⚡Work & the Work-Energy Theorem
What is Work?
In physics, work has a precise definition: it's the energy transferred to an object by a force acting over a displacement.
W = F · d · cos θ
Where: - F = force magnitude (N) - d = displacement magnitude (m) - θ = angle between force and displacement
Work is measured in Joules (J) = N·m.
Key cases: - Force parallel to motion (θ = 0°): W = Fd (maximum work) - Force perpendicular to motion (θ = 90°): W = 0 (no work done!) - Force opposite to motion (θ = 180°): W = −Fd (negative work)
Example: A waiter carrying a tray across a restaurant does zero work on the tray! The upward force is perpendicular to the horizontal motion.
Think About It
You push against a wall with all your might for 10 seconds but it doesn't move. Have you done any work on the wall (in the physics sense)?
The Work-Energy Theorem
The Work-Energy Theorem connects work to kinetic energy:
W_net = ΔKE = ½mv² − ½mv₀²
The net work done on an object equals the change in its kinetic energy. This is a powerful result that lets you solve problems without knowing the details of HOW the force was applied.
If positive work is done on an object: it speeds up (gains KE). If negative work is done: it slows down (loses KE). If zero net work: speed doesn't change.
Energy Skate Park
See the work-energy theorem in action: as the skater descends, net work is done and kinetic energy increases.
✏️ Worked Example
Problem: A 2 kg box starts at rest. A 10 N force pushes it 5 m horizontally on a frictionless surface. Find the final speed.
📐 Key Equations
Work and Work-Energy Theorem
W = Fd\cosθW_net = Δ KE = (1)/(2)mv² - (1)/(2)mv_0²⚠️ Common Mistakes
Misconception: Thinking that any force applied to an object automatically does work on it.
✓ Correct thinking: Work requires both a force AND displacement with a component in the direction of the force. If the displacement is zero, or the force is perpendicular to motion, W = 0.
Why: W = Fd cosθ: if θ = 90° then cosθ = 0. This is why the normal force and centripetal force never do work on a moving object.
Misconception: Forgetting that work can be negative (when force opposes motion).
✓ Correct thinking: When force and displacement point in opposite directions (θ = 180°), W = −Fd. Friction, for example, always does negative work on a sliding object.
Why: Negative work reduces kinetic energy. The work-energy theorem, W_net = ΔKE, uses the algebraic sum of all work done, so signs matter.
Misconception: Using the net force in W = Fd instead of the individual force components along displacement.
✓ Correct thinking: Calculate work done by each force separately using W = F·d·cosθ, then sum them for W_net. Or use the net force only if you want W_net directly.
Why: Mixing individual and net forces leads to double-counting. The work-energy theorem connects W_net (from all forces) to ΔKE.
📝 Practice Problems
Try these problems. Check your answer when ready.
A 5 kg crate is pushed 4 m along a frictionless floor by a horizontal 20 N force. How much work is done on the crate?
W = Fd\cosθA person carries a 15 kg box horizontally across a room (10 m). How much work does the person's upward support force do on the box? (g = 10 m/s²)
A 3 kg object starts at rest and a net force does 54 J of work on it. What is its final speed?
W_net = (1)/(2)mv² - 0A 10 N force is applied at 60° to the horizontal to push a 2 kg box 6 m along a frictionless floor. Find (a) work done by the applied force and (b) the final speed if it starts from rest.
W = Fd\cosθA 4 kg block is moving at 6 m/s. A friction force of 8 N acts on it over 6 m. Find the final speed.
W_net = Δ KEAn elevator motor lifts a 500 kg elevator 30 m at constant speed. How much work does the motor do? (g = 10 m/s²)
W = Fd\cosθA 2 kg ball is thrown horizontally at 10 m/s from a 5 m high cliff. Using the work-energy theorem, find its speed just before it hits the ground. (g = 10 m/s²)
W_gravity = mgh = Δ KEFinished reading through this lesson?