04Work, Energy, and Machines
Calculate work, kinetic and potential energy, and power; apply the work-energy theorem and conservation of mechanical energy; analyze simple machines and efficiency.
Energy is the currency of physics. Conservation of energy connects mechanics, thermodynamics, and every other branch of physics you will study.
Essential Question: How does energy transform between different forms, and how do simple machines allow us to do the same work with less force?
Lesson Overview
Work is done whenever a force moves an object through a displacement. The work-energy theorem tells us that the net work done on an object equals its change in kinetic energy. Energy can also be stored as gravitational potential energy (due to height) or elastic potential energy (due to compression or stretching). When only conservative forces act, the total mechanical energy — kinetic plus potential — remains constant. Power measures how quickly work is done. Simple machines like levers, pulleys, and inclined planes do not reduce the total work required, but they change the force needed by trading it for a longer distance. Mechanical advantage quantifies this trade-off, while efficiency tells us how much input work is actually converted to useful output work.
Simple Machines
Lever
MA = effort arm ÷ load arm
Fulcrum position determines MA; three classes based on fulcrum, effort, and load positions.
Pulley
MA = number of supporting rope segments
A single fixed pulley changes direction only (MA = 1); movable pulleys multiply force.
Inclined Plane
MA = length of slope ÷ height
A longer, gentler ramp gives greater mechanical advantage.
Wheel & Axle
MA = radius of wheel ÷ radius of axle
Effort applied to the larger wheel produces a greater force at the smaller axle.
Screw
MA = 2πr ÷ pitch
Converts rotational motion to linear; smaller pitch = greater MA.
Wedge
MA ≈ length ÷ thickness
Two inclined planes back-to-back; thinner wedge = greater MA.
Key Equations
Work
W = Fd cosθ
F = force (N), d = displacement (m), θ = angle between F and d
Kinetic Energy
KE = ½mv²
m = mass (kg), v = speed (m/s)
Gravitational PE
PE_g = mgh
g = 9.8 m/s², h = height above reference (m)
Elastic PE
PE_e = ½kx²
k = spring constant (N/m), x = compression/stretch (m)
Work-Energy Theorem
W_net = ΔKE
Net work equals change in kinetic energy
Conservation of Energy
KE₁ + PE₁ = KE₂ + PE₂
Valid when only conservative forces act
Power
P = W/t = Fv
P in watts (W), t in seconds, v = velocity (m/s)
Mechanical Advantage
MA = F_out / F_in = d_in / d_out
Ideal MA; no units
Efficiency
η = (W_out / W_in) × 100%
Always ≤ 100% due to friction
Worked Examples
A person pushes a 50 kg crate 10 m along a floor by applying a force of 200 N at an angle of 30° below the horizontal. How much work does the applied force do on the crate?
Identify: F = 200 N, d = 10 m, θ = 30°
Formula: W = Fd cosθ
W = 200 × 10 × cos(30°)
cos(30°) = √3/2 ≈ 0.866
W = 200 × 10 × 0.866
W = 1732 J
A 2 kg ball is dropped from rest at a height of 5 m. Using conservation of mechanical energy, find its speed just before it hits the ground. (Ignore air resistance.)
At the top: KE₁ = 0 (at rest), PE₁ = mgh = 2 × 9.8 × 5 = 98 J
At the bottom: PE₂ = 0 (reference level), KE₂ = ½mv²
Conservation: KE₁ + PE₁ = KE₂ + PE₂
0 + 98 = ½(2)v² + 0
98 = v²
v = √98 ≈ 9.9 m/s
A spring with k = 500 N/m is compressed 0.2 m. (a) Find the elastic potential energy stored. (b) A 0.5 kg block is attached. Find its speed when the spring returns to its natural length (released from rest on a frictionless surface).
Part (a): PE_e = ½kx² = ½ × 500 × (0.2)² = ½ × 500 × 0.04 = 10 J
Part (b): All PE converts to KE when spring is at natural length.
KE = ½mv² = 10 J
½ × 0.5 × v² = 10
0.25v² = 10
v² = 40
v = √40 ≈ 6.32 m/s
An electric motor lifts a 300 kg load to a height of 8 m in 12 seconds. What is the power output of the motor? Express your answer in watts and kilowatts.
Work done against gravity: W = mgh = 300 × 9.8 × 8 = 23 520 J
Power: P = W / t = 23 520 / 12
P = 1960 W
Convert: 1960 W ÷ 1000 = 1.96 kW
A lever has an effort arm of 2 m and a load arm of 0.5 m. (a) Find the ideal mechanical advantage. (b) What effort force is needed to lift an 800 N load? (c) If the lever is only 80% efficient, what is the actual effort force required?
Part (a): MA = effort arm / load arm = 2 / 0.5 = 4
Part (b) Ideal: F_in = F_out / MA = 800 / 4 = 200 N
Part (c) Efficiency: η = W_out / W_in × 100%
80% = W_out / W_in → W_in = W_out / 0.80
W_out = F_out × d_out = 800 × d_out
W_in = F_actual × d_in = F_actual × 4d_out
80% = (800 × d_out) / (F_actual × 4d_out)
0.80 = 800 / (4 × F_actual)
F_actual = 800 / (4 × 0.80) = 800 / 3.2 = 250 N
Guided Problems
A 10 kg box is pulled 6 m along a horizontal floor by a rope that makes a 45° angle with the floor. The tension in the rope is 80 N. How much work does the tension do on the box?
Hint: Use W = Fd cosθ. The angle is between the rope and the direction of motion (horizontal). cos(45°) ≈ 0.707.
A 5 kg object moving at 4 m/s is brought to rest by a braking force over a distance of 2 m. What is the magnitude of the braking force? Use the work-energy theorem.
Hint: W_net = ΔKE = KE_final − KE_initial. KE_initial = ½mv². The braking force does negative work: W = −F × d. Solve for F.
A 3 kg ball is thrown upward with an initial speed of 10 m/s. Using conservation of energy, find the maximum height it reaches. (Ignore air resistance.)
Hint: At maximum height, all kinetic energy converts to gravitational PE. Set ½mv² = mgh and solve for h. Notice that mass cancels.
A machine does 4500 J of useful work in 15 seconds. (a) What is its power output? (b) If the machine consumes 6000 J of input energy in that time, what is its efficiency?
Hint: (a) P = W/t. (b) η = (W_out / W_in) × 100%. Use W_out = 4500 J and W_in = 6000 J.
An inclined plane is 5 m long and 1.25 m high. (a) What is the ideal mechanical advantage? (b) What force is needed to push a 600 N crate up the ramp (assuming ideal conditions)?
Hint: MA = length of slope / height = 5 / 1.25. Then F_in = F_out / MA. The output force equals the weight of the crate.
Key Vocabulary
Work (W)
The product of the force applied to an object, the displacement of the object, and the cosine of the angle between them. Work is done only when a force causes displacement in the direction of the force.
Example: Pushing a box 5 m with 100 N parallel to the floor: W = 100 × 5 × cos0° = 500 J
Kinetic Energy (KE)
The energy an object possesses due to its motion. KE = ½mv², where m is mass in kg and v is speed in m/s. KE is always non-negative.
Example: A 2 kg ball moving at 3 m/s: KE = ½ × 2 × 9 = 9 J
Potential Energy (PE)
Stored energy due to an object's position or configuration. Gravitational PE = mgh; elastic PE = ½kx². Both can be converted to kinetic energy.
Example: A 1 kg book held 2 m high: PE_g = 1 × 9.8 × 2 = 19.6 J
Work-Energy Theorem
The net work done on an object equals its change in kinetic energy: W_net = ΔKE = KE_final − KE_initial. This links forces and motion through energy.
Example: A net force does 50 J of work on a stationary object → the object gains 50 J of KE.
Conservation of Energy
In a closed system with only conservative forces, the total mechanical energy (KE + PE) remains constant. Energy is neither created nor destroyed, only transformed.
Example: A pendulum swings: at the top, PE is maximum and KE = 0; at the bottom, KE is maximum and PE = 0.
Power (P)
The rate at which work is done or energy is transferred. P = W/t = Fv. The SI unit is the watt (W), where 1 W = 1 J/s.
Example: A motor doing 3000 J of work in 10 s: P = 3000/10 = 300 W
Mechanical Advantage (MA)
The ratio of output force to input force for a simple machine: MA = F_out / F_in = d_in / d_out. A MA > 1 means the machine multiplies force.
Example: A lever with effort arm 3 m and load arm 1 m: MA = 3/1 = 3 (triples the force)
Efficiency (η)
The ratio of useful output work to total input work, expressed as a percentage: η = (W_out / W_in) × 100%. Real machines are always less than 100% efficient due to friction.
Example: A pulley system that requires 500 J input to do 400 J of useful work: η = 400/500 × 100% = 80%
Workbook Check — Interactive Quiz
Interactive Practice — 5 Questions
A 50 N force is applied at 60° to the horizontal to push a box 4 m. How much work is done by this force?
A 4 kg object is moving at 6 m/s. What is its kinetic energy?
A ball is released from rest at a height of 10 m. What is its speed just before hitting the ground? (g = 9.8 m/s²)
A machine requires 800 J of input work to perform 600 J of useful output work. What is its efficiency?
A lever has an effort arm of 3 m and a load arm of 0.75 m. What is its ideal mechanical advantage?
Independent Practice
A 500 N force is applied at 30° above horizontal to push a crate 8.0 m across a floor. Calculate the work done by this force.
A 2.0 kg ball is dropped from a height of 5.0 m. Using conservation of energy, find its speed just before hitting the ground. (g = 9.8 m/s²)
A spring with k = 400 N/m is compressed 0.15 m. (a) How much elastic potential energy is stored? (b) If released, what is the maximum speed of a 0.50 kg block attached to it?
A 70 kg person climbs a 10 m staircase in 20 s. Calculate the power output. (g = 9.8 m/s²)
★ A 5.0 kg block starts from rest at the top of a 4.0 m high frictionless ramp, then slides onto a rough horizontal surface with μₖ = 0.25. (a) Find the speed at the bottom of the ramp. (b) How far does the block slide on the rough surface before stopping? (c) What percentage of the original potential energy is lost to friction? (g = 9.8 m/s²)
ChallengeCommon Mistakes
Calculating work as W = Fd without accounting for the angle between force and displacement
W = Fd cosθ where θ is the angle between the force vector and displacement vector. If F ⊥ d, then W = 0
Forgetting that the normal force and gravity do zero work on horizontal motion
Normal force is perpendicular to motion on a flat surface (W = 0). Gravity does work only when there is vertical displacement
Using conservation of energy when friction or air resistance is present without accounting for thermal energy
With friction: KE_i + PE_i = KE_f + PE_f + W_friction where W_friction = f_k × d (energy lost to heat)
Confusing power (rate of energy transfer, W) with energy (J)
Power P = W/t = Fv. A 100 W bulb uses 100 J every second. Energy = Power × time
Math Tips
Work-energy theorem: W_net = ΔKE = ½mv_f² − ½mv_i². This is the most powerful shortcut — use it whenever you need to find speed after a force acts over a distance
Gravitational PE: ΔPE = mgΔh. Only the vertical height change matters — the path taken (ramp, stairs, elevator) is irrelevant
Simple machine golden rule: work in = work out (ideal). F_in × d_in = F_out × d_out. Mechanical advantage MA = F_out/F_in = d_in/d_out
Spring PE = ½kx² where x is the compression/extension from equilibrium. At maximum compression/extension, all KE has converted to spring PE