PHYSICS · LESSON 06 OF 9
Work energy and power
Work transfers energy. Kinetic and potential energy trade places, and the total is conserved unless friction or other forces add or remove energy. Power is the rate of energy transfer.
What this lesson explains
Energy methods solve problems that are awkward with forces alone: the speed of a roller coaster at the bottom of a hill, the height a spring launches a ball, the power a motor needs to lift a load. Energy is also the currency of engineering: efficiency, losses, fuel and electricity costs are all energy bookkeeping.
Before you begin
Components and the cosine
The component of F along the displacement is F cos θ. cos 0° = 1, cos 90° = 0, cos 180° = −1.
Kinematics link
v² = v₀² + 2aΔx multiplied by m/2 gives ½mv² − ½mv₀² = maΔx = FΔx: the work–energy theorem for a constant force.
The idea, made visible
Work is how a force transfers energy to or from an object as it moves: W = Fd cos θ, where θ is the angle between the force and the displacement. Only the component of the force along the motion does work. A force perpendicular to the motion (like the normal force on a level floor) does no work. A force opposing the motion (friction) does negative work: it removes energy.
Kinetic energy K = 1 / 2mv2 is the energy of motion. The work–energy theorem says the net work done on an object equals its change in kinetic energy: Wnet = ΔK. Note the v²: doubling the speed quadruples the kinetic energy, which is why stopping distances grow so quickly with speed.
Some forces — gravity and ideal springs — store the work done against them as potential energy: Ug = mgh (height measured from any chosen reference level) and Us = 1 / 2kx2. These are conservative forces: the work they do depends only on the start and end positions, not the path.
If only conservative forces do work, mechanical energy is conserved: K + U stays constant. If friction or other non-conservative forces act, Ki + Ui + Wother = Kf + Uf; friction’s negative work becomes thermal energy. Energy is never destroyed, only transferred or transformed.
Power is the rate of doing work or transferring energy: P = W/Δt, measured in watts (1 W = 1 J/s). For a force along the velocity, P = Fv. Efficiency compares useful output with input: η = Pout/Pin, always less than 100% in real machines.
| height h (m) | U = mgh (J) | K = 98 − U (J) | speed v = √(2K/m) (m/s) |
|---|---|---|---|
| 5 | 98 | 0 | 0 |
| 2.5 | 49 | 49 | 7.0 |
| 1 | 19.6 | 78.4 | 8.85 |
| 0 | 0 | 98 | 9.90 |
Speed is not proportional to the distance fallen: halfway down the ball already has 71% of its final speed.
Key terms
- Work W
- Energy transferred by a force acting through a displacement: W = Fd cos θ (constant force). Unit: joule, 1 J = 1 N·m. Scalar, can be negative.
- Kinetic energy K
- K = 1 / 2mv2; never negative.
- Gravitational potential energy
- Ug = mgh near the Earth’s surface, with h measured from a chosen reference level. Only changes ΔU matter.
- Elastic potential energy
- Us = 1 / 2kx2 for an ideal spring stretched or compressed by x from its natural length.
- Mechanical energy
- E = K + U.
- Power P
- Rate of energy transfer, P = W / Δt; unit watt (W). 1 kW·h = 3.6 × 106 J.
The formulas and what they mean
| Symbol | Meaning | Unit |
|---|---|---|
| W | work | J |
| K, U | kinetic, potential energy | J |
| h | height above the reference level | m |
| k | spring constant | N/m |
| x | spring extension or compression | m |
| P | power | W = J/s |
| η | efficiency | % or fraction |
Work by a constant force
W = Fd cos θ
Conditions and limits: Force constant in size and direction over the displacement d; θ is the angle between F and the displacement. For a varying force, W is the area under the F–x graph.
Work–energy theorem
Wnet = ΔK = 1 / 2mvf2 − 1 / 2mvi2
Conditions and limits: Always true for a particle; W_net is the work of all forces together.
Conservation of mechanical energy
Ki + Ui = Kf + Uf
Conditions and limits: Only conservative forces (gravity, ideal springs) do work. Otherwise include Wother: Ki + Ui + Wother = Kf + Uf.
Spring force and energy
F = −kx; Us = 1 / 2kx2
Conditions and limits: Ideal spring within its elastic limit (Hooke’s law).
Power
P = W / Δt; P = Fv
Conditions and limits: P = Fv for a force parallel to the velocity; instantaneous power.
Why it works: Speed at the bottom of a frictionless slope does not depend on its shape
An object starts from rest at height h and slides without friction to the bottom.
Ki + Ui = Kf + Uf: 0 + mgh = 1 / 2mv2 + 0.
Only gravity does work (the normal force is perpendicular to the motion).
v2 = 2gh ⇒ v = √2gh.
The mass cancels.
A steep slope and a gentle one of the same height give the same final speed (the gentle one just takes longer). Energy methods ignore the path; that is their power.
A first worked example
Energy problem-solving
- Choose the system and the initial and final states. Sketch both.
- Choose a reference level for height (Ug = 0 there).
- List K and U in each state. Identify forces that do work but have no potential energy (friction, applied pushes): they enter as Wother.
- Write Ki + Ui + Wother = Kf + Uf and solve.
- Use forces and kinematics instead when you need time or acceleration; energy gives speeds and positions.
Work by a force at an angle
Problem. A 50 N pull at 37° above the horizontal drags a box 4.0 m across a floor. How much work does the pull do?
W = Fd cos θ = 50 × 4.0 × cos 37°.
Only the horizontal component does work.
= 200 × 0.799 = 160 J.
cos 37° ≈ 0.799.
Result: W ≈ 160 J.
What it means: The vertical component (≈ 30 N) does no work because the box does not move vertically.
A different case
Speed from energy conservation
Problem. A ball is dropped from 5.0 m. Find its speed just before hitting the ground (no air resistance).
mgh = 1 / 2mv2.
Potential energy becomes kinetic energy.
v = √2gh = √2 × 9.8 × 5.0 = √98 = 9.90 m/s.
Mass cancels.
Result: v ≈ 9.9 m/s.
What it means: Same result as kinematics (v² = 2gΔy), found without time.
More worked cases
Each case below uses a different skill. Every step and result is shown.
Friction removes energy
Problem. A 2.0 kg block slides down a 3.0 m high ramp from rest and reaches the bottom at 6.0 m/s. How much energy did friction remove?
Initial: Ui = mgh = 2.0 × 9.8 × 3.0 = 58.8 J; Ki = 0.
Reference level at the bottom.
Final: Kf = 1 / 2(2.0)(6.0)2 = 36 J; Uf = 0.
Kinetic energy at the bottom.
Wfriction = Ef − Ei = 36 − 58.8 = −22.8 J.
Non-conservative work equals the change in mechanical energy.
Result: Friction removed 22.8 J (it became thermal energy).
What it means: Without friction the block would have reached √(2 × 9.8 × 3) = 7.67 m/s.
A spring launcher
Problem. A spring with k = 400 N/m is compressed 0.10 m and launches a 0.050 kg ball vertically. How high does the ball rise above its launch point (no losses)?
Spring energy: 1 / 2kx2 = 1 / 2(400)(0.10)2 = 2.0 J.
Stored elastic energy.
At the top all of it is gravitational: mgh = 2.0 J.
K = 0 at the highest point.
h = 2.0 / 0.050 × 9.8 = 4.08 m.
Solve for h.
Result: h ≈ 4.1 m.
What it means: Doubling the compression would quadruple the stored energy and the height.
Power of a motor
Problem. A motor lifts a 100 kg load 12 m at constant speed in 20 s. Find the useful power. If the motor draws 800 W, what is its efficiency?
Work against gravity: W = mgh = 100 × 9.8 × 12 = 11 760 J.
Constant speed: no change in kinetic energy.
P = W / t = 11 760 / 20 = 588 W.
Rate of doing work.
η = 588 / 800 = 0.735.
Useful output over input.
Result: Useful power 588 W; efficiency ≈ 73.5%.
What it means: The other 212 W becomes heat in the motor and gearing.
Average power
Problem. A motor transfers 600 J in 5 s. Find its average power.
P = 600 J / 5 s = 120 W.
Definition of power.
Result: 120 W.
What it means: Energy and power are different: 600 J is an amount; 120 W is how fast it is delivered.
Common misunderstandings
Misunderstanding: “Any force on a moving object does work.”
Correct idea: Only the component along the displacement does work; perpendicular forces (normal force on a level floor, centripetal force) do none.
Misunderstanding: Forgetting the sign of work.
Correct idea: Friction and braking forces do negative work, reducing kinetic energy.
Misunderstanding: Using K = mv² or forgetting to square v.
Correct idea: K = ½mv².
Misunderstanding: Using conservation of mechanical energy when friction acts.
Correct idea: Include W_friction (negative) or the energy lost to heat.
Misunderstanding: Confusing energy (J) with power (W).
Correct idea: Power is energy per unit time. kW·h is a unit of energy.
Keep in mind
- W = Fd cos θWork by a constant force
- Wnet = ΔK = 1 / 2mvf2 − 1 / 2mvi2Work–energy theorem
- Ki + Ui = Kf + UfConservation of mechanical energy
- F = −kx; Us = 1 / 2kx2Spring force and energy
- P = W / Δt; P = FvPower
Scope of this lesson
- Gravitational potential energy mgh assumes heights small compared with the Earth’s radius.
- Work by variable forces (area under F–x graphs) is introduced; general line integrals are beyond this course.
Next: Impulse and momentum. Momentum p = mv is a vector. A force acting for a time changes it (impulse). In a collision with no external force, total momentum is conserved.
Original study text. Sources and credits.