Work, Energy and Power

Easy Recall (3)Direct Application (3)Concept Trap (2)

Elastic Collision 1D

8 MCQs1 PYQs1 revision cardWorked example

Easy Recall (3)Direct Application (3)Concept Trap (1)Calculation (1)

Elastic Collision 2D

Easy Recall (3)Direct Application (3)Concept Trap (1)Calculation (1)

Inelastic Collision 1D

8 MCQs1 PYQs1 revision cardWorked example

Easy Recall (3)Direct Application (3)Calculation (2)

Inelastic Collision 2D

8 MCQs2 revision cardsWorked example

Easy Recall (3)Direct Application (3)Concept Trap (1)Calculation (1)

Kinetic Energy

8 MCQs5 revision cardsWorked example

Easy Recall (3)Direct Application (3)Concept Trap (1)Calculation (1)

Motion Vertical Circle

8 MCQs2 revision cardsWorked example

Easy Recall (3)Direct Application (3)Concept Trap (1)Calculation (1)

Potential Energy Spring

8 MCQs1 PYQs2 revision cardsWorked example

Easy Recall (3)Direct Application (3)Concept Trap (1)Calculation (1)

Potential Energy

Easy Recall (3)Direct Application (3)Concept Trap (1)Calculation (1)

Power

8 MCQs3 PYQs1 revision cardWorked example

Easy Recall (3)Direct Application (3)Concept Trap (1)Calculation (1)

Work Constant Force

Easy Recall (3)Direct Application (3)Concept Trap (1)Calculation (1)

Work Energy Theorem

8 MCQs2 revision cardsWorked example

Easy Recall (3)Direct Application (3)Concept Trap (1)Calculation (1)

Work Variable Force

8 MCQs1 PYQs1 revision cardWorked example

Easy Recall (3)Direct Application (3)Concept Trap (1)Calculation (1)

Past-paper questions from this unit

8 questions from NEET 2021, 2022, 2023, 2024, 2025, 2026. Answers verified against NTA official keys.

By year in our set: 2021 (1) · 2022 (1) · 2023 (1) · 2024 (2) · 2025 (1) · 2026 (2)

Lesson: Work Variable Force

NEET 2026

A particle of mass M moves along a horizontal x axis from x=0 to x=L. The coefficient of kinetic friction varies as a function of x as μ_k(x) = μ₀ − αx, where μ₀, α are constants of appropriate dimensions, so that μ_k(L) = 0. The total work done by the frictional force during the motion is nμ₀MgL, where g is the acceleration due to gravity. The value of n is :

31/3

41/2

NTA Answer: Option 4(final)

Report an error · Every fix is public: corrections log

Lesson: Elastic Collision 1D

NEET 2026

Bob B of mass m at rest is hanging vertically from the ceiling via a massless string of length 10 m. Point mass A of mass m travelling horizontally with speed 10 ms⁻¹ hits bob B elastically. The bob B rises h meter after the collision. Taking the acceleration due to gravity g = 10 ms⁻² and neglecting the size of the bob, the value of h is :

42.5

NTA Answer: Option 3(final)

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NEET 2025

There are two inclined surfaces of equal length (L) and same angle of inclination 45° with the horizontal. One of them is rough and the other is perfectly smooth. A given body takes 2 times as much time to slide down on rough surface than on the smooth surface. The coefficient of kinetic friction (µ k) between the object and the rough surface is close to

10.75

20.25

30.40

40.5

NTA Answer: Option 1(final)

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Lesson: Power

NEET 2024Revised key

At any instant of time t, the displacement of any particle is given by 2t – 1 (SI unit) under the influence of force of 5N. The value of instantaneous power is (in SI unit):

110

NTA Answer: Option 1(revised_final)

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Two bodies A and B of same mass undergo completely inelastic one dimensional collision. The body A moves with velocity v1 while body B is at rest before collision. The velocity of the system after collision is v2. The ratio v1 : v2 is

NEET 2024Revised key

11 : 2

22 : 1

34 : 1

41 : 4

NTA Answer: Option 2(revised_final)

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The potential energy of a long spring when stretched by 2 cm is U. If the spring is stretched by 8 cm, potential energy stored in it will be

NEET 2023

14 U

28 U

316 U

42 U

NTA Answer: Option 3(final)

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Lesson: Power

NEET 2022

An electric lift with a maximum load of 2000 kg (lift + passengers) is moving up with a constant speed of 1.5 ms–1. The frictional force opposing the motion is 3000 N. The minimum power delivered by the motor to the lift in watts is : (g = 10 m s⁻²)

123500

223000

320000

434500

NTA Answer: Option 4(final)

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Lesson: Power

NEET 2021

Water falls from a height of 60 m at the rate of 15 kg/s to operate a turbine. The losses due to frictional force are 10% of the input energy. How much power is generated by the turbine? m/s2) (g = 10

17.0 kW

210.2 kW

38.1 kW

412.3 kW

NTA Answer: Option 3(final)

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Exam traps and common mistakes in this unit

Trap

Vertical circle: energy conservation alone is insufficient

Category: Overthinking

Student uses ½ m v₀² = m g (2L) (energy to reach top) and forgets the additional v_top² ≥ gL constraint for tension.

When it triggers

Question asks for minimum v₀ at lowest point so the bob can complete a full vertical circle.

How to avoid

TWO constraints: (1) energy: v_top² = v₀² - 4gL; (2) tension at top ≥ 0: v_top² ≥ gL. Combined: v₀² ≥ 5gL. Energy alone gives only v₀² ≥ 4gL which is insufficient.

Trap

Inelastic collision: defaulting to elastic

Category: Similar Terms

Student uses the elastic-collision velocity formulas (m1-m2)/(m1+m2) when the question explicitly says 'completely inelastic' (bodies stick).

When it triggers

Question says 'inelastic', 'stick together', 'after collision moves with common velocity'.

How to avoid

Perfectly inelastic 1D: v_common = (m₁ u₁ + m₂ u₂)/(m₁ + m₂). KE is NOT conserved; loss = ½ (m₁ m₂)/(m₁+m₂) × (u₁ - u₂)². Don't use elastic formulas.

Trap

Spring PE linear vs quadratic scaling

Category: Overthinking

Student treats spring PE as proportional to displacement (linear) instead of displacement-squared (quadratic). Common error: 'doubling the stretch doubles the PE'. Actual: doubling the stretch gives 4× the PE.

When it triggers

Question gives U at one stretch and asks for U at another. Distractors include linear-scaling answer (×2 instead of ×4 for double stretch).

How to avoid

U = ½ k x² is QUADRATIC. The PE-to-stretch ratio is the SQUARE of the stretch ratio: if stretch goes from x₁ to x₂, U_new / U_old = (x₂/x₁)².

Lesson: Power

Trap

Lift-power: friction term overlooked

Category: Overthinking

Student computes P = Mgv (just lifting against gravity) and ignores the friction-opposing-motion term.

When it triggers

Question describes a lift moving at constant speed with explicit friction force on cable or guides.

How to avoid

At constant speed, net force = 0, so cable tension T = Mg + f_friction. Power = T × v = (Mg + f) × v. Always add friction when stated.

Lesson: Power

Trap

Power problems: efficiency / loss factor dropped

Category: Overthinking

Student computes ideal power and forgets the (1 - loss_fraction) or efficiency multiplier.

When it triggers

Question gives turbine, motor, or transformer with stated efficiency or loss percentage.

How to avoid

Always read the question for efficiency η or loss%. Useful power P_useful = η × P_input or P_input × (1 - loss). Don't drop the factor even if the rest of the calc is in the unrelated parts of the problem.

Lesson: Power

Trap

Instantaneous power: x vs v

Category: Similar Terms

Student plugs displacement x into P = F·v formula instead of velocity v.

When it triggers

Question gives displacement x(t) explicitly and a constant force; asks for instantaneous power.

How to avoid

P = F · v where v = dx/dt. Compute v first (differentiate x(t) once), THEN plug into F·v. P is NOT F·x.

Lesson: Work Energy Theorem

Trap

Linear speed-distance ratio under uniform deceleration

Category: Overthinking

Student assumes proportionality of speed to remaining distance under uniform deceleration. In fact, KE drops linearly with distance (v² is the linear quantity, not v): v² = u² - 2as. Speed-vs-distance is a sqrt-curve, not a line.

When it triggers

Question describes a body decelerating through stages with given speed at one stage; asks for distance to stop or speed at another stage.

How to avoid

Always work with v², not v, when uniform deceleration is in play. The work-energy theorem gives the same answer faster: ½ m v² = work done against constant force over distance.

Unit-wide

Trap

Incline-with-friction: missing μ cos θ term

Category: Sign Convention

Student writes a = g sin θ for a rough incline (which is the smooth-incline answer); forgets to subtract μ g cos θ.

When it triggers

Question contrasts rough vs smooth inclines, or asks for acceleration on a rough incline.

How to avoid

On a rough incline (block sliding down): a = g(sin θ - μ_k cos θ). On a rough incline (block sliding up): a = -g(sin θ + μ_k cos θ). Smooth case (μ = 0): just g sin θ.

Mistake

Student applies K_i + U_i = K_f + U_f in a problem where friction or other non-conservative forces clearly do work.

Root cause: formula misuse

Correction

Mechanical-energy conservation requires that ONLY conservative forces do work. When friction or drag is present, use the work-energy theorem directly: K_f - K_i = W_conservative + W_non-conservative, where W_non-conservative is typically negative (energy goes to heat).

Wrong option pattern

Distractor sets m*g*h = (1/2)*m*v^2 for a block sliding down a rough incline.

Lesson: Elastic Collision 1D

Mistake

Student applies elastic-collision velocity formulas to an inelastic collision (or vice versa).

Root cause: concept gap

Correction

Elastic: BOTH momentum and kinetic energy conserved -> use the (m1-m2)/(m1+m2) form. Inelastic: ONLY momentum conserved; KE generally lost to heat. Perfectly inelastic: bodies stick together -> common velocity = (m1*u1 + m2*u2)/(m1+m2). Identify which type from the problem before choosing a formula.

Wrong option pattern

Distractor uses elastic-collision formulas for two bodies that the problem says stick together after impact.

Lesson: Potential Energy

Mistake

Student treats gravitational potential energy as having an absolute value, instead of being defined relative to a chosen reference level.

Root cause: concept gap

Correction

PE is defined up to an additive constant. Only DIFFERENCES in PE have physical meaning (they equal the negative work done by the conservative force between two points). Choosing the ground as zero is a convention, not a derivation.

Wrong option pattern

Distractor offers a numerical PE value where two different reference levels would give different correct numbers.

Lesson: Work Constant Force

Mistake

Student claims the normal reaction does positive (or negative) work on a block sliding on a horizontal surface.

Root cause: direction ignored

Correction

Work = F . s = F * s * cos(theta). For a block sliding horizontally, the normal force is perpendicular to displacement (theta = 90 deg, cos = 0), so the work done by N is ZERO. Always check the angle between force and displacement before computing work.

Wrong option pattern

Distractor reports W = N * s for a block sliding on a level surface.

Formulas in this unit

Conservation of mechanical energy

K_{i} + U_{i} = K_{f} + U_{f}

If only conservative forces do work on a system, the total mechanical energy E = K + U is constant in time.

Symbol	Quantity	SI Unit
K_i, K_f	Initial, final kinetic energy	J
U_i, U_f	Initial, final potential energy	J

Valid when

Only CONSERVATIVE forces do work (gravity, spring, electrostatic — not friction or drag)
Closed system; no energy exchange with surroundings

Do NOT use when

Friction, drag, or other non-conservative forces do work
External forces add energy to the system

Lesson: Elastic Collision 1D

Elastic collision — 1D final velocities

v_{1} = \frac{( m _{1} - m _{2} ) u _{1} + 2 m _{2} u _{2}}{m _{1} + m _{2}}, v_{2} = \frac{( m _{2} - m _{1} ) u _{2} + 2 m _{1} u _{1}}{m _{1} + m _{2}}

Final velocities of two bodies after an elastic head-on (1D) collision — momentum AND kinetic energy are both conserved. Special cases: equal masses exchange velocities; very heavy m2 at rest reflects m1 with reversed velocity.

Symbol	Quantity	SI Unit
m1, m2	Masses of the two bodies	kg
u1, u2	Initial velocities (signed, before collision)	m/s
v1, v2	Final velocities (signed, after collision)	m/s

Valid when

Collision is ELASTIC (kinetic energy conserved)
Head-on (1D) — for 2D collisions decompose along/perpendicular to line of impact

Do NOT use when

Inelastic collision (KE not conserved; use momentum-only + restitution)
Bodies stick together (perfectly inelastic case has its own formula)

Perfectly inelastic 1D collision — common final velocity and KE loss

v = \frac{m _{1} u _{1} + m _{2} u _{2}}{m _{1} + m _{2}}, Δ K = - \frac{1}{2} \frac{m _{1} m _{2}}{m _{1} + m _{2}} (u_{1} - u_{2})^{2}

When two bodies stick together after a 1D collision, the common velocity is given by momentum conservation. The KE lost is converted to internal energy (heat, deformation).

Symbol	Quantity	SI Unit
v	Common final velocity	m/s
m1, m2	Masses	kg
u1, u2	Initial velocities	m/s
Delta_K	Change in kinetic energy	J

Valid when

Bodies stick together immediately after collision (perfectly inelastic)
Net external force = 0 during the brief collision (momentum conservation)

Lesson: Kinetic Energy

Kinetic energy

K = \frac{1}{2} m v^{2}

Energy a body possesses by virtue of its motion. Always non-negative. Frame-dependent through v.

Symbol	Quantity	SI Unit
K	Kinetic energy	J
m	Mass	kg
v	Speed	m/s

Valid when

Non-relativistic speeds (v << c)
Translational motion only (rotational KE = (1/2) I omega^2 separately)

Spring potential energy (Hooke's law regime)

U = \frac{1}{2} k x^{2}

Elastic potential energy stored in an ideal spring of force constant k that has been displaced by x from its natural length. Independent of the sign of x (compression or extension).

Symbol	Quantity	SI Unit
U	Elastic PE	J
k	Spring constant (force per unit displacement)	N/m
x	Displacement from natural length	m

Valid when

Spring obeys Hooke's law (F = -k*x) over the displacement range
No internal damping / hysteresis assumed

Do NOT use when

Beyond elastic limit (Hooke's law fails)
Real springs with finite damping (some elastic PE goes to internal energy)

Lesson: Potential Energy

Gravitational potential energy (near Earth)

U = m g h

Potential energy of a body of mass m at height h above the chosen reference level, in a region where g is approximately constant.

Symbol	Quantity	SI Unit
U	Gravitational PE	J
m	Mass	kg
g	Gravitational acceleration	m/s^2
h	Height above reference level	m

Valid when

Region small enough that g is uniform (typically near Earth's surface)
Reference level is freely chosen — only DIFFERENCES in U have physical meaning

Do NOT use when

Large altitude changes (use U = -GMm/r general form)
Below the reference level — h is signed

Lesson: Power

Instantaneous power

P = \frac{d W}{d t} = F \cdot v

Instantaneous power is the time rate of doing work; equivalently, the dot product of the force and velocity. Average power over an interval is W/t.

Symbol	Quantity	SI Unit
P	Power (instantaneous)	W
F	Force	N
v	Velocity	m/s

Valid when

F.v form when force and velocity are both known at the instant

Lesson: Work Constant Force

Work done by a constant force

W = F \cdot s = F s cos θ

The work done by a constant force F on an object that undergoes a displacement s is the dot product F.s. Equivalently, W = (magnitude of F) * (magnitude of s) * cos(angle between them). Work is a scalar but has a sign.

Symbol	Quantity	SI Unit
W	Work (scalar, signed)	J
F	Constant force (vector)	N
s	Displacement (vector)	m
theta	Angle between F and s	rad/deg

Valid when

Force is CONSTANT in magnitude and direction over the displacement
Use the COMPONENT of force along the displacement, not magnitude alone

Do NOT use when

Force varies with position (use the variable-force integral)
Multiple forces act — apply this to each one separately or use net force

Lesson: Work Energy Theorem

Work-energy theorem

W_{net} = Δ K = \frac{1}{2} m (v^{2} - v_{0}^{2})

The net work done by all forces on a particle equals the change in its kinetic energy. Holds for both constant and variable forces, in 1D and higher dimensions.

Symbol	Quantity	SI Unit
W_net	Net work done by all forces	J
Delta_K	Change in kinetic energy	J
m	Mass of particle	kg
v, v0	Final, initial speed	m/s

Valid when

W_net is the NET (vector-sum) work, not work of any one force
Particle (point-mass) idealisation; for extended bodies handle internal energy separately

Lesson: Work Variable Force

Work done by a variable force

W = \int_{x_{i}}^{x_{f}} F (x) d x

When a force varies along the path, the work done is the line integral of force over displacement. In one dimension this is the area under the F-vs-x curve between the start and end positions.

Symbol	Quantity	SI Unit
W	Work done	J
F(x)	Force as a function of position	N
dx	Infinitesimal displacement	m
x_i, x_f	Initial and final positions	m

Valid when

Force may depend on position (e.g. spring force F = -k*x)
Generalises to W = integral F.dr in higher dimensions

NEET question patterns in this unit

Bob of mass m on string of length l given horizontal velocity v₀ at lowest point; find condition for string to remain taut throughout the vertical circle. Minimum: v₀² ≥ 5gl. At top: v_top² = v₀² - 4gl ≥ gl. Energy conservation + tension at any angle.

Multi StepHard

Common distractors

uses only energy not tension

Energy gives speed but tension constraint sets the floor

Two bodies of equal mass m, one moving at v₁ and one at rest, undergo completely inelastic 1D collision. Common velocity = v₁/2 (momentum conservation). KE loss = ½ m v₁² × ½ = ¼ m v₁².

Direct ApplicationEasy

Common distractors

no KE loss stated

Treats collision as elastic by default