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

Intuition

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.

By the work-energy theorem, the work done by friction equals the kinetic energy lost: $f\,d = \tfrac{1}{2}mv^2$ .
$\tfrac{1}{2}mv^2 = \tfrac{1}{2}(3)(6^2) = \tfrac{1}{2}(3)(36) = 54$ J.
$f \times 9 = 54 \Rightarrow f = 54/9 = 6$ N.

Answer:Frictional force = 6 N.

Practice this conceptself-check · 4 quick reps

Try it yourself

A 2 kg block sliding at 10 m/s on a rough surface comes to rest after 20 m. Find the frictional force using the work-energy theorem.

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

1.
State the work-energy theorem.
2.
A 1 kg body slows from 4 m/s to rest over 2 m. Find the retarding force.
3.
Does the work-energy theorem apply only to conservative forces?
4.
A 2 kg block has its speed raised from 0 to 5 m/s. Net work done on it?

From the bank · past-year question

Example 1Work, Energy and PowerMODERATE

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?

[Q65 · Apr · 2026]

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 1 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)

Intuition

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.

Work done against gravity: $W = mgh = 8 \times 10 \times 4 = 320$ J.
Power = work / time: $P = 320 / 2$ .
$P = 160$ W.

Answer:

P = 160

Practice this conceptself-check · 4 quick reps

Try it yourself

A constant-power engine pulls a block along a smooth horizontal surface from rest. How does the block's speed depend on time?

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

1.
What is the SI unit of power?
2.
A machine does 600 J of work in 3 s. Find its power.
3.
A 50 N force moves a body at 4 m/s. Find the power delivered.
4.
Under constant power on a smooth surface, the speed is proportional to which function of time?

From the bank · past-year question

Example 2Work, Energy and PowerEASY

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

) :

[Q123 · Apr · 2023]

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

Intuition

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

Try it yourself

Express one kilowatt-hour in joules.

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

1.
What is the commercial unit of electrical energy?
2.
1 kWh equals how many joules?
3.
Work is one joule when a force of 4 N moves an object through what distance?
4.
What is the SI unit shared by work and energy?

From the bank · past-year question

Example 3Work, Energy and PowerEASY

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

[Q61 · Sep · 2022]

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

Intuition

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

Use $U = -\int F\,dx = -\int(-kx)\,dx$ .
$U = \int kx\,dx = \tfrac{1}{2}kx^2 + C$ .
With $U(0) = 0$ , the constant $C = 0$ , so $U = \tfrac{1}{2}kx^2$ .

Answer:

U(x) = \tfrac{1}{2}kx^2

(the familiar spring potential energy).

Practice this conceptself-check · 4 quick reps

Try it yourself

A particle of mass m moves along the x-axis under a force

F(x) = Ax^2 - Bx

. Find its potential energy

U(x)

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

1.
How is potential energy obtained from a conservative force F(x)?
2.
If $U = \tfrac{1}{2}kx^2$ , what is the force?
3.
For $F = -mg$ (constant), find U(x) with U(0)=0.
4.
The force is the negative of which derivative of the potential energy?

From the bank · past-year question

Example 4Work, Energy and PowerHARD

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?

[Q70 · Sep · 2017]

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

Mastery check — 2 interleaved questions

Try each one before clicking. Questions are interleaved across the concepts above, not grouped — interleaving sharpens transfer.

Example 1Work, Energy and PowerMODERATE

A block of mass 2.0 kg slides on a rough horizontal surface at 10 m/s and comes to rest after 20 m. Which one of the following could be the magnitude of the frictional force ?

[Q123 · Sep · 2024]

Example 2Work, Energy and PowerMODERATE

A constant power machine pulls a block on a smooth horizontal surface. Which one correctly describes the relation between speed

(v)

and time

(t)

[Q60 · Apr · 2026]

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