NDA Physics · Work, Energy and Power

Work-Energy Theorem and Power

The work-energy theorem says the net work done on a body equals its change in kinetic energy. Power is the rate of doing work, P = W/t = Fv, measured in watts — and its commercial unit is the kilowatt-hour.

All Work, Energy and Power notes NDA Physics strategy

Why this matters

6 PYQs from 2017 to 2026, mostly MODERATE with one HARD (deriving potential energy from a force law). Two ideas dominate: the work-energy theorem (net work = ΔKE, used to find stopping forces and distances) and power (P = W/t = Fv, plus the watt-vs-joule and kilowatt-hour unit facts). The recurring trap is confusing power (rate) with energy (amount).

Concept 1 of 4: Work-energy theorem — net work equals change in kinetic energy

Whenever a net force acts on a moving body, it speeds it up or slows it down — and the total work that net force does is exactly the kinetic energy gained or lost. This holds for any force, conservative or not, along any path — which makes it the quickest route to stopping forces and braking distances.

Definition

The work-energy theorem states that the net work done by all forces on a body equals its change in kinetic energy: $W_\text{net} = \Delta KE = \tfrac{1}{2}mv_f^2 - \tfrac{1}{2}mv_i^2$ . It applies to any net force (conservative or non-conservative) over any path. A body slowing to rest loses kinetic energy $\tfrac{1}{2}mv^2$ , so the work done against it (e.g. by friction) equals that amount.

Work-energy theorem

W_\text{net} = \Delta KE = \tfrac{1}{2}mv_f^2 - \tfrac{1}{2}mv_i^2

$W_\text{net}$ net work done by all forces (J)
v_iinitial speed (m/s)
v_ffinal speed (m/s)

Worked example

A 3 kg block moving at 6 m/s is brought to rest by friction over a distance of 9 m. Find the magnitude of the frictional force.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

NDA · Apr 2026 · Q65Moderate

Example 1 · Work, Energy and Power · Work-Energy Theorem and Power

A particle moves from

X

Y

under a force field. The work done depends only on the initial and final speeds. Which one is correct?

The theorem uses NET work — not the work of one force

W_\text{net} = \Delta KE

sums the work of every force acting. If only friction acts on a sliding block, then friction's work equals the KE change; but when several forces act, add them all before equating to

\Delta KE

Drill 2 more on work-energy theorem — net work equals change in kinetic energy

Concept 2 of 4: Power — the rate of doing work (P = W/t = Fv)

Two cranes lift the same load to the same height and do the same work — but the faster one is more powerful. Power is how QUICKLY work is done, not how much. A 100 W bulb and a 100 J of energy are different ideas — one is a rate, the other an amount.

Definition

Power is the rate of doing work (or of transferring energy): $P = \dfrac{W}{t}$ . For a constant force moving a body at speed $v$ , $P = Fv$ . The SI unit is the watt (W), where 1 W = 1 J/s. A constant-power machine on a smooth surface gives $v \propto \sqrt{t}$ , because $P = mav = \text{const}$ integrates to $v^2 \propto t$ .

Power

P = \dfrac{W}{t} = Fv

Ppower (watts, W)
Wwork done (J)
ttime taken (s)
Fapplied force (N)
vspeed (m/s)

Worked example

A motor lifts an 8 kg mass through a vertical distance of 4 m in 2 s. Find the power. Take

g = 10

m/s².

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

NDA · Apr 2023 · Q123Easy

Example 2 · Work, Energy and Power · Work-Energy Theorem and Power

The power required to lift a mass of 8·0 kg up a vertical distance of 4 m in 2 s is (taking acceleration due to gravity as 10 m/s

^2

) :

Power is a RATE — do not confuse it with energy

Power (watt) is energy per unit time; energy (joule) is the total amount. Two machines that do the same work have the same energy output but different power if they take different times. "How much" is energy; "how fast" is power.

P = Fv uses the speed at that instant

When a force moves a body, the instantaneous power is

P = Fv

. At higher speed the same force delivers more power — which is why a constant-power engine cannot keep accelerating at the same rate.

Drill 1 more on power — the rate of doing work (p = w/t = fv)

Concept 3 of 4: Units of work, energy, and power

The NDA tests unit facts directly — what is one joule, what is a kilowatt-hour in joules, watt vs joule. These are pure recall marks: memorise the table and never lose them.

Definition

Work and energy share the unit joule (J); power uses the watt (W) = J/s. The kilowatt-hour (kWh) is the commercial unit of electrical energy — the energy used by a 1 kW device in 1 hour. Memorise the conversions below.

Quantity / unit	Definition	In SI base
Joule (J)	1 N acting through 1 m	work / energy unit
1 joule of work	force of 4 N over 0.25 m	$4 \times 0.25 = 1$ J
Watt (W)	1 joule per second	power unit, J/s
Kilowatt-hour (kWh)	energy of a 1 kW device in 1 hour	$3.6 \times 10^{6}$ J 1 kWh = 1000 W × 3600 s = 3.6 × 10⁶ J — the commercial unit of electrical energy.
Kilowatt (kW)	1000 watts	power unit

The two recall favourites: 1 J = 4 N over 0.25 m, and 1 kWh = 3.6 × 10⁶ J.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

NDA · Sep 2022 · Q61Easy

Example 3 · Work, Energy and Power · Work-Energy Theorem and Power

The commercial unit of electrical energy is kilowatt-hour (kWh), which is equal to

1 kWh is 3.6 × 10⁶ J — not 1000 or 3600

A kilowatt-hour combines 1000 W with 3600 s:

1000 \times 3600 = 3.6 \times 10^{6}

J. Multiplying only one of the two factors is the standard wrong answer.

Concept 4 of 4: Potential energy from a force — U = − ∫ F dx

For a conservative force, the force is the negative slope of the potential energy curve — so going the other way, the potential energy is the negative integral of the force over distance. Given a force law

F(x)

, you recover the potential energy by integrating and flipping the sign.

Definition

For a conservative one-dimensional force $F(x)$ , the potential energy is $U(x) = -\displaystyle\int F(x)\,dx$ (up to a constant). Equivalently $F(x) = -\dfrac{dU}{dx}$ : the force points in the direction of decreasing potential energy. So to get $U$ from a given force law, integrate $F$ with respect to $x$ and negate.

Potential energy from a conservative force

U(x) = -\int F(x)\,dx \quad\Longleftrightarrow\quad F(x) = -\dfrac{dU}{dx}

U(x)potential energy as a function of position (J)
F(x)conservative force along x (N)

Worked example

A particle on the x-axis feels a force

F(x) = -kx

(a spring). Find its potential energy

U(x)

, taking

U(0) = 0

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

NDA · Sep 2017 · Q70Hard

Example 4 · Work, Energy and Power · Work-Energy Theorem and Power

The force acting on a particle of mass m moving along the x-axis is given by

F(x) = Ax^2 - Bx

. Which one of the following is the potential energy of the particle?

Do not forget the MINUS sign when integrating

U = -\int F\,dx

— the negative sign is essential. Dropping it (writing

U = +\int F\,dx

) flips the sign of the whole potential energy and gives the wrong option.

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (3)

Work-energy theorem — net work equals change in kinetic energy
Work-energy theorem
$W_\text{net} = \Delta KE = \tfrac{1}{2}mv_f^2 - \tfrac{1}{2}mv_i^2$
Power — the rate of doing work (P = W/t = Fv)
Power
$P = \dfrac{W}{t} = Fv$
Potential energy from a force — U = − ∫ F dx
Potential energy from a conservative force
$U(x) = -\int F(x)\,dx \quad\Longleftrightarrow\quad F(x) = -\dfrac{dU}{dx}$

Reference tables (1)

Units of work, energy, and power5 rows

Quantity / unit	Definition	In SI base
Joule (J)	1 N acting through 1 m	work / energy unit
1 joule of work	force of 4 N over 0.25 m	$4 \times 0.25 = 1$ J
Watt (W)	1 joule per second	power unit, J/s
Kilowatt-hour (kWh)	energy of a 1 kW device in 1 hour	$3.6 \times 10^{6}$ J 1 kWh = 1000 W × 3600 s = 3.6 × 10⁶ J — the commercial unit of electrical energy.
Kilowatt (kW)	1000 watts	power unit

The two recall favourites: 1 J = 4 N over 0.25 m, and 1 kWh = 3.6 × 10⁶ J.

Watch out for (5)

The theorem uses NET work — not the work of one force
→ Work-energy theorem — net work equals change in kinetic energy
Power is a RATE — do not confuse it with energy
→ Power — the rate of doing work (P = W/t = Fv)
P = Fv uses the speed at that instant
→ Power — the rate of doing work (P = W/t = Fv)
1 kWh is 3.6 × 10⁶ J — not 1000 or 3600
→ Units of work, energy, and power
Do not forget the MINUS sign when integrating
→ Potential energy from a force — U = − ∫ F dx

Test yourself on Work, Energy and Power

10 past NDA questions from this chapter, timed at 10 minutes and marked the way the exam marks it. You see your score and every answer the moment you finish. Free to start.

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