W_{net}$ → object slows down, zero $W_{net}$ → constant speed.
Pause — copy the highlighted formulae and sign rules into your book before moving on.
A 5 kg object moving at 6 m/s is acted on by a net force that does 90 J of work. Its final speed is:
The normal force on a horizontally moving crate does ______ work because the angle between the force and displacement is ______.
Activities
Calculate the following:
- Work done by a 120 N force at 45° over 8 m
- KE of a 3 kg ball moving at 10 m/s
- Using the work-energy theorem: if a 2 kg object starts at rest and net work of 64 J is done on it, find the final speed
UnderstandBand 3(3 marks) 1. A person pushes a heavy box across a horizontal floor at constant speed. Explain why the person does work on the box but the box does not gain kinetic energy.
ApplyBand 4(3 marks) 2. A 1500 kg car accelerates from 10 m/s to 25 m/s. Calculate the work done on the car using the work-energy theorem.
AnalyseBand 5(4 marks) 3. A 60 N force acts on a 4 kg block at 37° to the horizontal. The block slides 10 m along a horizontal surface. A friction force of 12 N acts throughout. Calculate: (a) work done by applied force, (b) work done by friction, (c) net work, (d) final speed if block starts from rest. (cos 37° = 0.80)
Show all answers
Short Answer — Model Answers
Q1 (3 marks): The person does work because they exert a force on the box in the direction of its displacement. However, at constant speed the box is in dynamic equilibrium — the net force is zero. The work done by the push force is exactly cancelled by the negative work done by friction. The net work is zero, so $W_{net} = \Delta KE = 0$, meaning kinetic energy doesn't change. The energy input by the person is entirely dissipated as heat by friction.
Q2 (3 marks): $KE_i = \frac{1}{2} \times 1500 \times 10^2 = 75\,000\text{ J}$. $KE_f = \frac{1}{2} \times 1500 \times 25^2 = 468\,750\text{ J}$. $W = \Delta KE = 468\,750 - 75\,000 = 393\,750\text{ J} \approx 394\text{ kJ}$.
Q3 (4 marks): (a) $W_{applied} = 60 \times 10 \times \cos37° = 60 \times 10 \times 0.80 = 480\text{ J}$. (b) $W_{friction} = -12 \times 10 = -120\text{ J}$ (negative since friction opposes motion). (c) $W_{net} = 480 + (-120) = 360\text{ J}$. (d) $W_{net} = \frac{1}{2}mv^2 - 0$; $360 = \frac{1}{2} \times 4 \times v^2 = 2v^2$; $v^2 = 180$; $v = \sqrt{180} \approx 13.4\text{ m/s}$.
The waiter does zero physics-work on the tray — force is up, motion is horizontal, angle = 90°, $\cos90° = 0$. Felix Baumgartner's 2012 Red Bull Stratos jump reinforces the work-energy theorem: all 5.69 MJ of kinetic energy at peak speed came from the work done by gravity over 39,045 m of fall, with no horizontal force doing any work on him during free fall.