Gr 11 · Work, Energy & Power · Deep Study

Understanding
Energy

Energy is the capacity to do work. This guide builds from the definition of work through kinetic and potential energy, conservation principles (including the pendulum), and on to power and efficiency.

5 sections Worked examples with steps Checkpoint questions Intuition-first explanations
1
Work
How energy is transferred by a force acting over a distance

Work is how energy is transferred. Pushing a box horizontally while a force acts upward does no useful work — only the component along the direction of motion counts. This is why W = F·d·cosθ: cos θ extracts exactly the component of force that is parallel to the displacement.

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Why does only the parallel component matter?
Imagine pulling a suitcase at an angle. The upward component of your pull lifts the case slightly — it doesn't contribute to moving it forward. Only the horizontal component does work against friction and moves the suitcase along the floor. The perpendicular component merely changes the normal force.
Work = Force × Displacement × cos θ m F F cosθ θ d W = F·d·cosθ m F θ=90° d W = 0
Left: force at angle θ — only the horizontal component F cosθ does work along d. Right: force perpendicular to motion (θ = 90°) — W = 0.
Work formula
W = F · d · cos θ
θ = angle between the force vector and the displacement · units: Joules (J)
Work-energy theorem
W_net = ΔEk = Ek_f − Ek_i
Net work done on an object equals its change in kinetic energy
Easy
Horizontal force, flat surface
A 50 N force is applied horizontally, pushing a box 6 m on a flat floor. How much work is done?
Show solution
1
Identify the angle

Force is horizontal, displacement is horizontal. θ = 0°, so cos 0° = 1.

2
Apply the formula
W = F · d · cos θ = 50 × 6 × cos 0° = 50 × 6 × 1 = 300 J
Answer: W = 300 J
Intermediate
Force at an angle
An 80 N force is applied at 35° above horizontal. The box moves 10 m horizontally. How much work does this force do? What work does the normal force do?
Show solution
1
Work done by applied force
W = 80 × 10 × cos 35° = 800 × 0.819 = 655 J
2
Work done by normal force

The normal force acts vertically upward. The displacement is horizontal. θ = 90°, cos 90° = 0.

W_normal = F_N × d × cos 90° = 0 J
Answer: W_applied = 655 J · W_normal = 0 J
Checkpoint 1
a) A 40 N force is applied horizontally to push a box 8 m. What is the work done?
b) A 60 N force is applied at 45° above horizontal. The object moves 5 m. What is the work done by the force?
c) You carry a box horizontally at constant height across a room. How much work does gravity do on the box?

a) W = 40 × 8 × cos 0° = 320 J

b) W = 60 × 5 × cos 45° = 300 × 0.707 = 212 J

c) Zero. Gravity acts downward; displacement is horizontal. θ = 90°, so W = F·d·cos 90° = 0 J. No work is done by gravity when there is no vertical displacement.

2
Kinetic Energy
The energy of motion — and why speed matters so much

Kinetic energy depends on the square of speed — doubling speed quadruples Ek. This non-linear relationship has major consequences: a car at 100 km/h has four times the kinetic energy of one at 50 km/h, which is why stopping distances increase dramatically at higher speeds. The work-energy theorem connects work and kinetic energy directly.

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Why does Ek depend on v², not v?
When you accelerate an object, the distance it travels while accelerating also increases with speed. Work = force × distance — and since distance grows with v while force is constant, the total work (and thus energy stored) grows as v². This is a fundamental consequence of kinematics, not an arbitrary formula.
Kinetic Energy Scales as v² — Doubling Speed Quadruples Ek v = 0 Ek = 0 at rest v Ek Ek = ½mv² speed v 2v 4×Ek 4 × taller Ek = 4·½mv² speed 2v
Each bar represents Ek. At v the bar has height 1 unit; at 2v the bar is 4 units tall — the v² relationship makes kinetic energy grow much faster than speed.
Kinetic energy
Ek = ½mv²
m in kg, v in m/s, Ek in Joules
Easy
Finding kinetic energy
A 3 kg ball is moving at 8 m/s. What is its kinetic energy?
Show solution
1
Apply the formula
Ek = ½mv² = ½ × 3 × 8² = ½ × 3 × 64 = 96 J
Answer: Ek = 96 J
Intermediate
Work done by brakes
A 1200 kg car brakes from 20 m/s to 8 m/s. How much work did the brakes do?
Show solution
1
Use the work-energy theorem: W_net = ΔEk
ΔEk = ½m(v_f² − v_i²) = ½ × 1200 × (8² − 20²)
2
Calculate
= ½ × 1200 × (64 − 400) = 600 × (−336) = −201 600 J = −201.6 kJ
3
Interpret the sign

Negative work means the brakes removed energy from the car, converting it to heat. The brakes did −201.6 kJ of work.

Answer: W_brakes = −201.6 kJ (energy removed from car)
Checkpoint 2
a) A 5 kg box moves at 6 m/s. What is its kinetic energy?
b) An 80 J net work is done on a 2 kg object that starts from rest. What is its final speed?
c) Why does doubling speed quadruple kinetic energy?

a) Ek = ½ × 5 × 6² = ½ × 5 × 36 = 90 J

b) W = ΔEk = ½mv² − 0 → 80 = ½ × 2 × v² → v² = 80 → v = √80 ≈ 8.94 m/s

c) Ek = ½mv². If v doubles to 2v, then Ek = ½m(2v)² = ½m·4v² = 4(½mv²). The squaring of v means doubling v multiplies Ek by 4.

3
Potential Energy
Stored energy — gravitational and elastic

Potential energy is stored energy that can be converted to kinetic energy. Gravitational PE depends on height above a reference — you choose the reference. Spring PE stores energy when a spring is deformed from its natural length.

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Why is the reference level arbitrary?
Only differences in height matter, not absolute height. A book on a table has higher Ep relative to the floor, but lower Ep relative to the ceiling. Since we always use ΔEp = mgΔh in energy conservation, the reference cancels out. Pick the level that makes h = 0 at the most convenient point (usually the lowest point in the problem).
Gravitational PE (left) and Elastic PE (right) reference level (h = 0) m h Ep = mgh g ↓ m x F = kx Ee = ½kx² compressed spring
Left: mass m lifted height h above the reference — Ep = mgh. Right: spring compressed by x — elastic PE = ½kx², restoring force F = kx acts on the block.
Gravitational PE
Ep = mgh
Elastic PE (spring)
Ee = ½kx²
Hooke's Law
F = kx
Easy
Gravitational PE of a lifted book
A 4 kg book is lifted 2.5 m above the floor. What is its gravitational PE?
Show solution
1
Apply the formula (reference = floor, h = 2.5 m)
Ep = mgh = 4 × 9.8 × 2.5 = 98 J
Answer: Ep = 98 J
Intermediate
Spring elastic PE and force
A spring with k = 400 N/m is compressed 0.15 m. What is the elastic PE stored? What force is required to hold it compressed?
Show solution
1
Elastic PE
Ee = ½kx² = ½ × 400 × (0.15)² = 200 × 0.0225 = 4.5 J
2
Force required (Hooke's Law)
F = kx = 400 × 0.15 = 60 N
Answer: Ee = 4.5 J · F = 60 N
Checkpoint 3
a) A 70 kg person climbs 12 m. What is the gain in gravitational PE?
b) A spring with k = 250 N/m is compressed 0.2 m. What is the elastic PE?
c) If the reference level is the table top instead of the floor, how does the Ep of a book sitting on the table change?

a) Ep = mgh = 70 × 9.8 × 12 = 8232 J = 8.232 kJ

b) Ee = ½kx² = ½ × 250 × (0.2)² = 125 × 0.04 = 5 J

c) The book's Ep becomes zero. When the reference level is set at the table top, h = 0 for the book. Ep = mgh = mg × 0 = 0. Only differences in height between two positions matter in energy calculations, so the choice of reference doesn't affect those differences.

4
Conservation of Energy
The most powerful principle in physics — and the pendulum

In a closed system with no friction, total mechanical energy (Ek + Ep) is constant. Energy flows between kinetic and potential forms, but the total never changes. When friction exists, mechanical energy decreases — converted to thermal energy. The total energy of the universe is still conserved; we just lose useful mechanical energy.

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The pendulum is a perfect example: at the top of the swing, all energy is potential (v = 0). At the bottom, all energy is kinetic (h = 0). The total stays the same throughout. This is why mgh = ½mv² at those two extremes — mass cancels, giving v = √(2gh).
Conservation of Energy — Pendulum A B C h = 0 ref Ep Ek=0 E total all Ep Ep=0 Ek E total all Ek Ep Ek=0 E total all Ep E_total = Ep + Ek = constant at every point At A and C: v = 0, max Ep. At B: h = 0, max Ek.
Pendulum swings from A (all potential) through B (all kinetic) to C (all potential again). The dashed outline shows total mechanical energy is constant throughout — nothing is gained or lost.
No friction
Ek_i + Ep_i = Ek_f + Ep_f
With friction
E_final = E_initial − W_friction
Pendulum height
h = L(1 − cos θ)
Pendulum speed at bottom
v = √(2gh)
Easy — Conservation
Ball dropped from rest
A 0.5 kg ball is dropped from 3 m. What is its speed just before hitting the ground?
Show solution
1
Set up conservation of energy

At the top: Ek = 0 (dropped from rest), Ep = mgh. At the bottom: Ep = 0 (reference), Ek = ½mv².

mgh = ½mv²    (mass cancels)
2
Solve for v
gh = ½v² → v² = 2gh = 2 × 9.8 × 3 = 58.8
v = √58.8 ≈ 7.67 m/s
Answer: v ≈ 7.67 m/s
Intermediate — Pendulum
Pendulum speed at bottom
A 0.8 m pendulum is released from 25°. What is its speed at the bottom of the swing?
Show solution
1
Find the height h above the bottom
h = L(1 − cos θ) = 0.8 × (1 − cos 25°) = 0.8 × (1 − 0.906) = 0.8 × 0.094 = 0.0752 m
2
Apply mgh = ½mv² (mass cancels)
v = √(2gh) = √(2 × 9.8 × 0.0752) = √1.474 ≈ 1.21 m/s
Answer: v ≈ 1.21 m/s at the bottom
Intermediate — Friction
Energy lost to friction on a ramp
A 3 kg box slides 5 m down a ramp from h = 2 m and arrives at the bottom with v = 3.5 m/s. How much energy was lost to friction?
Show solution
1
Total energy at top
E_top = mgh = 3 × 9.8 × 2 = 58.8 J   (Ek = 0 at top)
2
Kinetic energy at bottom
Ek_bottom = ½mv² = ½ × 3 × (3.5)² = 1.5 × 12.25 = 18.375 J
3
Energy lost to friction
W_friction = E_top − Ek_bottom = 58.8 − 18.375 ≈ 40.4 J
Answer: ≈ 40.4 J converted to heat by friction
Checkpoint 4
a) A 2 kg ball is dropped from 5 m. What is its speed just before hitting the ground?
b) A pendulum with L = 1.2 m is released from 40°. What is its speed at the bottom?
c) Why does a real pendulum eventually stop even though energy is "conserved"?

a) mgh = ½mv² → v = √(2 × 9.8 × 5) = √98 ≈ 9.9 m/s

b) h = 1.2(1 − cos 40°) = 1.2(1 − 0.766) = 1.2 × 0.234 = 0.2808 m
v = √(2 × 9.8 × 0.2808) = √5.503 ≈ 2.35 m/s

c) A real pendulum experiences air resistance and friction at the pivot. These forces do negative work on the pendulum, converting mechanical energy to thermal energy. Energy is still conserved in the universe — it just leaves the mechanical system as heat. The total energy decreases with each swing until it reaches zero.

5
Power & Efficiency
How fast energy is transferred, and how much is actually useful

Power tells you how fast energy is transferred. Two workers can do the same work — the one who finishes faster has greater power. Efficiency measures how much input energy becomes useful output. No real machine is 100% efficient: friction, heat, and sound always claim some energy.

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Why can P = Fv?
P = W/t = (F·d)/t = F·(d/t) = F·v. This form is useful when you know a constant force and the speed at which it acts — for example, an engine maintaining a constant thrust at a given speed.
Power = Work ÷ Time | Efficiency = Useful Output ÷ Input mg h time = t P₁ = mgh/t Person 1 (slow) mg time = t/2 P₂ = 2P₁ Person 2 (fast) Same W = mgh — different time → different power E_in E_useful heat loss E_wasted eff = E_useful / E_in × 100% Always less than 100% — losses are unavoidable E_in = E_useful + E_wasted
Left: both people do the same work mgh, but Person 2 does it in half the time — twice the power. Right: efficiency funnel — input energy splits into useful output and unavoidable waste heat.
Power (basic)
P = W / t  (Watts)
Power (force)
P = Fv
Efficiency
eff = (E_useful / E_input) × 100%
Easy
Calculating power from work and time
300 J of work is done in 15 s. What is the power?
Show solution
1
Apply P = W/t
P = 300 / 15 = 20 W
Answer: P = 20 W
Intermediate
Power climbing stairs + efficiency
A 70 kg person runs up 4 m of stairs in 3.2 s. What is their mechanical power output? If the motor driving an equivalent lift is 85% efficient, what input power does the motor require?
Show solution
1
Work done against gravity
W = mgh = 70 × 9.8 × 4 = 2744 J
2
Mechanical power output
P_output = W / t = 2744 / 3.2 = 858 W
3
Required input power for 85% efficient motor
P_input = P_output / efficiency = 858 / 0.85 = 1009 W ≈ 1.01 kW
Answer: P_output = 858 W · P_input (motor) = 1009 W
Checkpoint 5
a) A motor does 6000 J of work in 40 s. What is its power in watts?
b) A 65 kg person climbs a 6 m ladder in 5 s. What is their power in watts and in horsepower (1 hp = 746 W)?
c) A machine receives 500 J of input energy and delivers 375 J of useful output energy. What is its efficiency?

a) P = W/t = 6000 / 40 = 150 W

b) W = mgh = 65 × 9.8 × 6 = 3822 J
P = 3822 / 5 = 764.4 W
In horsepower: 764.4 / 746 ≈ 1.02 hp

c) eff = (375 / 500) × 100% = 75%