NEET (UG) · Physics, Class 11
Work, Energy and Power: NEET PYQ, 2018–2026
Work, Energy and Power (Physics, Class 11) has been asked 10 times in the 7 NEET papers from 2018, 2020, 2021, 2023, 2024, 2025 and 2026: about 1.4 questions a paper, 3% of all Physics questions. It is the 18th most-asked of 20 Physics chapters. NEET 2026 had 1 from it. Last checked 28 October 2026.
Questions per paper
Year by year.
| Year | Questions |
|---|---|
| 2018 | 2 |
| 2020 | 0 |
| 2021 | 2 |
| 2023 | 1 |
| 2024 | 2 |
| 2025 | 2 |
| 2026 | 1 |
From 2021 to 2024 each paper had 200 questions (a Section B with a choice); the others had 180.
Inside the chapter
Topics asked most.
- Collisions3
- Conservation of energy2
- Motion in a vertical circle2
- Power2
- Work-energy theorem1
Try it
10 Work, Energy and Power questions to practise.
Our own questions, written in the NEET pattern and checked, each with an explanation. Pick an option to see the answer.
Q1 · Physics · Work, Energy and Power
A constant force of 10 N acts on a body at an angle of 60° to the direction of its velocity. The body moves with a steady speed of 4 m/s. What is the instantaneous power delivered by the force?
Answer: (A)
Instantaneous power is the dot product of force and velocity, P = F·v = F v cos θ, where θ is the angle between them. Here cos 60° = 0.5, so P = 10 × 4 × 0.5 = 20 W. Only the component of force along the velocity contributes.
Q2 · Physics · Work, Energy and Power
A 2 kg block is raised vertically from rest by a light rope, with a constant upward acceleration of 2 m/s². How much work, in joules, does the tension in the rope do on the block while it rises 5 m? (Take g = 10 m/s².)
Answer: (D)
Newton's second law gives T − mg = ma, so T = 2(10 + 2) = 24 N. Work by tension = T·d = 24 × 5 = 120 J. Check with the work-energy theorem: W_T + W_gravity = ΔKE, so 120 − 100 = 20 J, which equals ½mv² with v² = 2ad = 20.
Q3 · Physics · Work, Energy and Power
A body of mass 1.2 kg, at rest on a smooth horizontal surface, is pushed by a horizontal force that rises uniformly from zero to 12 N over the first 2 m, stays at 12 N for the next 3 m, and then falls uniformly to zero over the last 2 m. What is the speed of the body after these 7 m?
Answer: (C)
The work done is the area under the force–displacement graph: ½ × 2 × 12 + 3 × 12 + ½ × 2 × 12 = 12 + 36 + 12 = 60 J. By the work–energy theorem this equals the gain in kinetic energy, so ½ × 1.2 × v² = 60, v² = 100 and v = 10 m s⁻¹.
Q4 · Physics · Work, Energy and Power
A bob of mass 0.1 kg is attached to a light inextensible string and moves in a complete vertical circle about the fixed end of the string. Let T_L be the tension when the bob is at the lowest point and T_H the tension when it is at the highest point. Take g = 10 m/s². What is the difference T_L − T_H?
Answer: (D)
Principle: energy is conserved on the circle, and the height difference between the bottom and top is 2L. Energy gives v_L² = v_H² + 4gL. Writing the radial equations, T_L = mg + mv_L²/L and T_H = mv_H²/L − mg. Subtracting and substituting for v_L² gives T_L − T_H = 6mg, independent of L and of the speeds. With m = 0.1 kg and g = 10 m/s², the difference is 6 N.
Q5 · Physics · Work, Energy and Power
A 2 kg block moving at 10 m s⁻¹ on a rough horizontal surface comes to rest after sliding 20 m. What is the average friction force on it?
Answer: (B)
By the work–energy theorem, the work done by friction equals the change in kinetic energy: F × 20 = ½ × 2 × 10² = 100 J, so F = 5 N.
Q6 · Physics · Work, Energy and Power
A small block slides without friction from rest at the top of a fixed smooth hemispherical dome of radius R, after a negligible nudge. The block leaves the surface of the dome at some point. What is its speed at the instant it loses contact with the surface? (Neglect air resistance.)
Answer: (B)
Let θ be the angle of the radius to the block, measured from the vertical. The block has fallen a height R(1 − cosθ), so energy conservation gives v² = 2gR(1 − cosθ). Radially, mg cosθ − N = mv²/R. The block leaves when N = 0, so v² = gR cosθ. Equating: gR cosθ = 2gR(1 − cosθ), so 3cosθ = 2 and cosθ = 2/3. Then v² = gR·(2/3), so v = √(2gR/3).
Q7 · Physics · Work, Energy and Power
A projectile of mass 1 kg is launched from ground level with a speed of 20 m/s at an angle of 30° above the horizontal. Neglecting air resistance, what is its kinetic energy at the highest point of its trajectory? (Take cos 30° = 0.866.)
Answer: (A)
Principle: with no air resistance, mechanical energy is conserved. At launch, KE = ½(1)(20)² = 200 J and the height is zero. At the top, the vertical velocity is zero and the horizontal velocity is unchanged, so vₓ = 20 cos 30° = 17.32 m/s. The kinetic energy at the top is ½(1)(17.32)² = 150 J. The remaining 50 J has become potential energy, mgh = 50 J, so h = 5 m, which agrees with the vertical launch component.
Q8 · Physics · Work, Energy and Power
A lift of mass 500 kg, carrying no passengers, is moving upward with an acceleration of 2 m s⁻². At the instant its speed is 3 m s⁻¹, what power is the cable's tension delivering to it? (Take g = 10 m s⁻²; neglect friction.)
Answer: (D)
Newton's second law for the lift (upward positive) gives T − mg = ma, so T = m(g + a) = 500 × 12 = 6000 N. The tension acts along the velocity, so the instantaneous power it delivers is P = Tv = 6000 × 3 = 18 000 W = 18 kW.
Q9 · Physics · Work, Energy and Power
A pump raises water from a well in which the water surface is 5 m below the outlet, and discharges it from the outlet at a speed of 4 m/s. The mass of water delivered is 10 kg every second. Taking g = 10 m s⁻² and neglecting all losses, what is the minimum power the pump must deliver?
Answer: (B)
Power is the rate at which the pump does work, which equals the rate at which the water gains mechanical energy. Per second, 10 kg gains potential energy mgh = 10 × 10 × 5 = 500 J and kinetic energy ½mv² = ½ × 10 × 16 = 80 J. The total is 580 J per second, so P = 580 W. Taking only the potential term gives 500 W. Taking only the kinetic term gives 80 W. Using mv² in place of ½mv² gives 660 W.
Q10 · Physics · Work, Energy and Power
A 100 W electric bulb is switched on for 5 hours. How much energy does it consume, expressed in kilowatt-hours?
Answer: (A)
Energy consumed equals power multiplied by time. Convert 100 W to 0.1 kW, then multiply by 5 h to get 0.5 kWh. The kilowatt-hour is a unit of energy, not power, so the answer must be given in energy units.
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All 10 real NEET questions from Work, Energy and Power.
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About Work, Energy and Power in NEET.
How many questions come from Work, Energy and Power in NEET?
Work, Energy and Power (Physics, Class 11) has been asked 10 times in the 7 NEET papers from 2018, 2020, 2021, 2023, 2024, 2025 and 2026: about 1.4 questions a paper, 3% of all Physics questions. It is the 18th most-asked of 20 Physics chapters. NEET 2026 had 1 from it.
Which topics of Work, Energy and Power are asked most in NEET?
Collisions (3), Conservation of energy (2), Motion in a vertical circle (2), Power (2), counted over the papers from 2018–2026.
Is Work, Energy and Power important for NEET 2027?
It is in the syllabus in force, so it can be asked. By past papers it ranks 18 of 20 in Physics; use that to decide how much time it gets, not whether to study it.
Where can I solve Work, Energy and Power NEET previous year questions?
On Nikaalle, free: sign up and practise every past NEET question from Work, Energy and Power with an explanation for each, or sit the full papers timed. The ten questions on this page are our own, written in the same pattern.
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