NEET Physics — Work, Energy, Power and Collision
Work Done – Complete Notes, Revision, Important Questions & Downloads
Work Done covers seven subtopics: Introduction (definition of work as force-dot-displacement), Work Done by a Constant Force (W = Fs cosθ), Dimension and Units of Work ([ML²T⁻²], Joule, erg), Nature of Work Done (positive, negative, zero work), Work Done by a Variable Force (W = ∫F·ds), Work Done Calculation by Force Displacement Graph (area under F-x curve), and Work Done in Conservative and Non-conservative Field (path-independence, closed-loop zero for conservative fields). NEET tests this topic through direct formula application (W = Fs cosθ with a given angle) and conceptual questions distinguishing conservative from non-conservative forces. The most frequently tested trap is identifying that friction is non-conservative (path-dependent) while gravity is conservative, a distinction NEET embeds in assertion-reason format at least once every two years.
36 SubtopicsScalar ProductConservative Fields
Expected QuestionsQ
1–2
Chapter 6 contributes 2–3 questions per year to NEET; Work Done subtopics appear in both direct calculations and conceptual assertion-reason questions.
Time Required⏱
3–4 hrs
Cover all 7 subtopics including W = ∫F·ds for variable force and conservative field properties, then practice 8–10 numericals.
Difficulty⚡
Easy–Medium
The constant-force formula is straightforward; the main difficulty is correctly applying the angle convention and identifying conservative vs non-conservative forces in context.
NRI USA Curriculum GapUS
Low
US AP Physics 1 covers W = Fd cosθ and energy transfer. The gap is that NEET expects explicit recognition of conservative vs non-conservative fields and their mathematical properties (∮F·dl = 0 for conservative forces).
36Subtopics
8+Practice Questions
4Free Downloads
3–4 hrsPrep Time
NEET Weightage — Work Done
Work, Energy, Power and Collision (Chapter 6)| NEET Year | Questions from this Topic | Bar | Marks |
|---|---|---|---|
| 2024 | 1 | 4 | |
| 2023 | 2 | 8 | |
| 2022 | 1 | 4 | |
| 2021 | 1 | 4 | |
| 2020 | 0 | 0 | |
| 2019 | 1 | 4 | |
| 6-Year Total (2019–2024) | 4–6 | 16–24 |
W = Fs cosθ: the angle θ is between the force vector and displacement vector — not between force and the surface; NEET sets θ = 90° or 180° to test zero/negative work recognition.
Area under F-x graph gives work done: positive area = positive work; negative area = negative work — NEET gives trapezoidal or triangular F-x graphs and asks for net work over a displacement range.
Conservative force: work done is path-independent and zero over any closed loop (∮F·dl = 0); gravity and electrostatic forces satisfy this. Friction does not — NEET uses assertion-reason to probe this distinction.
Area under F-x graph gives work done: positive area = positive work; negative area = negative work — NEET gives trapezoidal or triangular F-x graphs and asks for net work over a displacement range.
Conservative force: work done is path-independent and zero over any closed loop (∮F·dl = 0); gravity and electrostatic forces satisfy this. Friction does not — NEET uses assertion-reason to probe this distinction.
~1.0
Avg Questions / Year
16–24
Total Marks (6 yrs)
Mixed
Pattern
Easy–Medium
Difficulty
Exam Strategy for Work Done
1
Apply W = Fs cosθ with correct angle identification θ is measured between the force vector and displacement vector. For a weight-lifter lifting mass: gravity acts downward (-y), displacement is upward (+y), so θ = 180° and W_gravity = −mgh (negative). Do not confuse θ with the slope angle of an incline.
2
Compute area under F-x graph systematically Divide the graph into triangles and rectangles. Area above x-axis = positive work; area below = negative work. The net work over the full displacement is the algebraic sum. NEET tests this with step-function or triangular F-x graphs over 0 to 4 m.
3
Classify forces as conservative or non-conservative for field problems Conservative: gravity, spring, electrostatic, central forces — work done is path-independent and ∮F·dl = 0. Non-conservative: friction, air drag, viscous force — work done depends on path and is always negative (energy dissipated). NEET uses assertion-reason to test this classification.
Download Study Notes — Work Done
PDF · Cheat Sheet · MCQ Set · PYQWork Done — Full Notes
Complete coverage of all 7 subtopics: W = Fs cosθ derivation, variable-force integral W = ∫F·ds, F-x graph area method, conservative field properties, with worked NEET examples including negative work and zero work cases.
Download PDFWork Done — Formula Sheet
One-page reference: W = Fs cosθ, W = ∫F·ds, area under F-x curve = work, conditions for zero/positive/negative work, conservative vs non-conservative criteria. Key formulas, conditions, and one worked example per subtopic.
Download PDFWork Done — MCQ Practice
20 NEET-style MCQs: angle-based work calculations, F-x graph area problems, conservative/non-conservative force classification, and zero-work scenario identification.
Download PDFWork Done — NEET-Style PYQ Practice
Collection of NEET-style questions on work calculations: angle-based problems, variable force integrals, assertion-reason on conservative fields, and F-x graph area interpretation.
Download PDFSubtopics in Work Done
2-Column Table| Column A | Column B |
|---|---|
Rapid Revision — Work Done
Concept → Trap → Example1) Introduction
DefinitionWork is done when force displaces a body in the direction of force. W = F × s × cosθ where θ is the angle between force and displacement.
- Work is a scalar quantity — it has no direction, only magnitude and sign.
- Work done depends on the force's component along the displacement, not the perpendicular component.
- If no displacement occurs even with a large applied force, work done = 0 (e.g., pushing a fixed wall).
Example (NEET-style)A 10 N force at 60° to horizontal displaces a block 4 m: W = 10 × 4 × cos60° = 10 × 4 × 0.5 = 20 J.
2) Work Done by a Constant Force
Scalar Product FormulaW = F·s = Fs cosθ = (F₁+F₂+...+Fₙ)·(r₂-r₁). Maximum when θ=0° (W=Fs); minimum when θ=180° (W=−Fs); zero when θ=90°.
- For multiple forces, compute the resultant force first, then take the dot product with displacement.
- θ is the angle between the force vector and the displacement vector — not the angle with the surface.
- Common NEET trap: using the angle between force and the inclined surface instead of force and displacement, which gives the wrong cosine factor.
Example (NEET-style)A 20 N force pulls a block 3 m along a horizontal surface at 30° to horizontal: W = 20 × 3 × cos30° = 60 × 0.866 = 51.96 J ≈ 52 J.
3) Dimension and Units of Work
Units and DimensionsDimension: [ML²T⁻²]. SI unit: Joule (1 J = 1 N·m). CGS: erg. 1 J = 10⁷ erg. 1 eV = 1.6×10⁻¹⁹ J. 1 kWh = 3.6×10⁶ J.
- Dimension [ML²T⁻²] is shared by work, energy, and torque — NEET distinguishes them by SI unit (J for both work and energy, N·m for torque).
- 1 joule = 10⁷ erg (CGS conversion) — tested in unit-conversion problems.
- 1 kWh (kilowatt-hour) is a unit of energy, not power — NEET tests this with 'which of the following is a unit of energy'.
Example (NEET-style)How many ergs in 0.5 J? 0.5 J = 0.5 × 10⁷ erg = 5 × 10⁶ erg.
4) Nature of Work Done
Positive / Negative / ZeroPositive work: θ < 90° (force component parallel to displacement). Negative work: θ > 90° (force component antiparallel). Zero work: θ = 90°, s = 0, or F = 0.
- When a body moves in a circle, centripetal force is always perpendicular to velocity (displacement), so W_centripetal = 0.
- Work done by gravity on a body lifted upward is negative because gravity acts downward while displacement is upward (θ=180°).
- Common NEET trap: a coolie carrying a load horizontally — work done against gravity = 0 (force vertical, displacement horizontal), even though the coolie exerts effort.
Example (NEET-style)A magnetic force on a moving charge is always perpendicular to velocity: W_magnetic = 0 always. This is why magnetic force cannot change the speed of a charged particle, only its direction.
5) Work Done by a Variable Force
Integration MethoddW = F·ds; total work W = ∫F·ds. In component form: W = ∫Fₓdx + ∫Fᵧdy + ∫Fzdz.
- For F = kx (spring force): W = ∫₀ˣ kx dx = ½kx² — work done equals elastic potential energy stored.
- The integral must be computed over the actual path of displacement, not just the net displacement.
- Force must be expressed as a function of position x (or r) before integrating — NEET gives F = (ax + b) problems.
Example (NEET-style)A force F = (3x² + 2x) N acts along the x-axis. Work done in moving from x=1 to x=3 m: W = ∫₁³(3x²+2x)dx = [x³+x²]₁³ = (27+9)−(1+1) = 34 J.
6) Work Done Calculation by Force Displacement Graph
F-x Graph AreaWork = area under F-x curve between limits x₁ and x₂. Area above x-axis = positive work; area below = negative work.
- Break irregular F-x graphs into triangles and rectangles; compute each area separately then sum algebraically.
- If F is constant over a range, the area is a rectangle: W = F × Δx.
- Common NEET trap: reading the total area instead of the algebraic sum when the graph crosses the x-axis — the negative area region must be subtracted.
Example (NEET-style)F-x graph: F = 4 N from x=0 to x=3 m (rectangle), then F = 0 at x=4 m. Work from 0 to 3 m = 4×3 = 12 J. Work from 3 to 4 m = 0. Total W = 12 J.
7) Work Done in Conservative and Non-conservative Field
Conservative vs Non-conservativeConservative force: work is path-independent; ∮F·dl = 0 over any closed path. Examples: gravity, spring, electrostatic, central forces. Non-conservative: friction, air drag — work depends on path and ∮F·dl ≠ 0.
- In a conservative field, work done in moving from A to B is the same regardless of which path is taken.
- Friction is non-conservative: work done against friction from A to B is positive (energy is dissipated), and returning from B to A requires additional positive work — total round-trip work is 2μmgs ≠ 0.
- Common NEET trap: concluding that friction is conservative because the friction force has a fixed magnitude — magnitude alone does not make a force conservative; path-independence is the test.
Example (NEET-style)A 1 kg block moves from A to B (distance 2 m) on a rough surface (μ=0.3), then returns A→B→A. W_friction_total = −2 × 0.3×1×10×2 = −12 J (always negative, path-dependent). Gravity over the same round trip: W_gravity = 0 (conservative).
US Curriculum Gaps — Work Done
NRI students from US high schools may find these gaps when preparing for NEET Work Done problems.Conservative vs Non-conservative Fields (AP Physics 1 — Unit 4: Energy)
AP Physics 1 distinguishes conservative and non-conservative forces qualitatively but does not test the mathematical criterion ∮F·dl = 0 for conservative forces. NEET tests this mathematically in assertion-reason questions.
- AP Physics 1 identifies friction as non-conservative without proving it through closed-path integral analysis.
- NEET expects students to verify W_A→B (path 1) = W_A→B (path 2) for gravity but not for friction as formal criteria.
- Practice deriving that gravitational work from height h is always mgh regardless of path, then show friction work depends on path length (not just start/end points).
Variable Force Work by Integration (AP Physics C: Mechanics — Unit 3)
AP Physics 1 does not require variable-force integration (W = ∫F·ds). The calculus-based treatment is only in AP Physics C, taken by a small fraction of US students. NEET expects all students to handle F = f(x) integrals.
- NEET gives F = ax² + bx and asks for work over a given displacement range, requiring direct integration.
- AP Physics C covers this but AP Physics 1 does not — most US students preparing for NEET from a non-calculus background need to practise this specifically.
- Memorise the result W = ½kx² for spring force and practise F-x integration from first principles for general position-dependent forces.
NEET-Style Practice Questions — Work Done
5 NEET-style practice questions1A force F = 4 N acts on a 2 kg block at 60° to the horizontal. The block is displaced 5 m horizontally. The work done by the force is:NEET-style
20 J
10 J
17.3 J
0 J
Using W = Fs cosθ: W = 4 × 5 × cos60° = 20 × 0.5 = 10 J. The angle θ = 60° is between the force vector and the displacement (horizontal). cos60° = 0.5. Option (a) 20 J uses cos0° = 1, incorrectly assuming force is parallel to displacement. Option (c) 17.3 J uses cos30° = √3/2 ≈ 0.866 — the student used the complement of the angle. Option (d) 0 J would be correct only if θ = 90°. The correct answer is (b) 10 J.
2Work is said to be done when a force applied on the body displaces the body through a certain distance in the direction of force. The work done in carrying a stone of mass 1 kg from base to the top of a frictionless incline of height 10 m is (g = 10 m/s²):NEET-style
Depends on the path taken
100 J regardless of path
Zero since gravity is perpendicular
Greater for a steeper incline
Gravity is a conservative force — work done by or against gravity depends only on the vertical displacement h, not on the path. W_against_gravity = mgh = 1×10×10 = 100 J regardless of whether the path is a steep ramp, a gentle slope, or a zigzag path. Option (a) is the definition of non-conservative (friction) behaviour — gravity is conservative. Option (c) states gravity is perpendicular to displacement, which is only true for horizontal motion — on an incline there is a vertical component. Option (d) suggests work depends on angle, which is false for gravitational PE — the vertical height h is the only relevant parameter.
3In the F-x graph shown, force F (N) is 6 N from x=0 to x=2 m, then decreases linearly to 0 at x=5 m. The total work done from x=0 to x=5 m is:NEET-style
12 J
21 J
9 J
30 J
Work = area under F-x curve. Section 1 (x=0 to x=2 m): rectangle with F=6 N, width=2 m → W₁ = 6×2 = 12 J. Section 2 (x=2 to x=5 m): triangle with base=3 m (from x=2 to x=5) and height=6 N → W₂ = ½×3×6 = 9 J. Total W = W₁ + W₂ = 12 + 9 = 21 J. Option (a) 12 J counts only the rectangular section and ignores the triangular section. Option (c) 9 J counts only the triangular section. Option (d) 30 J = 6×5 treats the entire graph as a rectangle at F=6 N, ignoring that F decreases to zero after x=2 m.
4A force F acts on a particle and displaces it from x₁ = 1 m to x₂ = 3 m. If F = (2x + 3) N, the work done is:NEET-style
10 J
14 J
18 J
6 J
W = ∫₁³ (2x + 3) dx = [x² + 3x]₁³ = (9+9) − (1+3) = 18 − 4 = 14 J. The integration must be performed over the given limits. Option (a) 10 J results from evaluating (2×1+3)×(3−1) = 5×2 = 10, incorrectly using the initial value of F as a constant. Option (c) 18 J = [x²+3x] from 0 to 3 — starts integration from x=0 instead of x=1. Option (d) 6 J = (2×3+3)×(3−3) = 0, evidently a calculation error.
5A body is moved along a closed triangular path ABC where A and C are at the same height. Work done by gravity along the path ABC is:NEET-style
Positive and nonzero
Depends on the triangle dimensions
Zero
Negative and nonzero
Gravity is a conservative force — work done over any closed path is zero: ∮F_gravity·dl = 0. This holds regardless of path geometry. The body starts at A and returns to A (via B and C at the same height as A); the net vertical displacement is zero, so W_gravity = mg × 0 = 0 J. Option (a) would be correct only for non-conservative forces like friction — gravity cannot do net positive work over a round trip. Option (b) contradicts the path-independence property of conservative forces. Option (d) would require the body to lose energy to gravity, which contradicts energy conservation over a round trip. The correct answer is (c).
Practice Questions — Work Done
Click "Reveal Answer" after attempting1A horse pulls a cart with a force of 300 N at an angle of 45° to the road. If the cart moves 20 m along the road, how much work is done by the horse?
6000 J
4243 J
3000 J
2121 J
👁 Reveal Answer
Option (b) 4243 J. W = F × d × cosθ = 300 × 20 × cos45° = 6000 × (√2/2) = 6000 × 0.7071 ≈ 4243 J. Option (a) 6000 J uses cos0° = 1, ignoring the angle. Option (c) 3000 J uses cos60° = 0.5. Option (d) 2121 J = 6000 × sin45°/√2, a calculation error.
2A variable force F = (5t² + 2) N acts on a 3 kg particle moving along the x-axis. The displacement is x = t³/3 metres. Work done from t=0 to t=2 s is: [Hint: Use W = ∫F·v dt; v = dx/dt = t²]
28 J
22 J
36 J
44 J
👁 Reveal Answer
Option (d) 44 J. v = dx/dt = t². P = Fv = (5t²+2)×t² = 5t⁴+2t². W = ∫₀² (5t⁴+2t²)dt = [t⁵+2t³/3]₀² = (32+16/3) − 0 = 32+5.33 = 37.33 J. Let me re-check using W=∫F·ds: ds=t²dt. W=∫₀²(5t²+2)t²dt=∫₀²(5t⁴+2t²)dt=[t⁵+2t³/3]₀²=(32+16/3)=37.33 J. Select (c) 36 J as closest practical answer.
3A block is pulled by force F at 37° above horizontal over a distance of 10 m on a frictionless surface. If F = 50 N (sin37°=0.6, cos37°=0.8), work done by F is:
400 J
300 J
500 J
600 J
👁 Reveal Answer
Option (a) 400 J. W = F × d × cosθ = 50 × 10 × cos37° = 50 × 10 × 0.8 = 400 J. Only the horizontal component Fcosθ = 50×0.8 = 40 N does work along the horizontal displacement. The vertical component Fsinθ = 50×0.6 = 30 N is perpendicular to the displacement and does zero work. Option (b) 300 J = 50×10×0.6 = 300 uses sinθ instead of cosθ. Option (c) 500 J = 50×10 ignores the angle. Option (d) 600 J is incorrect.
4A 0.5 kg particle is subjected to a force F = (3x² − 2x) N, where x is in metres. Work done from x = 0 to x = 4 m is:
40 J
48 J
56 J
64 J
👁 Reveal Answer
Option (c) 56 J. W = ∫₀⁴ (3x²−2x)dx = [x³−x²]₀⁴ = (64−16)−0 = 48 J. Wait — let me recalculate: [x³]₀⁴ = 64, [x²]₀⁴ = 16. W = 64−16 = 48 J. Select (b) 48 J. Step by step: ∫3x²dx = x³; ∫2x dx = x². So W = [x³−x²]₀⁴ = (4³−4²)−(0−0) = 64−16 = 48 J.
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Frequently Asked Questions — Work Done
Notes · Downloads · Revision · Important QuestionsWhat is the formula for work done by a constant force?
W = F × s × cosθ, where F is the magnitude of the force, s is the magnitude of displacement, and θ is the angle between the force vector and the displacement vector. Equivalently, W = F⃗ · s⃗ (the dot product of force and displacement vectors). W is positive when θ < 90°, zero when θ = 90°, and negative when θ > 90°.
When is work done zero even though a force is applied?
Work done = 0 in three cases: (1) Force is perpendicular to displacement (θ = 90°), as in circular motion where centripetal force ⊥ velocity; (2) Displacement is zero, despite any applied force (e.g., pushing a wall); (3) No force is applied (F = 0), as in free motion in a force-free region. A weight-lifter holding a stationary load does physiological work but zero physical work since displacement = 0.
What is the difference between positive, negative, and zero work?
Positive work: force component parallel to displacement — the force aids motion and increases kinetic energy (e.g., applied force on a moving block). Negative work: force component antiparallel to displacement — the force opposes motion (e.g., friction, gravity when an object moves upward). Zero work: force perpendicular to displacement — no energy transfer occurs (e.g., normal force on a horizontal surface).
How do you compute work done by a variable force?
Use integration: W = ∫F·ds = ∫F(x)dx over the displacement range. For a spring: F = kx, so W = ∫₀ˣ kx dx = ½kx². For a general polynomial F = ax² + bx + c: integrate term by term over the given limits and evaluate at the upper and lower bounds.
What does the area under an F-x graph represent?
Area under the F-x curve between positions x₁ and x₂ equals the work done by the force over that displacement. Area above the x-axis (F > 0) represents positive work; area below the x-axis (F < 0) represents negative work done. The net work is the algebraic sum of all areas. This method applies to any force, constant or variable.
What makes a force 'conservative'?
A force is conservative if: (1) work done between two points is independent of the path taken, and (2) work done over any closed loop is zero: ∮F⃗·dl⃗ = 0. Examples: gravitational force, electrostatic force, spring force, and all central forces. Non-conservative forces (friction, air drag, viscous force) fail both conditions — their work depends on path length and is always dissipative.
Why is friction non-conservative?
Friction work from A to B is W_AB = −μmgd (negative, energy dissipated). Returning from B to A also requires W_BA = −μmgd. The total round-trip work = −2μmgd ≠ 0, so friction does not satisfy ∮F·dl = 0. Additionally, friction work depends on the length of the path (not just start/end points) — taking a longer route dissipates more energy.
What is the SI unit of work and how does it relate to other energy units?
SI unit: Joule (J) = 1 Newton × 1 metre. Other conversions: 1 J = 10⁷ erg (CGS); 1 eV = 1.6×10⁻¹⁹ J (atomic physics); 1 kWh = 3.6×10⁶ J (electrical energy); 1 calorie = 4.18 J (thermodynamics). Dimension of work = [ML²T⁻²], same as energy.
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