JEE Mains Physics · Work, Energy and Power

Work-Energy Theorem and Kinetic Energy

The total work done by all the forces on a body equals its change in kinetic energy, and kinetic energy and momentum are linked by K = p²/2m.

All Work, Energy and Power notes JEE Mains Physics strategy

Why this matters

Twenty-eight PYQs, twenty of them multiple choice, and four from 2026, the largest page in the chapter. Ten find an unknown work or force from the theorem: air resistance on a falling body, sand stopping a ball, an engine at constant speed; five give the velocity or position as a formula and ask for the work; thirteen link kinetic energy and momentum, as ratios or percentage changes. None needs the acceleration; all of them need every force's work counted with its sign.

Concept 1 of 3: Work-energy theorem with several forces

Add up the work of every force on the body: gravity, friction, air resistance, an engine, the ground pushing back. The total equals the change in kinetic energy. So if every work but one is known, the last one follows, and an average force follows from its work divided by the distance.

Definition

$W_g + W_{\text{other}} = K_f - K_i$ , with each work carrying its own sign.
Falling h from rest and landing at speed v: $W_{\text{air}} = \tfrac12 mv^{2} - mgh$ (negative).
Falling h and then sinking a depth d before stopping: gravity works through $h + d$ , so $mg(h + d) = F_{\text{avg}}\,d$ .
Constant speed means $\Delta K = 0$ : lowering a load through h, the hand does $-mgh$ ; an engine moving a vehicle a distance d against friction does $\mu mgd$ .
Braking with a fixed force F: stopping distance $s = K/F$ . Equal kinetic energies and equal forces give equal distances, whatever the masses.
Energy to speed up from u to 2u is three times the energy from rest to u.

Work-energy theorem

W_{\text{net}} = \sum W_i = K_f - K_i

Worked example

A 0.5 kg stone is dropped from rest and falls 20 m. It lands at 18 m/s. Find the work done by air resistance. (

g = 10

m/s²)

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 2 Apr 2026 Shift 2 · Q11Moderate

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

A spherical ball of mass 2 kg falls from a height of 10 m and is brought to rest after penetrating 10 cm into sand. The average force exerted by sand on the ball is

\_\_\_\_

Gravity also works through the depth of penetration

A ball that falls h and then sinks d into sand is pulled down by gravity through h + d. Writing mgh = F·d leaves out mgd, and the options usually include both answers.

Constant speed does not mean every force does zero work

At constant speed the NET work is zero. The engine of a bus still does positive work, and friction does an equal negative work. The engine's work is μmgd, not zero.

Drill 9 more on work-energy theorem with several forces

Concept 2 of 3: Work from a given velocity or position law

If the speed at the start and at the end is known, the work done by the net force is the change in kinetic energy. There is no need to find the force. A velocity given as a function of x is put in at the two positions; a position given as a function of t is differentiated first.

Definition

Given $v(x)$ : find $v_1$ and $v_2$ at the two positions, then $W = \tfrac12 m(v_2^{2} - v_1^{2})$ .
Given $x(t)$ : $v = dx/dt$ , put in the two times, then the same formula.
If $x(t)$ is quadratic in t, the acceleration and the force are constant, so $W = F\,\Delta x$ gives the same answer: a quick check.
Check the starting speed: $v = \alpha\sqrt{x}$ is zero at the origin, but $v = 3x^{2} + 4$ is 4 m/s there.
The theorem gives the work done ON the body. The work done BY the body on its surroundings has the opposite sign.

Work from two speeds

W_{\text{net}} = \tfrac12 m\left(v_2^{2} - v_1^{2}\right), \quad v = \frac{dx}{dt}

Worked example

A 3 kg body moves along the x-axis with

x(t) = 2t^{2} + 3t

(x in m, t in s). Find the work done on it between

t = 1

s and

t = 2

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 21 Jan 2026 Shift 2 · Q19Moderate

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

A body of mass 2 kg is moving along x -direction such that its displacement as function of time is given by

x(t) = \alpha t^{2}+ \beta t + \gamma m

, where

\alpha = 1\text{ }m/s^{2}

\beta = 1\text{ }m/s

and

\gamma = 1\text{ }m

. The work done on the body during the time interval

t = 2\text{ }s

t = 3\text{ }s

, is

\_\_\_\_

Do not assume the body starts from rest

Put the starting position into the velocity law. For v = 3x² + 4 the speed at x = 0 is 4 m/s, so the starting kinetic energy is not zero. Dropping it makes the work too large.

Work done on the body, not by it

W = ΔK is the work done on the body by the forces acting on it. A question about the work done by the body on its surroundings wants the same size with the opposite sign.

Drill 4 more on work from a given velocity or position law

Concept 3 of 3: Kinetic energy and momentum, K = p²/2m

Kinetic energy and momentum are tied by K = p²/2m. For two bodies with the same kinetic energy, the heavier one has more momentum. For two bodies with the same momentum, the lighter one has more kinetic energy. For one body, the kinetic energy goes as the square of the momentum, and as the square of the speed.

Definition

$K = \dfrac{p^{2}}{2m}$ and $p = \sqrt{2mK}$ .
Same K: $\dfrac{p_1}{p_2} = \sqrt{\dfrac{m_1}{m_2}}$ .
Same p: $\dfrac{K_1}{K_2} = \dfrac{m_2}{m_1}$ .
Same body: p multiplied by n makes K multiplied by $n^{2}$ ; K multiplied by n makes p multiplied by $\sqrt n$ .
Percentage change: new over old, minus 1. Momentum up 20% means K multiplied by 1.44, up 44%.
Speed falling from 50 to 30 m/s keeps $(30/50)^{2} = 36\%$ of the kinetic energy, a loss of 64%.
A force F acting for time t changes p by Ft; then $\Delta K = \dfrac{p_f^{2} - p_i^{2}}{2m}$ .

Kinetic energy and momentum

K = \frac{p^{2}}{2m} \qquad p = \sqrt{2mK}

Worked example

(a) The momentum of a body is doubled. By what percentage does its kinetic energy increase? (b) The kinetic energy of a body is made 25 times larger. By what percentage does its momentum increase?

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2024 · 6 April 2024 · Q105Moderate

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

When kinetic energy of a body becomes 36 times of its original value, the percentage increase in the momentum of the body will be:

Square the factor, not the percentage

Momentum up by 50% means p is multiplied by 1.5, so K is multiplied by 2.25: a rise of 125%. Doubling the 50% to get 100% is wrong.

A percentage increase subtracts the starting value

If K becomes 25 times larger, p becomes 5 times larger. That is an increase of 400%, not 500%: the new value is 500% of the old one, and the increase is 100% less.

Drill 12 more on kinetic energy and momentum, k = p²/2m

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 with several forces
Work-energy theorem
$W_{\text{net}} = \sum W_i = K_f - K_i$
Work from a given velocity or position law
Work from two speeds
$W_{\text{net}} = \tfrac12 m\left(v_2^{2} - v_1^{2}\right), \quad v = \frac{dx}{dt}$
Kinetic energy and momentum, K = p²/2m
Kinetic energy and momentum
$K = \frac{p^{2}}{2m} \qquad p = \sqrt{2mK}$

Watch out for (6)

Gravity also works through the depth of penetration
→ Work-energy theorem with several forces
Constant speed does not mean every force does zero work
→ Work-energy theorem with several forces
Do not assume the body starts from rest
→ Work from a given velocity or position law
Work done on the body, not by it
→ Work from a given velocity or position law
Square the factor, not the percentage
→ Kinetic energy and momentum, K = p²/2m
A percentage increase subtracts the starting value
→ Kinetic energy and momentum, K = p²/2m

Test yourself on Work, Energy and Power

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