Work-energy theorem — derivation from Newton's second law: Question bank

The two ideas being probed everywhere on this page:

Work $W = \int F \cdot d r$ — force summed along the actual path, sign included.
Kinetic energy $K = \frac{1}{2} m v^{2}$ — a scalar built from speed squared, never negative, never directional.

True or false — justify

Every claim below is stated as fact. Decide true or false and give the one-line reason.

"If the net work on an object is zero, the object is not moving."

False — zero net work means

Δ K = 0

, so the speed is unchanged, not zero; a block dragged at constant velocity moves the whole time with

W_{n e t} = 0

"A force can act on an object over a large distance yet do zero work."

True — if the force is always perpendicular to the displacement (e.g. tension in circular motion, normal force on a slope),

F \cdot d r = 0

at every step, so the integral is zero.

"Kinetic energy can be negative if the object moves backwards."

False —

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

uses speed squared, so it is

\geq 0

regardless of direction; only work (which carries a sign) can be negative.

"Doubling an object's speed doubles its kinetic energy."

False —

K \propto v^{2}

, so doubling

v

quadruples

K

; this is why stopping distances grow so fast with speed.

"The work–energy theorem only works for constant forces."

False — the derivation integrates

\int F d x

, so any force with a known dependence on position works; the constant-force case

W = F d

is just the easiest special case.

"Two identical cars, one moving north and one moving south at the same speed, have opposite kinetic energies."

False —

K

depends on speed squared, a scalar; both have the same positive

K

, direction is irrelevant.

"If you push a box and it doesn't move, you've done work on it."

False — displacement is zero, so

\int F \cdot d r = 0

; your muscles get tired (biochemical cost) but the physics work on the box is zero.

"The work–energy theorem is a new physical law independent of Newton's laws."

False — it is

F = ma

integrated over distance; it contains no new physics, just a change of variable from time to position.

"Friction always does negative work."

False — kinetic friction on a sliding object does negative work, but friction can do positive work: e.g. friction from a conveyor belt speeding up a box, or the friction that accelerates a car's tyres forward.

Spot the error

Each item contains a subtle mistake in reasoning. Find it and state the fix.

"To find final speed, compute the work done by the applied force alone, then set it equal to

Δ K

Error: the theorem uses net work, not applied work; you must subtract friction and any opposing forces' work — otherwise

Δ K

is overstated.

"Since

W = F d

, a force at

6 0^{\circ}

to the motion over

4 m

with

F = 10 N

does

40 J

of work."

Error:

W = F d

needs force parallel to motion; here

W = F d cos 6 0^{\circ} = 10 \times 4 \times 0.5 = 20 J

, not

40 J

"The derivation writes

\frac{d v}{d t} d x = v d v

; this is just cancelling

d t

against

d x

like fractions."

Error (in justification): it isn't naive cancellation — we substitute

\frac{d x}{d t} = v

, i.e.

\frac{d v}{d t} d x = \frac{d x}{d t} d v = v d v

. The

d t

regroups via the chain rule, which is a real theorem, not fraction magic.

"A ball thrown straight up slows down, so gravity does zero work on the way up."

Error: gravity does negative work going up (force down, displacement up), which is exactly why

K

decreases; "zero" would mean no speed change.

"Since kinetic energy is conserved, the final speed equals the initial speed."

Error: kinetic energy is not generally conserved; only when net work is zero. The theorem says

Δ K = W_{n e t}

, and

W_{n e t}

is usually nonzero.

"For a variable force

F (x) = 6 x

, work is

F \times d = 6 (2) \times 2 = 24 J

Error: you can't use

F \times d

for a varying force; you must integrate:

\int_{0}^{2} 6 x d x = 12 J

. Using the final force value overcounts.

"The theorem applies to a system of many particles just by using the total mass and centre-of-mass displacement."

Error: for deformable/multi-particle systems,

W_{n e t} = Δ K

holds strictly for a single particle; for a system you must account for internal work and energy, so the naive extension can fail.

Why questions

Answer the "why" with the actual mechanism, not a restatement.

"Why do we multiply Newton's law by

d x

and not

d t

Because we want work

\int F d x

(force over distance); multiplying by

d t

instead gives impulse

\int F d t = Δ p

, the Impulse–Momentum Theorem — a different, time-based sibling.

"Why does the

v d v

trick 'eliminate time'?"

The substitution

\frac{d x}{d t} = v

absorbs the time derivative into the velocity itself, leaving

m v d v

with no

t

anywhere; time drops out because we integrate over position instead.

"Why is the theorem more convenient than

F = ma

when force depends on position?"

Solving

F = ma

requires finding

a (t)

then integrating twice; the theorem links

\int F d x

straight to speed change, skipping the messy time-dependence entirely.

"Why can the dot product

F \cdot d r

handle direction 'automatically'?"

It returns

F d r cos θ

, so it is positive when force aids motion, negative when it opposes, and zero when perpendicular — the sign of work falls out of the geometry with no extra rules.

"Why is

Δ K

a scalar even though force and displacement are vectors?"

The dot product turns two vectors into a single number, and

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

depends only on speed; direction information is 'used up' by the dot product and discarded.

"Why does braking correspond to negative work?"

The friction/braking force points opposite the displacement, so

F \cdot d r < 0

; this removes kinetic energy, which is exactly what slowing down means.

Edge cases

The boundary and degenerate scenarios the theorem must survive.

"What is

W_{n e t}

for an object moving at perfectly constant velocity?"

Exactly zero —

Δ K = 0

because speed is unchanged, so the net of all forces' work cancels (any applied work is balanced by opposing work).

"What does the theorem say if the object starts and ends at the same speed but takes a curved path?"

W_{n e t} = Δ K = 0

, even though individual forces may have done large positive and negative works along the way — only the net over the path matters.

"An object is momentarily at rest at the top of its rise. Is its kinetic energy zero? Is the net force zero?"

K = 0

(speed is zero) but the net force is not zero — gravity still acts, which is why it immediately starts moving again; zero

K

never implies zero force.

"A force is applied but the displacement is zero (pushing a wall). What is the work?"

Zero, since

\int F \cdot d r = 0

with

d r = 0

; no distance means no work no matter how large the force.

"If the net force is zero everywhere along the path, what is

Δ K

Zero —

W_{n e t} = \int F_{n e t} d x = 0

, so kinetic energy is unchanged; the object drifts at constant speed (Newton's first law, recovered as a special case).

"An object speeds up, then slows back to its original speed. What is the total net work?"

Zero overall, because

Δ K = K_{f} - K_{i} = 0

; the positive work during speed-up is exactly undone by negative work during slow-down.

Connections

Newton's Second Law — the single input every trap traces back to.
Kinetic Energy — why $K \geq 0$ and scales with $v^{2}$ , the source of several traps.
Work done by a variable force — the integral vs. $F d$ confusion lives here.
Conservation of Mechanical Energy — the "K is conserved" trap is really about when it is.
Power — rate of work, distinct from work itself.
Impulse–Momentum Theorem — the $d t$ -sibling contrasted with our $d x$ -theorem.