NEET Physics — Work Energy and Power Complete Chapter Guide

Chapter Overview & Weightage

Work, Energy, and Power is one of those chapters where understanding the concept saves you from brute-force calculations. The work-energy theorem is a shortcut machine — problems that would take 10 steps with Newton’s laws often collapse into 2 steps with energy methods.

This chapter carries 4-5% weightage in NEET with 2-3 questions. Conservation of energy and collision problems are the most frequent. Work-energy theorem questions are common too.

Year	NEET Q Count	Key Topics Tested
2025	2	Work-energy theorem, elastic collision
2024	2	Power, conservative forces
2023	3	Collision, KE-PE conversion
2022	2	Work by friction, spring energy
2021	2	Potential energy curve, collision

graph TD
    A[Work Energy Power] --> B[Work]
    A --> C[Energy]
    A --> D[Power]
    A --> E[Collisions]
    B --> F[Work by Constant Force]
    B --> G[Work by Variable Force]
    C --> H[KE and Work-Energy Theorem]
    C --> I[PE: Gravitational and Spring]
    C --> J[Conservation of Energy]
    E --> K[Elastic Collision]
    E --> L[Inelastic Collision]
    E --> M[Coefficient of Restitution]

Key Concepts You Must Know

Tier 1 (Always asked)

Work-energy theorem: $W_{net} = \Delta KE$
Conservation of mechanical energy (when only conservative forces act)
Elastic and inelastic collisions in 1D
Power = force times velocity

Tier 2 (Frequently asked)

Work done by spring force: $W = -\frac{1}{2}kx^2$
Work done by friction (negative work)
Potential energy curves and equilibrium points
Coefficient of restitution

Tier 3 (Occasional)

Oblique collisions (2D)
Variable force and work as area under F-x graph

Important Formulas

Work by constant force: $W = Fs\cos\theta$

Work-energy theorem:

W_{net} = \frac{1}{2}mv^2 - \frac{1}{2}mu^2

Work by a variable force:

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

When only conservative forces do work:

KE_i + PE_i = KE_f + PE_f

Spring PE: $U = \frac{1}{2}kx^2$

Gravitational PE: $U = mgh$

If non-conservative forces (like friction) also act:

KE_i + PE_i + W_{friction} = KE_f + PE_f

where $W_{friction}$ is negative.

Elastic collision (KE conserved):

v_1 = \frac{(m_1 - m_2)u_1 + 2m_2 u_2}{m_1 + m_2}

v_2 = \frac{(m_2 - m_1)u_2 + 2m_1 u_1}{m_1 + m_2}

Perfectly inelastic (bodies stick): $v = \dfrac{m_1 u_1 + m_2 u_2}{m_1 + m_2}$

Coefficient of restitution: $e = \dfrac{v_2 - v_1}{u_1 - u_2}$ (elastic: $e = 1$ , perfectly inelastic: $e = 0$ )

Average power: $P_{avg} = \dfrac{W}{t}$

Instantaneous power: $P = \vec{F} \cdot \vec{v} = Fv\cos\theta$

Special cases for elastic collisions: (1) Equal masses — velocities exchange. (2) Heavy hits light at rest — light flies off at $2u$ . (3) Light hits heavy at rest — light bounces back with nearly same speed. These three shortcuts appear in NEET repeatedly.

Solved Previous Year Questions

PYQ 1 — NEET 2023

Problem: A body of mass 2 kg moving with velocity 6 m/s collides head-on elastically with another body of mass 1 kg at rest. Find the velocity of each body after collision.

Solution:

Using elastic collision formulas:

v_1 = \frac{(m_1 - m_2)u_1}{m_1 + m_2} = \frac{(2 - 1) \times 6}{2 + 1} = \frac{6}{3} = \mathbf{2 \text{ m/s}}

v_2 = \frac{2m_1 u_1}{m_1 + m_2} = \frac{2 \times 2 \times 6}{3} = \mathbf{8 \text{ m/s}}

The heavier body slows down, the lighter one shoots ahead faster than the original speed. Momentum and KE are both conserved — you can verify.

PYQ 2 — NEET 2022

Problem: A block of mass 1 kg slides down a rough incline of height 5 m. If 10 J of energy is lost to friction, find the speed at the bottom. ( $g = 10$ m/s $^2$ )

Solution:

Using energy conservation with friction:

mgh = \frac{1}{2}mv^2 + W_{friction}

1 \times 10 \times 5 = \frac{1}{2}(1)v^2 + 10

50 = \frac{v^2}{2} + 10 \implies v^2 = 80 \implies v = \mathbf{4\sqrt{5} \approx 8.94 \text{ m/s}}

Students often forget to include friction work in the energy equation. If the surface is rough, mechanical energy is NOT conserved — you must account for the energy lost to friction. The modified equation is: $KE_f = PE_i - |W_{friction}|$ .

PYQ 3 — NEET 2021

Problem: A car of mass 1000 kg accelerates from rest. If the engine delivers constant power of 50 kW, find the maximum speed if the resistive force is 500 N.

Solution:

At maximum speed, acceleration = 0, so the entire engine power overcomes resistance:

P = F_{resistance} \times v_{max}

50000 = 500 \times v_{max}

v_{max} = \mathbf{100 \text{ m/s}}

Difficulty Distribution

Difficulty	% of Questions	What to Expect
Easy	35%	Work calculation, direct energy conservation
Medium	50%	Collision problems, energy with friction
Hard	15%	PE curves, oblique collisions, variable force work

Expert Strategy

Week 1: Master the work-energy theorem. Solve 20 problems where you find final speed using $W_{net} = \Delta KE$ instead of Newton’s laws. This builds the intuition for when energy methods are faster.

Week 2: Collision problems. Memorise the three special cases for elastic collisions. For inelastic collisions, always check — “is KE conserved?” — the answer is no, and the lost KE often appears as a follow-up question.

Week 3: Power problems and PE curves. The $P = Fv$ formula is the key to all power-related NEET questions. For PE curves, remember: force = negative slope of PE curve, and equilibrium points are where the slope is zero.

When a NEET problem says “find the speed at point B” and involves heights, springs, or friction — go straight to energy conservation. Do NOT use kinematics or Newton’s laws unless the problem specifically asks for time or acceleration.

Common Traps

Trap 1 — Applying conservation of KE in inelastic collisions. Only momentum is conserved in all collisions. KE is conserved only in elastic collisions. If the problem says “bodies stick together,” it is perfectly inelastic — KE is NOT conserved.

Trap 2 — Sign of work done by gravity. When an object moves downward, gravity does positive work ( $mgh$ ). When it moves upward, gravity does negative work ( $-mgh$ ). The sign depends on whether the force and displacement are in the same direction.

Trap 3 — Confusing PE at natural length of spring. The PE of a spring is $\frac{1}{2}kx^2$ where $x$ is the displacement from the natural length — not from any arbitrary point. If the spring is compressed by 3 cm and then released, the PE at natural length is zero.

Trap 4 — Power at maximum velocity. At maximum speed, acceleration is zero. So $P = Fv$ where $F$ equals the resistive force (not the engine force minus resistance). Students often set up the equation wrong by including $ma$ when $a = 0$ .

Trap 5 — Energy loss in perfectly inelastic collision. The KE lost is $\Delta KE = \dfrac{1}{2}\dfrac{m_1 m_2}{m_1 + m_2}(u_1 - u_2)^2$ . This is NOT zero. Many students assume “momentum conserved” means “no energy lost” — wrong.