JEE (Main) · Physics, Class 11
Work, Energy and Power: JEE Main PYQ, 2020–2025
Work, Energy and Power (Physics, Class 11) has been asked 115 times in the 111 JEE (Main) shift papers from 2020–2025: about 1 questions a paper, 4% of all Physics questions. It is the 15th most-asked of 20 Physics chapters. The 2025 shifts had 12 from it in all. Last checked 28 October 2026.
Questions per year, all shifts
Year by year.
| Year | Questions |
|---|---|
| 2020 | 17in 14 papers |
| 2021 | 28in 26 papers |
| 2022 | 25in 24 papers |
| 2023 | 15in 12 papers |
| 2024 | 18in 18 papers |
| 2025 | 12in 17 papers |
JEE (Main) is sat in many shifts a year, each with its own paper; the counts add up every shift. 2021–2024 papers had 30 questions a subject, the others 25.
Inside the chapter
Topics asked most.
- Work-energy theorem44
- Collisions23
- Conservation of energy22
- Power16
- Motion in a vertical circle10
Try it
10 Work, Energy and Power questions to practise.
Our own questions, written in the JEE Main 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 1 kg ball on a string 2.5 m long just completes a vertical circle. What is the tension in the string at the lowest point? (Take g = 10 m/s².)
Answer: (A)
At the top, T = 0 so v_top² = gr = 25. At the bottom, v² = v_top² + 4gr = 125, and T = mg + mv²/r = 10 + 125/2.5 = 60 N, which is 6mg.
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 115 real JEE Main questions from Work, Energy and Power.
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About Work, Energy and Power in JEE Main.
How many questions come from Work, Energy and Power in JEE Main?
Work, Energy and Power (Physics, Class 11) has been asked 115 times in the 111 JEE (Main) shift papers from 2020–2025: about 1 questions a paper, 4% of all Physics questions. It is the 15th most-asked of 20 Physics chapters. The 2025 shifts had 12 from it in all.
Which topics of Work, Energy and Power are asked most in JEE Main?
Work-energy theorem (44), Collisions (23), Conservation of energy (22), Power (16), counted over the papers from 2020–2025.
Is Work, Energy and Power important for JEE Main 2027?
It is in the syllabus in force, so it can be asked. By past papers it ranks 15 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 JEE Main previous year questions?
On Nikaalle, free: sign up and practise every past JEE Main 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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