JEE Physics — Work Energy and Power Complete Chapter Guide

Chapter Overview & Weightage

Work, Energy and Power sits in the sweet spot of JEE Physics — moderate difficulty, high conceptual connectivity, and consistent presence every year. Questions here often combine with Rotational Mechanics, Centre of Mass, or even Modern Physics (photoelectric effect uses work function).

Weightage pattern (JEE Main): 1–2 questions per paper, occasionally 3. That’s 4–8 marks per attempt — enough to swing your percentile by 3–5 points. JEE Advanced typically places one question in the comprehension or integer type format, testing potential energy curves or collision types.

Year	JEE Main (Questions)	JEE Advanced	Key Topics Tested
2024	2	1	Work-energy theorem, elastic collision
2023	2	1	Conservative forces, PE curves
2022	1	2	Power, non-conservative work
2021	2	1	Variable force work, collision
2020	2	1	Spring potential energy, power
2019	1	2	WE theorem with friction, head-on collision

The chapter has been present in every JEE Main session from 2019–2024 without exception. Treat it as a guaranteed 4 marks.

Key Concepts You Must Know

Ranked by exam frequency (high to low):

Work-Energy Theorem — The net work done on an object equals its change in kinetic energy: $W_{net} = \Delta KE$ . This is the most-tested idea in the chapter.
Conservative vs Non-conservative forces — Gravity and spring force are conservative (path-independent); friction is non-conservative. The distinction matters for when you can apply conservation of energy.
Potential Energy Curves — Reading $U$ vs $x$ graphs: finding equilibrium points, determining stability, extracting force via $F = -\frac{dU}{dx}$ .
Conservation of Mechanical Energy — Valid only when no non-conservative forces do work (or their work is zero).
Elastic and Inelastic Collisions — Coefficient of restitution, velocity expressions post-collision, and the special case of equal masses.
Work done by variable forces — $W = \int F \cdot dx$ . Usually appears as a graph (area under $F$ - $x$ curve) or polynomial force.
Power — Instantaneous: $P = Fv\cos\theta$ ; Average: $P_{avg} = W/t$ . Frequently combined with Newton’s second law.
Spring Potential Energy — $U = \frac{1}{2}kx^2$ and work done compressing/stretching: $W_{spring} = -\Delta U_{spring}$ .

Important Formulas

W = \vec{F} \cdot \vec{d} = Fd\cos\theta

When to use: Force and displacement are both constant. Remember $\theta$ is the angle between $\vec{F}$ and $\vec{d}$ , not the incline angle. Many students confuse these in inclined plane problems.

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

When to use: Any time you know forces and need to find velocity (or vice versa), especially when the path is curved or acceleration is variable. This avoids kinematics entirely.

W = \int_{x_1}^{x_2} F(x)\, dx

When to use: Force depends on position. In JEE, this appears as a polynomial $F = ax + bx^2$ or as an $F$ - $x$ graph where you find the area geometrically.

KE_i + PE_i = KE_f + PE_f

\frac{1}{2}mv_i^2 + U_i = \frac{1}{2}mv_f^2 + U_f

When to use: Only when friction and air resistance are absent (or their work is specified as zero). When friction is present, use: $KE_i + PE_i = KE_f + PE_f + W_{friction}$ where $W_{friction}$ is the magnitude of energy lost.

F_x = -\frac{dU}{dx}, \quad F_y = -\frac{dU}{dy}, \quad F_z = -\frac{dU}{dz}

When to use: Given $U(x)$ , find force at any point. Equilibrium exists where $F = 0$ , i.e., $\frac{dU}{dx} = 0$ . Stable equilibrium: $\frac{d^2U}{dx^2} > 0$ (potential energy minimum).

v_1' = \frac{(m_1 - m_2)}{m_1 + m_2}v_1 + \frac{2m_2}{m_1 + m_2}v_2

v_2' = \frac{2m_1}{m_1 + m_2}v_1 + \frac{(m_2 - m_1)}{m_1 + m_2}v_2

Special case (equal masses): Velocities exchange. This is the “Newton’s cradle” result — appears directly in PYQs.

e = \frac{v_2' - v_1'}{v_1 - v_2} = \frac{\text{relative velocity of separation}}{\text{relative velocity of approach}}

Elastic: $e = 1$ . Perfectly inelastic: $e = 0$ . JEE Advanced often gives $e$ between 0 and 1 and asks for post-collision velocities.

P = \vec{F} \cdot \vec{v} = Fv\cos\theta

When to use: Problems giving force and velocity at an instant. Common in engine-power problems where $P$ is constant and you need to find maximum velocity (when acceleration = 0, so driving force = friction).

Solved Previous Year Questions

PYQ 1 — Work-Energy Theorem with Friction (JEE Main 2023, Jan Session)

Question: A block of mass 2 kg is pulled up a rough incline of angle 30° and length 4 m by a force parallel to the incline. The coefficient of kinetic friction is 0.2. If the block starts from rest and reaches the top with velocity 4 m/s, find the work done by the applied force. Take $g = 10$ m/s².

Solution:

Apply the work-energy theorem — it’s the cleanest approach here because we have multiple forces acting.

W_{net} = \Delta KE = \frac{1}{2}(2)(4)^2 - 0 = 16 \text{ J}

Now, $W_{net} = W_{applied} + W_{gravity} + W_{friction}$ .

Work done against gravity (displacement along incline = 4 m, height gained = $4\sin30° = 2$ m):

W_{gravity} = -mgh = -(2)(10)(2) = -40 \text{ J}

Normal force on incline: $N = mg\cos30° = 2 \times 10 \times \frac{\sqrt{3}}{2} = 10\sqrt{3}$ N

Work done by friction (opposing motion, so negative):

W_{friction} = -\mu N \cdot d = -(0.2)(10\sqrt{3})(4) = -8\sqrt{3} \approx -13.86 \text{ J}

From $W_{net} = W_{applied} + W_{gravity} + W_{friction}$ :

16 = W_{applied} - 40 - 8\sqrt{3}

W_{applied} = 56 + 8\sqrt{3} \approx 69.86 \text{ J}

Students often forget to include $W_{friction}$ or calculate normal force as $mg$ instead of $mg\cos\theta$ on an incline. On an incline, $N \neq mg$ .

PYQ 2 — Potential Energy Curve (JEE Advanced 2022, Paper 1)

Question: The potential energy of a particle moving along the x-axis is $U(x) = 20 + (x-2)^2$ joules, where $x$ is in metres. The particle is released from $x = 5$ m. Find (a) the equilibrium position, (b) whether it is stable/unstable, and (c) the speed at $x = 2$ m. Mass of particle = 2 kg.

Solution:

(a) Equilibrium position:

F = -\frac{dU}{dx} = -2(x-2)

Setting $F = 0$ : $x = 2$ m is the equilibrium position.

(b) Stability check:

\frac{d^2U}{dx^2} = 2 > 0

Positive second derivative means $U$ has a minimum at $x = 2$ m → stable equilibrium.

(c) Speed at $x = 2$ m:

At $x = 5$ m: $U_i = 20 + (5-2)^2 = 20 + 9 = 29$ J

At $x = 2$ m: $U_f = 20 + (2-2)^2 = 20$ J

The particle is released from rest, so $KE_i = 0$ .

Conservation of energy:

KE_i + U_i = KE_f + U_f

0 + 29 = \frac{1}{2}(2)v^2 + 20

v^2 = 9 \implies v = 3 \text{ m/s}

In PE curve questions, always compute $U$ at both points first, then use energy conservation. Don’t integrate force — it’s circular and slower.

PYQ 3 — Elastic Collision (JEE Main 2024, April Session)

Question: A ball of mass 4 kg moving at 6 m/s collides head-on elastically with a stationary ball of mass 2 kg. Find the velocities of both balls after collision.

Solution:

Use the elastic collision formulas directly. $m_1 = 4$ kg, $m_2 = 2$ kg, $v_1 = 6$ m/s, $v_2 = 0$ .

v_1' = \frac{m_1 - m_2}{m_1 + m_2} \cdot v_1 = \frac{4-2}{4+2} \times 6 = \frac{2}{6} \times 6 = 2 \text{ m/s}

v_2' = \frac{2m_1}{m_1 + m_2} \cdot v_1 = \frac{2 \times 4}{4+2} \times 6 = \frac{8}{6} \times 6 = 8 \text{ m/s}

Verification:

Momentum: $4(6) = 4(2) + 2(8) = 8 + 16 = 24$ ✓
KE: $\frac{1}{2}(4)(36) = 72$ J; $\frac{1}{2}(4)(4) + \frac{1}{2}(2)(64) = 8 + 64 = 72$ J ✓

Always verify elastic collisions by checking both momentum AND kinetic energy. In a 4-mark Advanced question, the verification itself can earn you method marks.

Difficulty Distribution

For JEE Main (per question, based on 2019–2024 analysis):

Difficulty	% of Questions	What It Tests
Easy	30%	Direct formula: $W = Fd\cos\theta$ , KE calculation, $P = Fv$
Medium	50%	Work-energy theorem with friction/inclines, spring-mass problems
Hard	20%	PE curves with stability analysis, collision + other mechanics combined

For JEE Advanced, the split shifts: roughly 20% Easy / 40% Medium / 40% Hard, with Hard questions usually appearing as multi-concept problems (collision + rotational mechanics, or PE curve + oscillations).

Expert Strategy

Week 1 — Foundation (if you’re starting fresh):

Master work-energy theorem first, before anything else. Solve 20 problems where you identify all forces, calculate each one’s work, and apply $W_{net} = \Delta KE$ . This single tool handles 60% of the chapter’s JEE questions.

Week 2 — Concept deepening:

Take PE curves seriously — they’re a favourite in JEE Advanced and connect directly to Simple Harmonic Motion. Practice sketching $F$ vs $x$ from $U$ vs $x$ graphs by hand. Recognise that where $U$ is minimum, $F = 0$ and motion is SHM-like.

For collisions, don’t memorise the formulas — derive them once from momentum conservation + coefficient of restitution. That way you’ll never mix up which mass goes in the numerator.

PYQ targeting: After concept clarity, solve the last 5 years of JEE Main from this chapter (that’s ~15 questions). You’ll notice the same 4–5 problem patterns recycled. Pattern recognition here directly translates to marks.

The power question trap: Problems with engines at constant power often ask for “time to reach maximum velocity” or “velocity at time $t$ .” These require differential equation setup: $P = mav = m\frac{dv}{dt}v$ . Separate variables and integrate. Practice this specific setup at least three times.

Common Traps

Trap 1 — Wrong angle in $W = Fd\cos\theta$

The angle is between $\vec{F}$ and $\vec{d}$ (displacement), not between the force and the horizontal. On an inclined plane with a horizontal force, the angle is the incline angle. Sketch the vectors every time.

Trap 2 — Using energy conservation when friction is present

$\frac{1}{2}mv_i^2 + mgh_i = \frac{1}{2}mv_f^2 + mgh_f$ is only valid without friction. With friction, the correct relation is $\Delta KE = W_{gravity} + W_{friction} + W_{applied}$ . Missing this costs 4 marks every time.

Trap 3 — Sign error in $F = -dU/dx$

Students often forget the negative sign. If $U$ increases with $x$ (positive slope), the force points in the $-x$ direction. Physically: force pushes the particle toward lower potential energy. Always draw the $U$ curve first, then reason about force direction before calculating.

Trap 4 — Perfectly inelastic collision: only momenta add, not velocities

When two objects stick together, use $m_1v_1 + m_2v_2 = (m_1 + m_2)v_f$ . The KE is NOT conserved — always calculate the KE lost separately if the question asks. $(KE_{lost} = KE_i - KE_f)$

Trap 5 — Power = $Fv$ assumes F and v are parallel

The full expression is $P = Fv\cos\theta$ . In problems where a vehicle moves horizontally but a force acts at an angle (like a towrope), you must include $\cos\theta$ . Forgetting this is a standard 2-mark trap in JEE Main.

Examiner’s favourite trick: Give you a PE function like $U = ax^2 - bx^4$ and ask for the stable equilibrium position AND the speed at that point when released from a given $x$ . This combines $F = -dU/dx$ with energy conservation in two steps. Practise this exact format from Cengage or DC Pandey before your exam.