Starting from Newton's second law, derive the work-energy theorem for a particle moving under a variable force acting along a straight line.

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Accepted Answer

Concept understanding — Work Energy Theorem

The Work-Energy Theorem: From Intuition to Precision

Imagine pushing a heavy box across the floor. The harder you push and the farther it slides, the faster it moves when you let go. That connection — the push (force) over a distance (displacement) changing the box's speed — is exactly what the Work-Energy Theorem captures.

The Intuition First

Think of work as the "currency" that buys motion. When you do work on an object, you transfer energy to it. That energy shows up as kinetic energy — the energy of motion. The more work you do, the more the object's kinetic energy changes.

If you push a stationary ball, it starts moving. If you push a moving ball in the same direction, it speeds up. If you push against its motion, it slows down. In every case, the work done equals the change in the ball's kinetic energy.

Note

Work is done by a force on an object. The object's kinetic energy changes by exactly that amount (assuming no other forces do work).

The Precise Statement

W_{\text{net}} = \Delta K = K_f - K_i

Where:

$W_{\text{net}}$ is the net work done on the object (the total work from all forces combined)
$K_f$ is the final kinetic energy
$K_i$ is the initial kinetic energy

And kinetic energy is defined as:

K = \frac{1}{2}mv^2

So the theorem can also be written as:

W_{\text{net}} = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2

Why "Net" Work Matters

This is the most common point of confusion. The theorem uses net work — the work done by the net force (the vector sum of all forces). If you push a box and friction opposes it, the net work is the work you do minus the work friction does. Only that net amount changes the kinetic energy.

Watch out

If you push a box at constant speed, your work is positive, but friction does equal negative work. The net work is zero, so kinetic energy doesn't change — the box keeps moving at the same speed. Your work didn't "disappear"; it was dissipated as heat by friction.

A Simple Derivation (for constant force)

Consider a constant net force $F_{\text{net}}$ acting on an object of mass $m$ over a displacement $s$ . From Newton's second law:

F_{\text{net}} = ma

From kinematics (constant acceleration):

v_f^2 = v_i^2 + 2as

Multiply both sides by $\frac{1}{2}m$ :

\frac{1}{2}mv_f^2 = \frac{1}{2}mv_i^2 + mas

But $mas = F_{\text{net}} s = W_{\text{net}}$ , so:

\frac{1}{2}mv_f^2 = \frac{1}{2}mv_i^2 + W_{\text{net}}

Rearranging:

W_{\text{net}} = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2 = \Delta K

Tip

The theorem holds even for variable forces and curved paths — the derivation uses calculus then, but the result is the same.

What It Tells You (and What It Doesn't)

It tells you: How much the speed changes when you know the net work done. Or, how much net work is needed to achieve a certain speed change.

It doesn't tell you: The direction of motion, the time taken, or the path followed. Work and kinetic energy are scalars — they have no direction.

A Quick Example

A 2 kg block initially at rest is pulled by a net force of 10 N over 4 m. Find its final speed.

Solution:

Net work: $W = Fs = 10 \times 4 = 40 \text{ J}$
Initial kinetic energy: $K_i = 0$
By the theorem: $40 = \frac{1}{2}(2)v_f^2 - 0$
So: $40 = v_f^2$
Therefore: $v_f = \sqrt{40} \approx 6.32 \text{ m/s}$

Important

The Work-Energy Theorem is a scalar alternative to Newton's laws for problems involving speed changes. It often simplifies calculations because you don't need to find acceleration or time — just work and kinetic energy.

Looking up "Work Energy Theorem: definition, formula & real-world examples" is a good habit before an exam, and it is worth knowing that Work Energy Theorem is drawn directly from the Work, Energy and Power coverage of the NCERT/CBSE Class 11 Physics syllabus and recurs often in JEE Main and NEET papers. Cross-checking this explanation against the relevant NCERT Physics chapter and solving a few past-year questions will round out your preparation.

Consider a particle of constant mass $m$ moving along a straight line under a net force $F(x)$ that may vary with position $x$ . By Newton's second law, $F = m\frac{dv}{dt}$ The work done as the particle moves from $x_1$ (where its speed is $v_1$ ) to $x_2$ (where its speed is $v_2$ ) is, by definition, $W_{\text{net}} = \int_{x_1}^{x_2} F\,dx = \int_{x_1}^{x_2} m\frac{dv}{dt}\,dx$ The key step is to rewrite $\dfrac{dv}{dt}$ using the chain rule, treating $v$ as a function of $x$ : $\frac{dv}{dt} = \frac{dv}{dx}\cdot\frac{dx}{dt} = v\frac{dv}{dx}$ (since $dx/dt = v$ by definition). Substituting this back: $W_{\text{net}} = \int_{x_1}^{x_2} m\left(v\frac{dv}{dx}\right)dx = \int_{v_1}^{v_2} mv\,dv$ where the limits have now been converted from positions to the corresponding speeds. This integral is elementary: $W_{\text{net}} = \left[\frac{1}{2}mv^2\right]_{v_1}^{v_2} = \frac{1}{2}mv_2^2 - \frac{1}{2}mv_1^2 = K_f - K_i = \Delta K$

This derivation shows the work-energy theorem is not an independent postulate, but follows directly and generally from Newton's second law alone, for any net force -- constant or variable -- acting along a straight line.

✓Final answer

W_{\text{net}} = \Delta K

Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

Showing the 12 most recent of 16 on this concept.

CBSE 2026Set ANNUAL1 markMCQ
$Q. Match the column - select the correct definition (from Column B) for the term 'Energy' (Column A):$
(a) change in linear momentum
(b) motion opposing force
(c) loss of energy
(d) rate of change of momentum
(e) ability of doing work
(f) rate of doing work
›Reveal solutionSolution

Energy is defined as the capacity of a body to do work.

In mechanics, the energy possessed by a body (whether kinetic, due to its motion, or potential, due to its position/configuration) is precisely defined as its capacity to perform work — a body with more energy can do more work before that capacity is exhausted. This directly matches column B option (e), 'ability of doing work'.

✓Final answer
$Energy (Column A) matches (e) ability of doing work (Column B).$
CBSE 2026Set ANNUAL1 markMCQ
$Q. A particle moves under the effect of a force F = αx from x = 0 to x = d. The work done in the process is (α = constant)$
(a) αd
(b) αd^2
(c) (1/2)αd^2
(d) zero
›Reveal solutionSolution

W = integral of alpha x dx from 0 to d = ½ alpha d^2. Answer (C).

For a position-dependent force, work done = integral of F dx.

W = integral from x=0 to x=d of (alpha x) dx = alpha [x^2/2] from 0 to d = alpha (d^2/2 - 0) = ½ alpha d^2.

✓Final answer
$(C) (1/2) alpha d^2.$
CBSE 2025Set ANNUAL1 markMCQ
$Q. A force of 40 N acts on body of mass 5 kg which is initially at rest. What is the amount of work done in the first 10 s?$
(a) 1600 J
(b) -1600 J
(c) 400 J
(d) -400 J
›Reveal solutionSolution

Work done by a constant force on a body starting from rest is $W = F^2t^2/(2m)$ . With the numbers as printed the arithmetic gives 16{,}000 J (ten times option (a)); using $m = 50$ kg — the value consistent with the printed options, and a very plausible OCR/typo of '5 kg' for '50 kg' in the source paper — gives exactly 1600 J, so that is the answer selected here.

Acceleration: $a = F/m$

Distance covered from rest in time $t$ : $s = (1/2)at^2$

Work done by the applied force: $W = F \times s = F \times (1/2)(F/m)t^2 = F^2t^2/(2m)$

With $F = 40$ N, $t = 10$ s, and $m = 50$ kg (see note above):

$a = 40/50 = 0.8$ m/s²

$s = (1/2)(0.8)(10)^2 = 40$ m

$W = 40 \times 40 = 1600$ J

Honest note: taking the mass literally as printed (5 kg) gives $a = 8$ m/s², $s = 400$ m, and $W = 16{,}000$ J — exactly ten times larger than any listed option, which strongly suggests a digit was dropped from '50 kg' in scanning/transcribing this paper. Option (a) 1600 J is selected as the intended answer.

✓Final answer
$= 1600$
CBSE 2024Set ANNUAL1 markMCQ
$Q. The dimensional formula of work done is the same as the dimensional formula of$
(a) Momentum
(b) Power
(c) Energy
(d) Torque
›Reveal solutionSolution
Work and energy are dimensionally identical because work done equals the change in energy (work-energy theorem).

Work done, W = Force x displacement = [MLT^-2][L] = [M L^2 T^-2].

Energy (kinetic or potential) has the same dimensional formula [M L^2 T^-2], since energy is literally defined as the capacity to do work.

Check the other options:

Momentum = mass x velocity = [M][LT^-1] = [MLT^-1] (different).

Power = work/time = [ML^2T^-3] (different).

Torque = force x perpendicular distance = [MLT^-2][L] = [ML^2T^-2] — interestingly, torque works out to the SAME dimensional formula as work/energy too, but it is a physically different quantity (a vector, measured in N·m, never expressed in joules by convention). Among the choices, Energy is the direct, conventional pairing being asked for here.

✓Final answer
$(c) Energy.$
CBSE 2024Set ANNUAL1 markMCQ
$Q. Energy is$
(a) capacity of doing work
(b) rate of doing work
(c) change of work
(d) none of these
›Reveal solutionSolution

Energy is defined as the capacity of a body or system to do work.

By definition, energy is the ability of a system to perform work; a body with more energy can do more work. This is why work and energy share the same SI unit, the joule, and the same dimensional formula [ML^2T^-2]. 'Rate of doing work' actually describes power, not energy.

✓Final answer
$(a) capacity of doing work.$
CBSE 2024Set ANNUAL1 markMCQ
$Q. Area under force-displacement curve represents$
(a) Velocity
(b) Acceleration
(c) Impulse
(d) Work done
›Reveal solutionSolution

Work done = force x displacement (for a varying force, the integral of F dx), which is exactly the area enclosed under an F-x graph.

For a constant force, W = F x s, which is literally the area of the rectangle under a horizontal F-x line. For a varying force, W = ∫ F dx, the area under the (possibly curved) F-x graph, found by summing up thin rectangular strips (F dx) — this is the standard way of computing work done from a force-displacement graph.

✓Final answer
$(d) Work done.$
CBSE 2024Set ANNUAL1 markMCQ
$Q. Which of the following is a unit of energy?$
(a) Horse power
(b) Joule second
(c) Kilowatt hour
(d) Watt.
›Reveal solutionSolution

Kilowatt hour = power × time = energy, unlike horse power and watt (power) or joule-second (action).

Horse power and watt are units of power (rate of doing work), not energy. Joule second is the unit of a quantity called 'action' (energy multiplied by time), not energy itself. A kilowatt hour is the energy delivered by a power of 1 kilowatt sustained for 1 hour: $1\,kWh = 1000\,W \times 3600\,s = 3.6\times10^{6}\,J$ , which is indeed a unit of energy (commonly called one 'unit' of electricity).

✓Final answer
$The correct option is (c) Kilowatt hour.$
CBSE 2024Set ANNUAL1 mark
$Q. State whether true or false: Work is a vector quantity.$
›Reveal solutionSolution

False. Work W = F.d (dot product of force and displacement) is a scalar quantity.

Work done by a force F causing a displacement d is defined as W = F . d = |F||d| cos(theta), the scalar (dot) product of the two vectors.

A dot product always yields a scalar result (it has magnitude only, no direction), so work is a scalar quantity, even though both force and displacement individually are vectors.

✓Final answer
$False — work is a scalar quantity, not a vector quantity.$
CBSE 2023Set ANNUAL1 markMCQ
$Q. The work done by a body against friction always results in$
(1) loss of kinetic energy
(2) loss of potential energy
(3) gain of kinetic energy
(4) gain of potential energy
›Reveal solutionSolution

Friction is dissipative: work done against it always drains kinetic energy from the system, converting it irreversibly into heat.

Friction opposes relative motion, so when a body moves against friction, the friction force does negative work on the body. By the work-energy theorem, negative work done on a body reduces its kinetic energy. This lost mechanical energy is not stored as potential energy (friction is not a conservative force, so no potential energy can be associated with it) - it is dissipated as heat at the surfaces in contact.

So work done against friction always results in a loss of kinetic energy of the body (and, correspondingly, generation of heat), not a gain of KE/PE or a loss of PE.

✓Final answer
$(1) loss of kinetic energy.$
CBSE 2023Set ANNUAL1 markMCQ
$Q. A raindrop of mass 1 g falling from a height of 1 km hits the ground with a speed of 50 ms^-1. If the resistive force is proportional to the speed of the drop, then the work done by the resistive force is (Take g = 10 ms^-2)$
(1) 10 J
(2) -10 J
(3) 8.75 J
(4) -8.75 J
›Reveal solutionSolution

Apply the work-energy theorem: total work done (by gravity and the resistive force together) equals the change in kinetic energy; solve for the unknown resistive-force work.

Given:

Mass, m = 1 g = 0.001 kg

Height fallen, h = 1 km = 1000 m

Final speed, v = 50 m/s

g = 10 m/s^2

Work done BY gravity (positive, since displacement is in the direction of gravity):

W_gravity = mgh = 0.001 x 10 x 1000 = 10 J

Change in kinetic energy (drop starts from rest):

ΔKE = (1/2)mv^2 - 0 = (1/2)(0.001)(50)^2 = (1/2)(0.001)(2500) = 1.25 J

By the work-energy theorem, the net work done by ALL forces equals ΔKE:

W_gravity + W_resistive = ΔKE

10 + W_resistive = 1.25

W_resistive = 1.25 - 10 = -8.75 J

The negative sign confirms the resistive (drag) force does negative work, as expected since it opposes the fall.

✓Final answer
$(4) -8.75 J.$
CBSE 2023Set ANNUAL1 mark
$Q. Mention two physical quantities which have same dimensions as that of work.$
›Reveal solutionSolution

Work has dimensional formula $[M L^2 T^{-2}]$ ; energy and torque share exactly this formula.

Work is defined as $W = F\cdot d$ , giving dimensions $[MLT^{-2}][L] = [M L^2 T^{-2}]$ . By the work-energy theorem, energy (kinetic energy $\tfrac12mv^2$ , potential energy $mgh$ , heat, etc.) has exactly the same dimensional formula, since energy is measured in the same unit, the joule. Torque is defined as $\tau = F\times r$ (force times perpendicular distance), which also gives $[MLT^{-2}][L] = [ML^2T^{-2}]$ — the same dimensions as work, even though torque is conceptually a different (vector) quantity measured in N·m rather than J.

✓Final answer
$[ML^2T^{-2}]$
CBSE 2023Set ANNUAL1 markMCQ
$Q. Which of the following quantities represents the change in kinetic energy of any object?$
(a) Force
(b) Mass
(c) Linear momentum
(d) Work
›Reveal solutionSolution

The work-energy theorem directly identifies work as the physical quantity equal to the change in kinetic energy of a body.

Starting from Newton's second law, F = ma = m(dv/dt). The work done by this force over a displacement ds is:

dW = F ds = m (dv/dt) ds = m v dv (since ds/dt = v)

Integrating from initial speed u to final speed v:

W = integral of m v dv from u to v = (1/2) m v^2 - (1/2) m u^2 = KE_final - KE_initial = Delta KE

So the work done on an object exactly equals its change in kinetic energy. Force, mass, and linear momentum are all separate physical quantities that do not, by themselves, represent this change.

✓Final answer
$The correct option is (d) Work.$

This derivation shows the work-energy theorem is not an independent postulate, but follows directly and generally from Newton's second law alone, for *any* net force -- constant or variable -- acting along a straight line.

> [!ANSWER]
> $W_{	ext{net}} = \Delta K$: the net work done on a particle always equals the change it produces in the particle's kinetic energy, a direct consequence of integrating Newton's second law using $dv/dt = v\,dv/dx$.

Q.Starting from Newton's second law, derive the work-energy theorem for a particle moving under a variable force acting along a straight line.

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