Power — average and instantaneous, units: Question bank

Before we start, one reminder of the three tools these traps exploit, so no symbol appears unexplained:

Recall The three ideas every trap below leans on

Average power $P_{avg} = \frac{W}{Δ t}$ — total work $W$ (in joules, J) shared over the whole stretch of time $Δ t$ (in seconds, s). Read it as "job done, spread over the wait."
Instantaneous power $P = F \cdot v = F v cos θ$ — the rate right now. Here $F$ is the force, $v$ the velocity, and $θ$ the angle between them. The dot $\cdot$ is the dot product: it keeps only the part of the force that lies along the motion.
$cos θ$ is the switch. $θ = 0^{\circ}$ (push along motion) gives $cos θ = 1$ ; $θ = 9 0^{\circ}$ (push sideways) gives $cos θ = 0$ ; $θ = 18 0^{\circ}$ (push against motion) gives $cos θ = - 1$ . This one factor decides whether power is positive, zero, or negative.

Figure — Power — average and instantaneous, units

The four figures that follow anchor the four groups of traps. Refer back to them line by line: the amber force arrow, the cyan velocity arrow, and the sign of their overlap are what every trap is really testing.

True or false — justify

(Look at figure s02: the crate rises the same height either way — the red "job done" area is identical; only the clock differs.)

Two cranes lift identical crates to the same height; the faster one does more work

False — same force through the same height is the same work

W = m g h

(mass

m

, gravity

g

, height

h

, as defined above); the faster crane does the same work in less time, so it has more power, not more work.

A 100 W bulb always uses more energy than a 60 W bulb

False — energy is power × time. A 60 W bulb left on all day uses more energy than a 100 W bulb flicked on for one second; wattage is a rate, not a total.

If instantaneous power is zero, the force must be zero

False — power

F v cos θ

is also zero if

v = 0

(pushing a wall) or if

θ = 9 0^{\circ}

(force perpendicular to motion, like tension in circular motion), even with a large force. See the middle panel of figure s03.

Negative power is physically impossible

False — when

θ > 9 0^{\circ}

the force opposes the motion (

cos θ < 0

), so

P < 0

: energy is being taken out of the object, e.g. brakes or drag removing kinetic energy. This is the right panel of figure s03.

kWh and kW measure the same kind of quantity

False — kW is power (rate, J/s) while kWh is power × time = energy (J). One is a speed of spending, the other a total amount spent.

Average power over an interval always equals the value of instantaneous power at the midpoint of that interval

False — that only holds when power varies linearly in time (constant force from rest). For a general curve the average and the midpoint value differ — compare the two curves in figure s04.

Doubling an object's speed while keeping the driving force constant doubles the power

True — with

θ

fixed,

P = F v cos θ

is linear in

v

, so twice the speed is twice the power (this is why cars need far more power to hold high speeds).

A satellite in a circular orbit has power delivered to it by gravity

False — gravity points to the centre while velocity is tangential, so

θ = 9 0^{\circ}

cos θ = 0

, and

P = 0

; speed and kinetic energy stay constant (see Kinetic Energy and Work-Energy Theorem, and the perpendicular case in figure s03).

Spot the error

(Figure s03 is your reference: the sign of power is just the overlap of the amber force arrow with the cyan velocity arrow.)

"Power is force times distance,

P = F d

That is work, not power. Power is work per unit time,

P = W /Δ t

, or in instantaneous form

P = F v

— force times speed, not distance.

"The engine delivers 30 kW, so after 2 s it has delivered 30 kW of energy."

Units clash: kW is power, energy is kW × time. It delivers

30 kW \times 2 s = 60 kJ

of energy, not "30 kW of energy."

"At constant velocity the net power on the car is

F v

, so kinetic energy keeps rising."

Here the

F

F v

is the engine's driving force, not the net force — the engine's power is indeed

F v

, but at constant speed the drag force delivers an equal negative power, so the net power (sum over all forces) is zero and kinetic energy is constant.

"1 horsepower is bigger than 1 kilowatt because horses are strong."

Wrong ordering:

1 hp \approx 746 W

, which is less than

1 kW = 1000 W

. Naming has nothing to do with size.

"Since

P = F \cdot v

uses a dot product, power is a vector."

A dot product of two vectors returns a scalar (a plain number). Power has magnitude and sign but no direction — it is a scalar.

"To find average power just average the starting and ending instantaneous powers."

Only valid if power is linear in time. In general, average power is total work over total time, which need not equal the mean of two endpoint values (figure s04).

"Base units of the watt are

kg m^{2} s^{- 2}

That is the joule (energy). The watt is a joule per second:

kg m^{2} s^{- 3}

— divide the joule by one more second. The cancellation is drawn out in figure s02 (right panel).

Why questions

Why does pushing hard on a stationary wall deliver zero power

Because

v = 0

, so

P = F v cos θ = 0

; no displacement means no work per second, no matter how hard you strain (see Work — definition and W = F·d cosθ).

Why does a car engine need more power to hold 100 km/h than 50 km/h even at "steady speed"

Drag grows with speed, so the force needed rises, and

P = F v

multiplies a bigger force by a bigger speed — power climbs much faster than speed itself.

Why do we define instantaneous power with a limit

lim_{Δ t \to 0} Δ W /Δ t

instead of just

W /Δ t

Because the rate can change moment to moment; shrinking the interval to an instant captures the rate right now rather than a smeared-out average — the same logic as instantaneous velocity.

Why is

P = F \cdot v

more useful than

P = W /Δ t

for a machine running continuously

It gives the rate at any instant directly from force and speed, without needing to measure total work over an interval — ideal for engines whose output varies as speed changes.

Why does the dot product (not ordinary multiplication) appear in

P = F \cdot v

Only the component of force along the motion transfers energy; the dot product

F v cos θ

automatically keeps that component and discards the perpendicular part — see the projection drawn in figure s03 (left panel).

Why can output power never exceed input power for a real machine

Some input energy is always lost to friction and heat, so efficiency

= P_{out} / P_{in} < 1

; a value above 1 would create energy from nothing (see Energy conservation and efficiency).

Why is average power equal to

\frac{1}{2}

the final power for a body starting from rest under constant force

Force is constant but

v = a t

rises linearly from zero, so

P = F v

rises linearly from 0 to

P_{final}

; the average of a straight line from 0 is its midpoint,

\frac{1}{2} P_{final}

— this is the straight line in figure s04.

Edge cases

(Each of these is a force–velocity arrow picture; figure s03 shows the three governing geometries — aligned, perpendicular, opposed.)

A body moves at constant velocity with net force zero — what is the net power

Zero, because net force is zero so

F \cdot v = 0

; individual forces (engine, drag) may deliver equal-and-opposite non-zero powers that cancel.

A force acts but the object hasn't started moving yet (

v = 0

t = 0

) — instantaneous power at that instant

Exactly zero, since

P = F v = F \cdot 0 = 0

; power only becomes non-zero once the object gains speed, even though work will accumulate.

Force is perpendicular to velocity for the entire motion — total work and average power

Both zero:

cos 9 0^{\circ} = 0

throughout, so no work is done and average power over any interval is zero (e.g. magnetic force on a charge, or circular-motion centripetal force) — the middle panel of figure s03.

A car brakes to a stop — sign of the power delivered by the braking force

Negative throughout, because friction opposes velocity (

θ = 18 0^{\circ}

cos θ = - 1

); this negative power drains kinetic energy until

v = 0

, at which instant the power reaches zero.

Over one full oscillation of an ideal pendulum, what is the average power delivered by gravity

Zero — gravity does positive work descending and equal negative work ascending, so net work per cycle is zero and average power over the cycle is zero.

A ball is thrown straight up — track the sign of the power delivered by gravity through the flight

Gravity points down throughout. During the rise velocity is upward, so force and velocity are opposed (

θ = 18 0^{\circ}

) and power is negative (gravity drains kinetic energy). At the highest point

v = 0

, so power is exactly zero for that instant. During the fall velocity is downward, aligned with gravity (

θ = 0^{\circ}

), so power is positive. The sequence is: negative → zero → positive.

A machine rated "2 kW" is switched on for 0 seconds — energy delivered

Zero, because energy = power × time and

Δ t = 0

; the rating is a capacity, and no time means no energy transferred.