NEET Work, Energy and Power PYQs & Practice Questions

10 free NEET Physics questions on Work, Energy and Power, starting with previous year questions (2016–2026). Tap “Show answer” to check each one. No login needed.

Topics: Work-Energy Theorem · Work Done by a Constant Force · Potential Energy and Conservative Forces · Conservation of Mechanical Energy · Power · Elastic and Inelastic Collisions · Spring Potential Energy · Coefficient of Restitution and Inelastic Collisions · Work Done by Variable Force · Work Done by a Force · Conservation of Energy · Collisions

Q1NEET 2026Work Done by Position-Dependent Friction
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 has magnitude nμ₀MgL, where g is the acceleration due to gravity. The value of n is:
- A. 1
- B. 1/3
- C. 1/2
- D. 3
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Answer: B — 1/3
Since μ_k(L)=0, α = μ₀/L. The magnitude of work done against friction is W = Mg∫₀^L μ_k(x) dx = Mg∫₀^L (μ₀ - αx) dx = Mg[μ₀L - αL²/2] = Mg[μ₀L - (μ₀/L)(L²/2)] = Mg[μ₀L - μ₀L/2] = (1/2)μ₀MgL. So n = 1/2.
Q2NEET 2026Elastic Collision and Pendulum Rise
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 metre after the collision. Taking g = 10 ms⁻² and neglecting the size of the bob, the value of h is:
- A. 7
- B. 5
- C. 2.5
- D. 8
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Answer: A — 7
In an elastic collision between two equal masses where one is initially at rest, the velocities are completely exchanged: A comes to rest and B moves off with the full initial speed, 10 ms⁻¹. B then rises as a pendulum, converting kinetic energy to potential energy: (1/2)mv² = mgh, so h = v²/(2g) = 100/20 = 5 m.
Q3NEET 2025Work-Energy Theorem
The kinetic energies of two similar cars A and B are 100 J and 225 J respectively. On applying breaks, car A stops after 1000 m and car B stops after 1500 m. If FA and FB are the forces applied by the breaks on cars A and B, respectively, then the ratio FA/FB is
- A. 3/2
- B. 2/3
- C. 1/3
- D. 1/2
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Answer: A — 3/2
ΔKE = F·S. So KEA/KEB = (FA·SA)/(FB·SB) ⟹ 100/225 = (FA/FB)×(1000/1500) ⟹ FA/FB = (100/225)×(1500/1000) = 150/225 = 2/3.
Q4NEET 2024Instantaneous Power from Displacement-Time Relation
At any instant of time t, the displacement of any particle is given by 2t - 1 (SI unit) under the influence of force of 5 N. The value of instantaneous power is (in SI unit):
- A. 7
- B. 6
- C. 10
- D. 5
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Answer: B — 6
x = 2t - 1, so v = dx/dt = 2 m/s (constant). Power P = F·v = 5 × 2 = 10 Watt.

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Q5NEET 2024Perfectly Inelastic Collision - Velocity Ratio
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:
- A. 4 : 1
- B. 1 : 4
- C. 1 : 2
- D. 2 : 1
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Answer: C — 1 : 2
By conservation of momentum: mv1 + 0 = (m+m)v2, so mv1 = 2mv2, giving v1/v2 = 2/1, i.e., v1:v2 = 2:1.
Q6NEET 2023Spring Potential Energy Scaling
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:
- A. 4 U
- B. 8 U
- C. 16 U
- D. 2 U
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Answer: B — 8 U
U = (1/2)Kx². Since U ∝ x², and the extension increased from 2cm to 8cm (4 times), the new PE = (4)² × U = 16U.
Q7NEET 2022Power Delivered by Motor
An electric lift with a maximum load of 2000 kg (lift + passengers) is moving up with a constant speed of 1.5 m/s. 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)
- A. 23000
- B. 20000
- C. 34500
- D. 23500
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Answer: B — 20000
Weight = 2000x10 = 20000 N. Total force needed = weight + friction = 20000 + 3000 = 23000 N. Power = Force x velocity = 23000 x 1.5 = 34500 W.
Q8NEET 2021Energy Conservation in Free Fall
A particle is released from height S from the surface of the Earth. At a certain height its kinetic energy is three times its potential energy. The height from the surface of earth and the speed of the particle at that instant are respectively:
- A. S/4, √(3gS/2)
- B. S/4, 3gS/2
- C. S/4, √(3gS)/2
- D. S/2, √(3gS)/2
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Answer: —
PE + KE = mgS at the given point, and KE = 3PE, so 4PE = mgS, giving height from ground H = S/4. From KE = 3mgS/4 = (1/2)mV², V = √(3gS)/2 = √(3gS/2). Per the official answer key, option (1).

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Q9NEET 2019Elastic Collision
Body A of mass 4m moving with speed u collides with another body B of mass 2m, at rest. The collision is head on and elastic in nature. After the collision the fraction of energy lost by the colliding body A is:
- A. 4/9
- B. 5/9
- C. 1/9
- D. 8/9
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Answer: C — 1/9
Using conservation of momentum and elastic collision equations (e=1): v1 = u/3. Initial KE of A = 2mu^2. Final KE of A = (1/2)(4m)(u/3)^2 = 2mu^2/9. Fraction of energy lost = (2mu^2 - 2mu^2/9)/2mu^2 = 8/9.
Q10NEET 2019Work Done by Variable Force
A force F = 20 + 10y acts on a particle in y-direction where F is in newton and y in meter. Work done by this force to move the particle from y=0 to y=1 m is
- A. 25 J
- B. 20 J
- C. 30 J
- D. 5 J
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Answer: —
W = integral of F dy from 0 to 1 = integral of (20+10y) dy = [20y + 5y^2] from 0 to 1 = 20(1) + 5(1) = 25 J.

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