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Work by constant and variable forces: NEET MCQs with answers

51 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 force F = (3î + 4ĵ - 2k̂) N acts on a particle and displaces it from point A(1, 2, 3) m to point B(2, 4, 1) m. The work done by the force on the particle is:

  1. 7 J
  2. 15 J
  3. 11 J
  4. 19 J
Show answer and explanation

Answer: (B) 15 J

The displacement vector is Δr = r_B - r_A = (2 - 1)î + (4 - 2)ĵ + (1 - 3)k̂ = (1î + 2ĵ - 2k̂) m. The work done is W = F · Δr = (3)(1) + (4)(2) + (-2)(-2) = 3 + 8 + 4 = 15 J.

medium

2. Under the action of a force, a 2 kg body moves such that its position x as a function of time t is given by x = t³ / 3, where x is in metres and t is in seconds. The work done by the force in the first 2 seconds is:

  1. 8 J
  2. 16 J
  3. 32 J
  4. 4 J
Show answer and explanation

Answer: (B) 16 J

Velocity is v = dx/dt = t². At t = 0, v₁ = 0 m/s. At t = 2 s, v₂ = 2² = 4 m/s. By the work-energy theorem, work done equals change in kinetic energy: W = (1/2) m (v₂² - v₁²) = (1/2)(2)(4² - 0) = 16 J.

hard

3. A small box of mass m is placed on a horizontal turntable at a distance r from the axis of rotation. The turntable starts from rest and rotates with a constant angular acceleration α. If the box does not slip relative to the turntable, the work done by the total frictional force on the box in time t is:

  1. m α² r² t²
  2. (1/2) m α² r² t²
  3. (1/2) m α r² t
  4. 0
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Answer: (B) (1/2) m α² r² t²

The speed of the box at time t is v = ω r = (α t) r = α r t. Since the turntable starts from rest, initial kinetic energy is zero. By the work-energy theorem, the net work done on the box (which is solely by the friction force, as gravity and normal force are vertical) equals ΔK = (1/2) m v² - 0 = (1/2) m α² r² t².

medium

4. A constant force F acts on a particle of mass m initially at rest. The work done by the force during the first t seconds is directly proportional to:

  1. t²
  2. t
  3. t³
  4. t^(1/2)
Show answer and explanation

Answer: (A) t²

Since the force is constant, acceleration a = F/m is constant. The displacement starting from rest is d = (1/2) a t². The work done is W = F · d = F(1/2 a t²) = (F² / 2m) t², which is directly proportional to t².

easy

5. A force F acting on a particle varies with displacement x according to the relation F = (8 - 4x) N, where x is in metres. The particle moves from x = 0 to x = 3 m. The net work done by the force is:

  1. 12 J
  2. 6 J
  3. -6 J
  4. 24 J
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Answer: (B) 6 J

Work done is given by W = ∫ F dx from 0 to 3. W = ∫(8 - 4x) dx = [8x - 2x²] from 0 to 3 = (8 × 3 - 2 × 3²) - 0 = 24 - 18 = 6 J.

hard

6. A body of mass 2 kg rests on a rough horizontal surface with coefficient of kinetic friction μ = 0.2. A horizontal force F varies with distance x as F(x) = (10 + 2x) N. Taking g = 10 m/s², the net work done on the body as it moves from x = 0 to x = 5 m is:

  1. 35 J
  2. 45 J
  3. 55 J
  4. 75 J
Show answer and explanation

Answer: (C) 55 J

The frictional force is constant: f = μmg = 0.2 × 2 × 10 = 4 N. The net force acting on the body in the direction of motion is F_net(x) = F(x) - f = (10 + 2x) - 4 = (6 + 2x) N. The net work done is W_net = ∫₀⁵ (6 + 2x) dx = [6x + x²]₀⁵ = 6(5) + 5² = 30 + 25 = 55 J.

easy

7. A gardener pushes a lawn roller of mass 50 kg with a force of 100 N inclined at an angle of 60° to the horizontal ground. The work done by the pushing force in moving the roller through a horizontal distance of 15 m is:

  1. 1500 J
  2. 1500√3 J
  3. 750√3 J
  4. 750 J
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Answer: (D) 750 J

The work done by a constant force inclined at angle θ to displacement is W = F d cos θ. Here, W = 100 × 15 × cos(60°) = 1500 × 0.5 = 750 J.

easy

8. A uniform chain of mass M and length L is held vertically such that its lower end just touches a horizontal table. It is released from rest. The work done by the downward gravitational force during the complete fall of the chain onto the table is:

  1. MgL
  2. 2MgL
  3. MgL/4
  4. MgL/2
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Answer: (D) MgL/2

The center of mass of the uniform chain initially lies at a height L/2 above the table. When the chain completely rests on the table, its center of mass is at height 0 (assuming negligible thickness). Thus, the vertical displacement of the center of mass is h = L/2. The work done by gravity is W = Mg(Δh) = Mg(L/2) = MgL/2.

medium

9. A particle of mass m moves along the x-axis under the action of a position-dependent force F(x) = (6x² - 4x + 2) N. The total work done by this force in moving the particle from x = 1 m to x = 3 m is:

  1. 52 J
  2. 38 J
  3. 40 J
  4. 36 J
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Answer: (C) 40 J

Work done by a variable force is W = ∫ F(x) dx from x = 1 to 3. W = [2x³ - 2x² + 2x] from 1 to 3. At x = 3: 2(27) - 2(9) + 2(3) = 54 - 18 + 6 = 42 J. At x = 1: 2(1) - 2(1) + 2(1) = 2 J. Thus, W = 42 - 2 = 40 J.

easy

10. A force acting on a particle moving along the x-axis varies as F(x) = (3x² - 2x) N, where x is in metres. The work done by this force in moving the particle from x = 1 m to x = 3 m is:

  1. 18 J
  2. 26 J
  3. 20 J
  4. 16 J
Show answer and explanation

Answer: (A) 18 J

Work done is given by W = ∫ F(x) dx from x = 1 to 3. Integrating: W = [x³ - x²] from 1 to 3 = (3³ - 3²) - (1³ - 1²) = (27 - 9) - (1 - 1) = 18 J.

easy

11. A force F = (5 + 3x²) N acts on a particle in the x-direction, where x is in metres. The work done by this force during a displacement from x = 1 m to x = 3 m is:

  1. 42 J
  2. 26 J
  3. 36 J
  4. 16 J
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Answer: (C) 36 J

The work done by a variable force along the x-axis is W = ∫ F dx from x = 1 to 3. W = [5x + x³] from 1 to 3 = (5(3) + 27) - (5(1) + 1) = 42 - 6 = 36 J.

medium

12. A particle is displaced from position vector r₁ = (2î + 3ĵ) m to r₂ = (4î + 7ĵ) m under the simultaneous action of two constant forces F₁ = (3î - 2ĵ) N and F₂ = (î + 5ĵ) N. The total work done by the two forces is:

  1. 8 J
  2. 14 J
  3. 26 J
  4. 20 J
Show answer and explanation

Answer: (D) 20 J

The resultant constant force is F_net = F₁ + F₂ = (3 + 1)î + (-2 + 5)ĵ = (4î + 3ĵ) N. The displacement vector is Δr = r₂ - r₁ = (4 - 2)î + (7 - 3)ĵ = (2î + 4ĵ) m. The total work done is W = F_net · Δr = (4)(2) + (3)(4) = 8 + 12 = 20 J.

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