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Kinetic and potential energy; work-energy theorem and power: NEET MCQs with answers

60 questions on this topic of Work, Energy and Power (NEET 2027 syllabus). Here are 12; tap one to see its answer and explanation.

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easy

1. A motor lifts a 10 kg load through a height of 5 m in 10 s at constant speed. Taking g = 10 m/s², the power delivered is:

  1. 50 W
  2. 500 W
  3. 5 W
  4. 5000 W
Show answer and explanation

Answer: (A) 50 W

P = work/time = mgh/t = 10 × 10 × 5/10 = 50 W.

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2. A body of mass 2 kg initially at rest is acted upon by a horizontal force F = (3x² + 2) N, where x is displacement in metres. What is the kinetic energy of the body when it reaches x = 2 m?

  1. 8 J
  2. 10 J
  3. 12 J
  4. 16 J
Show answer and explanation

Answer: (C) 12 J

By the work-energy theorem, ΔK = W = ∫ F dx from 0 to 2. Here, W = ∫₀² (3x² + 2) dx = [x³ + 2x]₀² = (8 + 4) = 12 J. Since the body starts from rest, final K = 12 J.

hard

3. The potential energy function for the force between two atoms in a diatomic molecule is given by U(x) = A/x¹² - B/x⁶, where A and B are positive constants and x is the interatomic distance. The dissociation energy of the molecule (energy required to separate the atoms from equilibrium to infinity) is:

  1. B² / (2A)
  2. B² / (4A)
  3. A² / (4B)
  4. B² / A
Show answer and explanation

Answer: (B) B² / (4A)

At equilibrium, dU/dx = 0 implies -12A/x¹³ + 6B/x⁷ = 0, so x⁶ = 2A/B. Substituting this into U(x) gives the equilibrium potential energy U_min = A/(2A/B)² - B/(2A/B) = B²/(4A) - B²/(2A) = -B²/(4A). The energy required to dissociate the molecule to infinity (where U = 0) is ΔU = 0 - U_min = B²/(4A).

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4. An object of mass 2 kg moves at 36 km/h. Its kinetic energy is:

  1. 20 J
  2. 200 J
  3. 100 J
  4. 50 J
Show answer and explanation

Answer: (C) 100 J

Unit check: Convert to SI: m = 2 kg, v = 10 m/s. KE = ½mv² = 100 J.

easy

5. A motor lifts a 90 kg load through a height of 30 m in 20 s at constant speed. Taking g = 10 m/s², the power delivered is:

  1. 135 W
  2. 1350 W
  3. 27000 W
  4. 5.4 × 10⁵ W
Show answer and explanation

Answer: (B) 1350 W

P = work/time = mgh/t = 90 × 10 × 30/20 = 1350 W.

hard

6. A vehicle of mass M is accelerated on a horizontal road by an engine delivering a constant power P. Neglecting friction, the distance covered by the vehicle in terms of its velocity v starting from rest is proportional to:

  1. v
  2. v²
  3. v⁴
  4. v³
Show answer and explanation

Answer: (D) v³

Power P = F · v = (M v dv/dx) v = M v² (dv/dx). Rearranging gives dx = (M/P) v² dv. Integrating both sides from 0 to v yields x = (M / 3P) v³, so x is directly proportional to v³.

easy

7. A motor lifts a 40 kg load through a height of 25 m in 25 s at constant speed. Taking g = 10 m/s², the power delivered is:

  1. 10000 W
  2. 40 W
  3. 400 W
  4. 2.5 × 10⁵ W
Show answer and explanation

Answer: (C) 400 W

P = work/time = mgh/t = 40 × 10 × 25/25 = 400 W.

easy

8. A motor lifts a 10 kg load through a height of 25 m in 5 s at constant speed. Taking g = 10 m/s², the power delivered is:

  1. 12500 W
  2. 500 W
  3. 50 W
  4. 2500 W
Show answer and explanation

Answer: (B) 500 W

P = work/time = mgh/t = 10 × 10 × 25/5 = 500 W.

easy

9. A motor lifts a 30 kg load through a height of 10 m in 5 s at constant speed. Taking g = 10 m/s², the power delivered is:

  1. 15000 W
  2. 600 W
  3. 3000 W
  4. 60 W
Show answer and explanation

Answer: (B) 600 W

P = work/time = mgh/t = 30 × 10 × 10/5 = 600 W.

easy

10. A motor lifts a 40 kg load through a height of 10 m in 5 s at constant speed. Taking g = 10 m/s², the power delivered is:

  1. 20000 W
  2. 800 W
  3. 4000 W
  4. 80 W
Show answer and explanation

Answer: (B) 800 W

P = work/time = mgh/t = 40 × 10 × 10/5 = 800 W.

medium

11. An engine pumps water continuously through a hose with a cross-sectional area A. If the water leaves the hose with a speed v, what is the rate at which kinetic energy is imparted to the water (density of water = ρ)?

  1. (1/2) ρ A v²
  2. ρ A v³
  3. (1/2) ρ A v³
  4. (1/2) ρ A² v²
Show answer and explanation

Answer: (C) (1/2) ρ A v³

Mass of water exiting per second is dm/dt = ρ A v. The rate of kinetic energy imparted is Power = dK/dt = (1/2)(dm/dt)v² = (1/2)(ρ A v)v² = (1/2) ρ A v³.

easy

12. A motor lifts a 70 kg load through a height of 20 m in 5 s at constant speed. Taking g = 10 m/s², the power delivered is:

  1. 2800 W
  2. 280 W
  3. 70000 W
  4. 14000 W
Show answer and explanation

Answer: (A) 2800 W

P = work/time = mgh/t = 70 × 10 × 20/5 = 2800 W.

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