Power — average and instantaneous, units: Worked examples

This is a companion to Power — average and instantaneous, units. If any symbol here feels unfamiliar, it was built in the parent note; we only use the tools here.

The scenario matrix

Power via $P = F v cos θ$ has exactly a few "knobs": the sign of $cos θ$ (the angle between force and velocity), whether $v = 0$ (degenerate), whether power is constant or changing in time, and whether the problem is dressed as units/billing or an exam twist. Here is every cell we must cover.

Cell	What makes it special	Sign / behaviour of $P$	Covered by
A. $θ = 0$	force parallel to motion	$P = + F v$ , maximum positive	Ex 1
B. $0 < θ < 90°$	force partly along motion	$P > 0$ but reduced	Ex 2
C. $θ = 90°$	force perpendicular	$P = 0$ (circular motion!)	Ex 3
D. $90° < θ \leq 180°$	force opposes motion	$P < 0$ (braking)	Ex 4
E. $v = 0$ degenerate	nothing is moving	$P = 0$ regardless of $F$	Ex 5
F. Time-varying $P$	$v$ grows, so $P$ grows	$P_{inst} \neq = P_{avg}$	Ex 6
G. Limiting behaviour	max speed of an engine	$P$ fixed $\Rightarrow v_{m a x} = P / F$	Ex 7
H. Units / real-world	billing in kWh	energy = power × time	Ex 8
I. Exam twist	efficiency + two powers	useful/input ratio	Ex 9

The single most important geometric idea is the angle $θ$ , so we look at it first as a picture.

Figure — Power — average and instantaneous, units

Why $cos$ and not $sin$ ? Because we want the component along $v$ . On a right triangle, the side adjacent to $θ$ (the shadow lying flat along $v$ ) over the hypotenuse ( $F$ ) is exactly $cos θ$ . The perpendicular part ( $F sin θ$ ) does no work per second — it only bends the path.

Cell A — force fully along motion ( $θ = 0$ )

Cell B — force at an oblique angle ( $0 < θ < 90°$ )

Worked example Ex 2 · Child pulling a sled by a rope

A child pulls a sled with a $50$ N force along a rope tilted $60°$ above the horizontal ground. The sled slides horizontally at $2$ m/s. Power delivered to the sled? (Cell B)

Forecast: Only the horizontal shadow of the force does work, so power should be less than $50 \times 2 = 100$ W. Guess by how much.

The velocity is horizontal; the force is $60°$ above it, so $θ = 60°$ . Why this step? $θ$ is measured between $F$ and $v$ , not from the ground arbitrarily — but here the ground is the velocity direction.
Component of force along motion: $F cos θ = 50 cos 60° = 50 \times 0.5 = 25$ N. Why this step? The vertical part ( $F sin 60°$ ) only lifts against the ground normal — it moves nothing along $v$ , so it does no work per second.
$P = (F cos θ) v = 25 \times 2 = 50 W$ .

Verify: $50$ W is exactly half of the "naive" $100$ W, matching $cos 60° = 0.5$ . ✓ Positive and reduced, as forecast.

Cell C — force perpendicular to motion ( $θ = 90°$ )

Worked example Ex 3 · Ball on a string (circular motion)

A ball whirls in a horizontal circle at constant speed $3$ m/s. The string tension is $12$ N, always pointing to the centre. What power does the tension deliver? (Cell C)

Forecast: The string pulls sideways to the motion at every instant. Does it speed the ball up? Guess the power.

Velocity is tangent to the circle; tension points to the centre — these are perpendicular, so $θ = 90°$ . Why this step? In uniform circular motion the centripetal force is always at right angles to velocity.
$cos 90° = 0$ , so $P = F v cos 90° = 0$ . Why this step? A perpendicular force changes direction only, never speed — no energy per second is added.
$P = 12 \times 3 \times 0 = 0 W$ .

Verify: Consistent with constant speed: constant speed ⇒ constant kinetic energy ⇒ zero net work ⇒ zero power. ✓ In the first figure, imagine the blue force turned to point straight up (perpendicular to the orange velocity): its dashed shadow onto $v$ collapses to zero length — that zero-length shadow is exactly this zero power.

Cell D — force opposes motion ( $90° < θ \leq 180°$ )

Cell E — velocity is zero (degenerate)

Cell F — power changing in time

Worked example Ex 6 · Constant force, growing speed

A $2$ kg trolley starts from rest under a constant $6$ N force. Find (a) the instantaneous power at $t = 4$ s and (b) the average power over $0$ – $4$ s. (Cell F)

Forecast: Force is constant — is the power constant too? Guess whether $P_{avg}$ equals $P_{inst} (4)$ .

Acceleration: $a = F / m = 6/2 = 3 m/s^{2}$ . Why this step? We need $v (t)$ , which comes from acceleration.
Velocity at $t = 4$ : $v = a t = 3 \times 4 = 12$ m/s (started from rest). Why this step? Instantaneous power uses the current velocity.
$P_{inst} (4) = F v = 6 \times 12 = 72 W$ . Why this step? Force and motion are parallel here, so $P = F v$ .
Total work via KE: $W = \frac{1}{2} m v^{2} = \frac{1}{2} (2) (1 2^{2}) = 144$ J. Why this step? Average power needs total work over the interval; the work–energy theorem gives it cleanly.
$P_{avg} = W /Δ t = 144/4 = 36 W$ .

Verify: $P$ rose linearly from $0$ to $72$ W, so its time-average is the midpoint $\frac{1}{2} (72) = 36$ W. ✓ Constant force does not mean constant power — that is the whole point of this cell.

The next figure makes this concrete: it plots the instantaneous power $P = F v$ against time. Because $v = a t$ climbs steadily from rest, the blue power line is a straight ramp rising from $0$ to the peak $72$ W at $t = 4$ s. The green dashed line marks the average, $36$ W — sitting exactly halfway up the ramp, which is why $P_{avg} = \frac{1}{2} P_{peak}$ here.

Cell G — limiting behaviour (top speed)

Worked example Ex 7 · Car's maximum speed at rated power

A car engine delivers a maximum constant power of $60$ kW. On a level road the total resistance (drag + friction) is $1500$ N at high speed. What is the car's top speed? (Cell G)

Forecast: At top speed the car cannot accelerate any more. What balances what? Guess $v_{m a x}$ .

At top speed, acceleration $= 0$ , so the engine's drive force $F_{drive}$ just equals the resistance: $F_{drive} = 1500$ N. Why this step? No acceleration ⇒ net force zero ⇒ the forward drive force equals the backward resistive force in magnitude.
The engine's power moves the car forward, so it is the drive force (parallel to motion) that we put into $P = F_{drive} v$ , giving $v = P / F_{drive}$ . Why this step? We must use the force that does positive work along the motion — the drive force, not the resistance. They happen to be equal in size here, but they are different forces pointing opposite ways.
$v_{m a x} = \frac{60000}{1500} = 40 m/s$ .

Verify: $P = F_{drive} v = 1500 \times 40 = 60000$ W = $60$ kW ✓. Limiting insight: since $P$ is capped, the faster the car goes the less drive force it can muster — $F_{drive} = P / v$ shrinks as $v$ grows until it just matches drag. That is why top speed exists.

Cell H — units and billing (real-world word problem)

Cell I — exam twist (efficiency of two powers)

Worked example Ex 9 · Pump with efficiency

A water pump is rated at $5$ kW input power but is only $80%$ efficient. It lifts water of density $1000 kg/m^{3}$ to a height of $10$ m. How many litres per second can it raise? Use $g = 10 m/s^{2}$ (the acceleration due to gravity, defined in the symbol list above). (Cell I)

Forecast: The useful power is less than $5$ kW. Which power do we use to compute the water raised? Guess.

Useful (output) power: $P_{out} = η P_{in} = 0.80 \times 5000 = 4000$ W. Why this step? Efficiency $η = P_{out} / P_{in}$ — only the useful part lifts water.
Useful power raises mass $m$ per second against gravity: $P_{out} = \frac{m g h}{t}$ , so the mass rate is $\frac{m}{t} = \frac{P _{out}}{g h}$ . Why this step? Work to lift = $m g h$ (weight $m g$ times height $h$ ); power = that work per second.
$\frac{m}{t} = \frac{4000}{10 \times 10} = 40 kg/s$ .
Since $1000$ kg of water $= 1000$ L, $40$ kg/s $= 40$ L/s. Why this step? Density $1000 kg/m^{3}$ means $1$ L $= 1$ kg for water.

Verify: Check power back: lifting $40 kg/s$ to $10$ m needs $40 \times 10 \times 10 = 4000$ W of useful power ✓, and $4000/0.8 = 5000$ W input ✓, matching the rating. See Energy conservation and efficiency.

Active Recall

Recall Cover the answers

Sign of power when a force opposes motion? ::: Negative — energy is being removed. Power delivered by a centripetal force in uniform circular motion? ::: Zero, because $θ = 90°$ . If force is constant but object accelerates, is power constant? ::: No — $P = F v$ grows as $v$ grows. Formula for a car's top speed at fixed engine power $P$ against resistance $F$ ? ::: $v_{m a x} = P / F$ . Is kWh energy or power, and why? ::: Energy — it is power multiplied by time. Which power (input or output) do you use to find useful work rate? ::: Output (useful) power = $η \times$ input.

Connections

Power — average and instantaneous, units — the parent note with the core formulas.
Work — definition and W = F·d cosθ — the $cos θ$ projection reused here.
Kinetic Energy and Work-Energy Theorem — total work for average power (Ex 6).
Velocity and instantaneous rate (Kinematics) — velocity that grows in time (Cell F).
Dot product of vectors — the geometry behind $F \cdot v$ (Cell A–D).
Energy conservation and efficiency — input vs output power (Ex 9).

Worked examples — Power — average and instantaneous, units