Work — definition, dot product F·d, sign convention: Question bank

Reminder of the machine we are testing, so no symbol is unexplained:

$W = F \cdot d = F d cos θ$ — work equals (force magnitude) $\times$ (displacement magnitude) $\times$ ( $cos$ of the angle $θ$ between them, drawn tail-to-tail).
$θ$ is always the angle between $F$ and $d$ ; the sign of $W$ is decided by $cos θ$ alone.
$cos θ$ : positive for $0° \leq θ < 90°$ , zero at $90°$ , negative for $90° < θ \leq 180°$ .

The one picture behind every trap

Before the questions, fix the geometry in your eyes. Draw $F$ and $d$ from the same starting point (tail-to-tail) — the angle they open up is $θ$ . Now drop $F$ straight down onto the line of $d$ : the shadow it casts along $d$ has length $F cos θ$ (this is the parallel part $F_{∥}$ ), and what sticks out sideways is $F sin θ$ (the perpendicular part $F_{⊥}$ ).

Figure — Work — definition, dot product F·d, sign convention

Notice the three regimes at a glance: when $F$ leans toward the motion the shadow points the same way as $d$ (positive work); when $F$ is straight across, the shadow shrinks to a point (zero work); when $F$ leans back against the motion the shadow points opposite $d$ (negative work). That single shadow is the whole sign convention.

And here is the mathematical reason the perpendicular part vanishes, not just "feels useless". Work is really the running sum (integral) of tiny force-dots-displacement steps, $W = \int F \cdot d r$ . Split each tiny step's force into $F_{∥}$ (along the path) and $F_{⊥}$ (across it). Since $d r$ points along the path, $F_{⊥} \cdot d r = F_{⊥} d r cos 90° = 0$ in every step — the perpendicular piece multiplies by $cos 90° = 0$ and drops out of the sum term by term. Only $\int F_{∥} d r$ survives.

True or false — justify

Each of these is a one-sentence claim. Decide true or false, then give the reason.

A force with large magnitude always does large work.

False. Work is

F d cos θ

; if

θ = 90°

then

cos θ = 0

and

W = 0

no matter how large

F

is, and if

θ > 90°

the work is negative.

If an object returns to its starting point, the total work done on it by every force must be zero.

False. Only forces whose work depends solely on start/end points (like gravity) give zero over a closed loop; friction always opposes motion, so it does negative work on every leg and totals to a nonzero (negative) value.

Zero net work on an object means the object did not move.

False. It can move a lot yet gain no speed: e.g. a satellite in a circular orbit moves constantly but gravity is always perpendicular to its velocity, so net work is zero and its kinetic energy is unchanged.

Negative work means the force is small or weak.

False. Negative work only means

cos θ < 0

, i.e. the force has a component against the motion (

90° < θ \leq 180°

); a huge braking force does large-magnitude negative work.

Work done by the normal force on a box sliding along a flat floor is zero.

True. The normal force points perpendicular to the floor (perpendicular to the horizontal displacement), so

θ = 90°

cos 90° = 0

, and

W = 0

A force perpendicular to velocity can never change an object's kinetic energy.

True. Perpendicular force does zero work (

cos 90° = 0

), and by the work–energy idea (see Kinetic Energy & Work-Energy Theorem) no work means no change in

\frac{1}{2} m v^{2}

; it only redirects motion.

If two forces act and one does

+ 8 J

while the other does

- 8 J

, the object's speed is unchanged.

True. Net work is

8 + (- 8) = 0 J

, and net work equals the change in kinetic energy, so the speed at the end equals the speed at the start.

Doing positive work on an object always speeds it up.

Careful — False as stated. Positive work by one force adds energy via that force, but another force may remove more; only the net work sets whether speed rises. A single positive contribution can be outweighed.

The unit of work, the joule, is the same as the unit of force times distance.

True.

1 J = 1 N\cdotm

by definition, since

W = F d cos θ

multiplies a force (N) by a length (m) and a dimensionless cosine.

Spot the error

Each line contains a piece of reasoning with a mistake buried in it. Name the error and correct it.

"The rope pulls at

30°

, so the useful force is

F sin 30°

Error: wrong trig function.

θ

is the angle between

F

and the horizontal displacement, and work needs the component along displacement, which is

F cos 30°

sin

would give the vertical part, which does no horizontal work.

"I walked a longer, curvier path carrying the same box the same height, so I did more work against gravity."

Error: path length is not displacement. Gravity's work depends only on the vertical displacement (start-to-end height change), not the path taken; same height change means same gravitational work.

"Friction did

+ 15 J

on the sliding block because friction is a strong real force."

Error: sign ignored. Kinetic friction points opposite the motion (

θ = 180°

cos 180° = - 1

), so its work is negative; strength affects magnitude, never the sign. See Friction.

"The angle in

F d cos θ

30°

because the ramp is inclined at

30°

Error: wrong angle.

θ

is the angle between the specific force and the specific displacement drawn tail-to-tail, not automatically the ramp's tilt; you must identify each vector's direction first (see Vectors & Components).

"Since I held the dumbbell perfectly still for a minute, I did a lot of work — my arm was exhausted."

Error: effort is not work. Displacement is zero, so

W = F d cos θ = 0

; muscular fatigue is biological (fibres repeatedly firing), not physics work.

"Gravity does zero work on a horizontally thrown ball because it moves horizontally."

Error: displacement isn't purely horizontal. A projectile also falls, so there is downward displacement aligned with gravity;

cos θ

for that vertical part is positive, giving positive gravity work.

"Work is a vector because it is built from two vectors

F

and

d

Error: dot product output. The scalar (dot) product of two vectors yields a scalar; work therefore has magnitude and sign but no direction. See Dot Product (Scalar Product).

Why questions

Answer with the underlying mechanism, not just the rule.

Why does only the component of force along the displacement contribute to work?

Because work is the integral

W = \int F \cdot d r

; splitting

F

into

F_{∥}

and

F_{⊥}

, each perpendicular term becomes

F_{⊥} d r cos 90° = 0

and drops out term by term, leaving only

\int F_{∥} d r

. Physically,

F_{⊥}

produces no displacement in its own direction, so it transfers no energy.

Why is

cos θ

(not

sin θ

) the right factor in

W = F d cos θ

cos θ

extracts the projection of

F

onto the displacement direction (adjacent side of the angle), which is exactly the along-motion share that does work;

sin θ

would give the useless perpendicular share.

Why does the sign of work come entirely from the angle and not the magnitudes

F

d

F

and

d

are magnitudes and are never negative, so the only factor that can flip sign is

cos θ

, which is negative precisely when the force opposes the motion (

θ > 90°

Why is the component form

W = F_{x} d_{x} + F_{y} d_{y} + F_{z} d_{z}

often more convenient than

F d cos θ

It computes work directly from the vectors' components without ever needing to find the angle

θ

, and it delivers the correct sign automatically. See Vectors & Components.

Why does centripetal (string) tension do zero work in uniform circular motion?

Tension always points toward the centre, perpendicular to the velocity (and hence the instantaneous displacement), so

θ = 90°

cos 90° = 0

, and

W = 0

— which is why the speed stays constant.

Why can two people both "push the wall hard" yet the physics work on the wall be zero?

The wall's displacement is zero, so

W = F d cos θ = 0

regardless of force; without motion of the point of application there is no energy transfer.

Why does negative work correspond to the object losing energy?

Negative work means the force has a component opposite the motion, so it removes kinetic energy from the object; by the work–energy connection this shows up as a drop in

\frac{1}{2} m v^{2}

Why is power related to but distinct from work?

Work is the total energy transferred, while power

P = F \cdot v

is the rate of that transfer per unit time; the same work done faster means more power. See Power.

Edge cases

The boundary and degenerate situations the definition must still handle.

What is the work if the displacement

d

is exactly zero (object doesn't move)?

W = F d cos θ = F \cdot 0 \cdot cos θ = 0

; with no displacement no work is done regardless of how large the force is.

What is the work at exactly

θ = 90°

, and why is this the knife-edge case?

W = F d cos 90° = 0

; it is the boundary separating energy-adding forces (

θ < 90°

cos > 0

) from energy-removing forces (

θ > 90°

cos < 0

What is the maximum possible work for a given

F

and

d

, and at what angle?

The maximum is

W = F d

, occurring at

θ = 0°

where

cos 0° = 1

and the force points entirely along the displacement.

What is the most negative possible work for a given

F

and

d

W = - F d

θ = 180°

where

cos 180° = - 1

, i.e. the force points exactly opposite the displacement (like friction directly opposing motion).

Can a

1000 N

force do exactly zero work?

Yes, whenever it is perpendicular to the displacement (

θ = 90°

) or the displacement is zero; large magnitude does not force nonzero work.

If a box moves right but the force points up-and-left at

135°

to the motion, what sign is the work?

Negative, because

θ = 135°

lies in

(90°, 180°]

cos 135° < 0

; the force has a leftward component opposing the rightward displacement.

For a ball thrown straight up then caught at the same height, what is the total work done by gravity over the whole trip?

Zero: gravity does negative work going up (

θ = 180°

) and equal positive work coming down (

θ = 0°

), and they cancel because the net vertical displacement is zero (path-independence of gravity — see Potential Energy & Conservative Forces).

Connections

Parent: Work definition & sign convention
Kinetic Energy & Work-Energy Theorem — net work sets the change in speed
Dot Product (Scalar Product) — why work is a scalar
Friction — the classic negative-work force
Potential Energy & Conservative Forces — closed-loop and path-independence traps
Power — work per unit time
Vectors & Components — resolving forces, picking the right angle