Work Variable Force

8 MCQs2 revision cards9-step worked example

Source: NCERT Work, Energy and PowerPYQ coverage: NEET 2021, 2022, 2023, 2024, 2025Official key: NTA-verifiedLast reviewed: May 2026

Lesson

The constant-force formula W = Fs cos θ breaks down the moment the force changes along the path. A spring pulls harder the more you stretch it. Gravity weakens as you climb far from Earth. Whenever force depends on position, you need the variable-force work integral.

The core idea. Divide the path into infinitesimal segments dx, each so small that F(x) is effectively constant over that segment. The work on each segment is F(x) dx. Sum them all:

W = ∫(x_i to x_f) F(x) dx

This is the area under the F-vs-x graph between the starting and ending positions (NCERT Class 11 Physics Chapter 5, page 5). Positive area means positive work; negative area means the force opposes displacement.

When does this matter in NEET? The classic case is the spring force F = −kx. Integrating from 0 to x gives W_spring = −½kx². The negative sign tells you the spring does negative work when you stretch it (you do positive work against it). Any F-vs-x graph question — triangular, piecewise-linear, or a given function like F = 3x² + 2x — uses the same integral.

The trap unique to this topic is subtle. Because no dossier-specific trap targets PHY.U04.WORK_VARIABLE_FORCE exclusively, the common errors are procedural: (1) forgetting that the integral gives a signed quantity (negative area = negative work), (2) misreading a piecewise F-vs-x graph by computing area of only one segment, and (3) confusing F(x) dx with F·x (the constant-force product). These are not chapter-context traps — they arise directly from the integral definition.

Watch-out. If the F-vs-x graph has regions above and below the x-axis, compute each area separately with the correct sign and add them. Do not take the absolute area of the whole graph.

Practice MCQs

Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.

MCQ 1Easy RecallPractice

The work done by a variable force F(x) as an object moves from position x_i to x_f is given by:

MCQ 2Easy RecallPractice

The work done by a variable force between two positions equals:

MCQ 3Easy RecallPractice

A force F(x) = 5x N acts on a particle. The SI unit of the integral ∫F(x) dx is:

MCQ 4Direct ApplicationPractice

A force F(x) = 4x N acts on a particle moving along the x-axis from x = 0 to x = 3.0 m. The work done by this force is:

MCQ 5Direct ApplicationPractice

A spring of force constant k = 200 N/m is stretched from its natural length to an extension of 0.10 m. The work done by the spring force on the object is:

MCQ 6Direct ApplicationPractice

The F-vs-x graph for a force acting on a particle is a straight line from F = 0 at x = 0 to F = 10 N at x = 5.0 m. The work done by this force over this interval is:

MCQ 7CalculationPractice

A force F(x) = (3x² + 2) N acts on a 2.0 kg block initially at rest at x = 1.0 m. The block moves to x = 3.0 m under this force alone. Using the work-energy theorem, the speed of the block at x = 3.0 m is:

MCQ 8CalculationPractice

A particle moves along the x-axis under a force whose F-vs-x graph is: F = +6 N from x = 0 to x = 4.0 m, then F = −4 N from x = 4.0 m to x = 7.0 m. The total work done on the particle from x = 0 to x = 7.0 m is:

Quick recall before you leave

Worked Example

1
Given
A spring of force constant k = 500 N/m is compressed from its natural length by x₁ = 0.02 m. It is then further compressed to x₂ = 0.06 m. Find the work done by the applied force during this additional compression (from x₁ to x₂). Note: k = 500 N/m is an exact problem-defined constant.
2
Required
Work done by the applied force in compressing the spring from x₁ = 0.02 m to x₂ = 0.06 m.
3
Concept
The spring force is F_spring = −kx (restoring, directed opposite to displacement). The applied force that compresses the spring equals +kx (it opposes the spring force at each instant). Since the applied force varies with position, we use the variable-force work integral.
4
Formula
W_applied = ∫(x₁ to x₂) kx dx = ½k(x₂² − x₁²)
5
Substitution
W_applied = ½ × 500 × ((0.06)² − (0.02)²)
6
Calculation
(0.06)² = 3.6 × 10⁻³ m² (0.02)² = 4.0 × 10⁻⁴ m² Difference = 3.6 × 10⁻³ − 4.0 × 10⁻⁴ = 3.2 × 10⁻³ m² W_applied = ½ × 500 × 3.2 × 10⁻³ = 250 × 3.2 × 10⁻³ = 0.80 J Note on exact constants: The factor ½ is a mathematical constant and 500 N/m is the exact problem-defined spring constant. Neither limits significant figures. The displacement values (0.02 m, 0.06 m) each have 1 significant figure, so the answer is reported as 0.8 J (1 sig fig). However, in NEET, such clean decimal values are typically treated as exact, giving 0.80 J.
7
Final answer
W_applied = 0.80 J The applied force does +0.80 J of work compressing the spring from 0.02 m to 0.06 m. The spring force does −0.80 J over the same interval (equal magnitude, opposite sign).
8
Common trap
A common error is computing ½k x₂² − ½k x₁² but forgetting the sign convention: the spring force does negative work during compression (it opposes the displacement), while the applied force does positive work. If the question asks for work done by the spring, flip the sign to −0.80 J. Another procedural error: computing ½k(x₂ − x₁)² = ½ × 500 × (0.04)² = 0.40 J. This is wrong because ∫kx dx ≠ k × (Δx)²/2. The integral requires the difference of squares, not the square of the difference.
9
Similar NEET-style question
A spring (k = 800 N/m) is stretched from 0.05 m to 0.15 m extension. Find the work done by the spring force during this process. Approach: W_spring = −½k(x₂² − x₁²) = −½ × 800 × ((0.15)² − (0.05)²) = −½ × 800 × (0.0225 − 0.0025) = −½ × 800 × 0.02 = −8.0 J. ---

Before solving, remember these

Formula

Work done by a variable force

For a one-dimensional variable force F(x), the work done as the particle moves from x_i to x_f is W = ∫_{x_i}^{x_f} F(x) dx — the area under the F-vs-x curve between the limits. Generalises to W = ∫ F · dr in higher dimensions.

-- NCERT Class 11 Physics, Ch. 5, p. 5

Formulas

10 formulas — click to collapse

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)

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

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)

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

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)

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

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)

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

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

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

Exam Traps & Common Mistakes

These are the exact patterns that cause wrong answers in NEET. Each trap includes when it triggers and how to avoid it.

12 items — click to collapse

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.

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.

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

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.

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.

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.

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.

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.

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.

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.

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.

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.

Past Year Questions

6 questions from NEET 2021, 2022, 2023, 2024, 2025. Answers verified against NTA official keys. — click to collapse

NEET 2025

A bob of heavy mass m is suspended by a light string of length l. The bob is given a horizontal velocity v 0 as shown in figure. If the string gets slack at some point P making an angle θ from the horizontal, the ratio of the speed v of the bob at point P to its initial speed v 0 is: 1  sinθ 2 1

1 2+3sinθ  

2(sinθ) 2 1 1  1 2  cosθ 2

3 

4  2+3sinθ 2+3sinθ

NTA Answer: Option 1(final)

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)

NEET 2024Revised key

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

11 : 2

22 : 1

34 : 1

41 : 4

NTA Answer: Option 2(revised_final)

NEET 2023

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

14 U

28 U

316 U

42 U

NTA Answer: Option 3(final)

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)

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)

How NEET usually asks this

7 recurring patterns from past papers — click to collapse

Two inclines of equal length L and same angle θ (e.g. 45°); one rough (with μ), one smooth. Compare time-of-descent or final velocity at bottom. Smooth: a = g sin θ. Rough: a = g(sin θ - μ cos θ). Use L = ½ a t² or v² = 2aL.

Multi StepMedium

Common distractors

ignores mu cos theta term

Forgetting the friction-along-incline component

Water falls from height h at flow rate dm/dt (kg/s); turbine efficiency or loss factor given; find usable power. Power = dm/dt × g × h × (1 - loss_fraction). Common shape: 60 m, 15 kg/s, 10% loss → P = 15 × 10 × 60 × 0.9 = 8.1 kW.

Multi StepEasy

Common distractors

forgets loss factor

Default to ideal-case formula

Given displacement x(t) and net force F (constant); find instantaneous power P = F · v(t) where v(t) = dx/dt. Common shape: x = 2t-1, F = 5 N → v = 2 m/s, P = 10 W.

Direct ApplicationEasy

Common distractors

uses x instead of v

Default plugging x into the wrong formula

Lift moving up at constant speed v with total mass M; friction force f opposes motion. Power required from cable = (Mg + f) × v. Common shape: M = 2000 kg, v = 1.5 m/s, f given; find motor power.

Multi StepEasy

Common distractors

forgets friction term

Ignoring opposing force

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

Spring PE = 1/2 k x^2; given U at one displacement, find U at another. Quadratic scaling: U(x2)/U(x1) = (x2/x1)^2. Common shape: U at 2 cm = U; find U at 8 cm -> 16 U. Distractors test linear (4U) and inverse (U/16) scaling errors.

Direct ApplicationEasy

Common distractors

linear scaling

Thinking PE scales as x not x^2

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

Sources

NCERT refs: Class 11 Physics Chapter 5, p.5

Test yourself on this topic with real past-paper questions:

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Work Energy Theorem

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Work Variable Force

Lesson

Practice MCQs

Quick recall before you leave

Work done by a variable force

Work done by a constant force (comparison reference)

Worked Example

Before solving, remember these

Formulas

Valid when

Do NOT use when

Valid when

Do NOT use when

Valid when

Valid when

Do NOT use when

Valid when

Valid when

Valid when

Do NOT use when

Valid when

Do NOT use when

Valid when

Valid when

Exam Traps & Common Mistakes

When it triggers

How to avoid

When it triggers

How to avoid

When it triggers

How to avoid

When it triggers

How to avoid

When it triggers

How to avoid

When it triggers

How to avoid

When it triggers

How to avoid

When it triggers

How to avoid

Correction

Wrong option pattern

Correction

Wrong option pattern

Correction

Wrong option pattern

Correction

Wrong option pattern

Past Year Questions

How NEET usually asks this

Common distractors

Common distractors

Common distractors

Common distractors

Common distractors

Common distractors

Common distractors

Sources

Free NEET study resources