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NEET Work, Energy and Power MCQs with answers

309 questions on Work, Energy and Power for NEET 2027, 1 of them from previous NEET papers. Below are 15 to try; tap a question to see its answer and explanation.

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15 sample questions

easyNEET 2021

1. Work done by centripetal force on circular motion:

  1. Maximum
  2. Minimum
  3. Zero
  4. Depends on speed
Show answer and explanation

Answer: (C) Zero

Centripetal force is always perpendicular to the displacement, so W = Fd cos 90° = 0. Speed stays constant in uniform circular motion.

Quick tip: Force perpendicular to displacement → W=0. Centripetal, Normal on horizontal → zero work.

NCERT Class 11, Chapter 6, page 118

easy

2. A force F→ = (k₁ x î + k₂ y² ĵ) acts on a particle, where k₁ = 4 N/m and k₂ = 3 N/m². Find the work done by this force in moving the particle from the point (1 m, 1 m) to (2 m, 3 m).

  1. 18 J
  2. 38 J
  3. 32 J
  4. 26 J
Show answer and explanation

Answer: (C) 32 J

Since F_x depends only on x and F_y depends only on y, the force is conservative. The work done is W = ∫ (4x dx) from 1 to 2 + ∫ (3y² dy) from 1 to 3 = [2x²]₁² + [y³]₁³ = 2(4 - 1) + (27 - 1) = 6 + 26 = 32 J.

medium

3. A force F = F₀ sin(πx / L) acts along the x-direction on a particle moving from x = 0 to x = 2L. The total work done by this force during the complete displacement is:

  1. 2 F₀ L / π
  2. 0
  3. F₀ L / π
  4. 4 F₀ L / π
Show answer and explanation

Answer: (B) 0

W = ∫ F₀ sin(πx / L) dx from 0 to 2L = F₀ [ - (L/π) cos(πx / L) ] from 0 to 2L = -(F₀ L / π) [cos(2π) - cos(0)] = -(F₀ L / π) [1 - 1] = 0 J.

hard

4. A bullet of mass m moving horizontally with speed v embeds itself into a stationary block of mass M attached to a horizontal spring of stiffness k on a frictionless table. The maximum compression of the spring is:

  1. v √[ (M + m) / k ]
  2. v √[ m / (k(M + m)) ]
  3. v √[ mM / (k(M + m)) ]
  4. v √[ m² / (k(M + m)) ]
Show answer and explanation

Answer: (D) v √[ m² / (k(M + m)) ]

During the inelastic impact, momentum is conserved: mv = (M + m)V, so the common velocity immediately after impact is V = mv / (M + m). After the collision, mechanical energy is conserved: (1/2)(M + m)V² = (1/2)k x_max². Substituting V gives x_max = V √[(M + m)/k] = [mv / (M + m)] √[(M + m)/k] = v √[ m² / (k(M + m)) ].

medium

5. A conservative force field has a potential energy function given by U(x, y) = 4xy - 5x + 2y (in Joules). What is the net force vector F→ acting on a particle when it is at the position (2 m, -1 m)?

  1. (9 î + 6 ĵ) N
  2. (-î - 6 ĵ) N
  3. (9 î - 10 ĵ) N
  4. (-9 î - 10 ĵ) N
Show answer and explanation

Answer: (C) (9 î - 10 ĵ) N

The force components are F_x = -∂U/∂x = -(4y - 5) and F_y = -∂U/∂y = -(4x + 2). At (2, -1): F_x = -[4(-1) - 5] = -(-9) = 9 N, and F_y = -[4(2) + 2] = -10 N. Hence, F→ = (9 î - 10 ĵ) N.

easy

6. A particle of mass m is suspended from a rigid support by a string of length L. It is projected horizontally with a speed u = √(3gL) from the lowest point. What is the tension in the string at the instant when the string becomes horizontal?

  1. mg
  2. 2mg
  3. 3mg
  4. Zero
Show answer and explanation

Answer: (A) mg

By conservation of mechanical energy between the lowest point and the horizontal position: (1/2)m u² = (1/2)m v² + mgL. Given u² = 3gL, we have v² = u² - 2gL = 3gL - 2gL = gL. When the string is horizontal, the tension provides the necessary centripetal force: T = m v² / L = m(gL) / L = mg.

hard

7. A small object of mass m tied to a light string of length L is moving in a vertical circle. The ratio of the maximum tension to the minimum tension in the string is 4. Assuming the string does not slacken at the top, what is the speed of the object at the highest point of the circle?

  1. √(2gL)
  2. √(5gL)
  3. 2√(gL)
  4. √(3gL)
Show answer and explanation

Answer: (D) √(3gL)

At the bottom, T_max = m u² / L + mg. At the top, T_min = m v² / L - mg. By energy conservation, u² = v² + 4gL. Therefore, T_max - T_min = 6mg. We are given T_max / T_min = 4, which gives T_max = 4 T_min. Thus, 4 T_min - T_min = 6mg ⇒ 3 T_min = 6mg ⇒ T_min = 2mg. Substituting T_min into its formula: m v² / L - mg = 2mg ⇒ m v² / L = 3mg ⇒ v = √(3gL).

medium

8. A particle of mass m moves under the action of a force F→ = -k·v→, where k is a positive constant and v→ is the instantaneous velocity. Which of the following assertions about this force is true?

  1. It is a conservative force because it always opposes motion.
  2. A unique scalar potential energy function U(r→) can be defined for this force.
  3. The work done by this force along any path where the particle moves is always negative.
  4. The mechanical energy of the particle is conserved during its motion.
Show answer and explanation

Answer: (C) The work done by this force along any path where the particle moves is always negative.

The force F→ = -k·v→ is a velocity-dependent viscous drag force, which is inherently non-conservative, so no scalar potential energy function exists. The work done is dW = F→ · dr→ = (-k·v→) · (v→ dt) = -k v² dt, which is always negative for any non-zero motion, continuously dissipating mechanical energy.

easy

9. Which of the following physical quantities is directly related to the negative of the work done by a conservative internal force during an arbitrary displacement?

  1. Change in kinetic energy
  2. Change in potential energy
  3. Change in total mechanical energy
  4. Rate of heat dissipation
Show answer and explanation

Answer: (B) Change in potential energy

By fundamental definition in mechanics, the change in potential energy ΔU of a system is equal to the negative of the work done by the conservative internal force, i.e., ΔU = -W_c. Hence, work done by conservative forces directly quantifies the change in potential energy.

easy

10. A small block slides down a curved frictionless track from height h and then travels along a rough horizontal surface with coefficient of kinetic friction μ before coming to rest. If W_g is the work done by gravity and W_f is the work done by friction over the entire journey, which relation is correct?

  1. W_g + W_f > 0
  2. W_g + W_f = 0
  3. W_g = W_f
  4. W_g - W_f = 0
Show answer and explanation

Answer: (B) W_g + W_f = 0

By the work-energy theorem, the total work done by all forces equals the change in kinetic energy: W_total = ΔK. Since the block starts from rest and ends at rest, ΔK = 0. Therefore, W_total = W_g + W_N + W_f = 0. Normal force does zero work throughout, so W_g + W_f = 0 (where W_g = mgh and W_f = -mgh).

medium

11. A block of mass m rests on a smooth horizontal surface and is connected to two identical horizontal springs of force constant k each, arranged in series on one side and anchored to a wall. The block is pulled by a distance x from equilibrium and released from rest. Its speed when passing through the equilibrium position is:

  1. x √(k / 2m)
  2. x √(k / m)
  3. x √(2k / m)
  4. 2x √(k / m)
Show answer and explanation

Answer: (A) x √(k / 2m)

For two identical springs connected in series, the equivalent spring constant is k_eq = (k × k)/(k + k) = k / 2. By conservation of mechanical energy, (1/2)k_eq x² = (1/2)mv² ⇒ (1/2)(k/2)x² = (1/2)mv². Solving for v gives v = x √(k / 2m).

easy

12. A spring of force constant 1000 N/m is compressed by 10 cm. The elastic potential energy stored in it is:

  1. 5 J
  2. 50000 J
  3. 50 J
  4. 10 J
Show answer and explanation

Answer: (A) 5 J

U = ½kx² with x = 0.1 m: U = ½ × 1000 × (0.1)² = 5 J.

medium

13. Two springs S₁ and S₂ have spring constants k₁ and k₂ respectively, where k₁ > k₂. If both springs store equal elastic potential energy, which of the following statements about the stretching forces F₁ and F₂ applied to them is correct?

  1. F₁ = F₂
  2. F₁ < F₂
  3. F₁ > F₂
  4. F₁ and F₂ cannot be compared without knowing the extensions
Show answer and explanation

Answer: (C) F₁ > F₂

The elastic potential energy stored in a spring can be written as U = F²/(2k). Since U₁ = U₂, we have F₁²/(2k₁) = F₂²/(2k₂), which implies F₁/F₂ = √(k₁/k₂). Because k₁ > k₂, it follows that F₁ > F₂.

easy

14. A light spring of force constant k is cut into two equal halves. One half is connected to a block of mass m on a frictionless horizontal plane. If the block is pulled to extend this spring half by x, the elastic potential energy stored in it is:

  1. (1/4) kx²
  2. (1/2) kx²
  3. kx²
  4. 2 kx²
Show answer and explanation

Answer: (C) kx²

When a spring of constant k is cut into two equal halves, the spring constant of each half becomes k' = 2k. The potential energy stored when one half is extended by x is U = (1/2)k'x² = (1/2)(2k)x² = kx².

medium

15. A 3 kg body moving at 8 m/s collides with a stationary 3 kg body and they stick together. Their common velocity after collision is:

  1. 4 m/s
  2. 2 m/s
  3. 8 m/s
  4. 6 m/s
Show answer and explanation

Answer: (A) 4 m/s

Momentum is conserved in a perfectly inelastic collision: v = m₁v₁/(m₁ + m₂) = 3 × 8/6 = 4 m/s.

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