NEETAceToday's NEET Daily

Spring potential energy and conservation of mechanical energy: NEET MCQs with answers

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

Practise Work, Energy and Power freeAll Work, Energy and Power questions

Free with a Google sign-in; NEET Daily needs no sign-in at all.

easy

1. Given below are two statements: Statement I: The work done by an ideal spring force on a connected body depends only on the initial and final positions of the body and is independent of the path taken. Statement II: Spring force is a non-conservative force because it changes direction whenever the spring crosses its mean position. In light of the above statements, choose the correct answer:

  1. Both Statement I and Statement II are correct
  2. Statement I is correct but Statement II is incorrect
  3. Statement I is incorrect but Statement II is correct
  4. Both Statement I and Statement II are incorrect
Show answer and explanation

Answer: (B) Statement I is correct but Statement II is incorrect

Statement I is correct because the spring force is a conservative force, meaning work done depends solely on initial and final extensions/compressions. Statement II is incorrect because an ideal spring force is strictly conservative; reversing direction across the equilibrium position is characteristic of a restoring force, not an indicator of non-conservativeness.

medium

2. A uniform spring of spring constant k and natural length L is cut into two pieces whose lengths are in the ratio 1 : 2. What is the ratio of the spring constant of the shorter piece to that of the longer piece?

  1. 3 : 2
  2. 1 : 2
  3. 3 : 1
  4. 2 : 1
Show answer and explanation

Answer: (D) 2 : 1

The spring constant of a spring is inversely proportional to its length (k ∝ 1/L). If the lengths are L₁ = L/3 and L₂ = 2L/3, their spring constants are k₁ = 3k and k₂ = (3/2)k. Thus, the ratio k₁ : k₂ is 3 : (3/2) = 2 : 1.

hard

3. A particle of mass m moves along the x-axis under the influence of a conservative force whose potential energy is given by U(x) = (1/2) k x² - (1/4) b x⁴, where k and b are positive constants. If the particle is released from rest at x = 0, which of the following statements about its equilibrium at x = 0 is correct?

  1. It is in stable equilibrium with small-amplitude oscillation frequency ω = √(k / m)
  2. It is in unstable equilibrium because U(0) = 0
  3. It is in neutral equilibrium because dU/dx = 0 at x = 0
  4. It cannot be in equilibrium since the force depends on x⁴
Show answer and explanation

Answer: (A) It is in stable equilibrium with small-amplitude oscillation frequency ω = √(k / m)

At x = 0, dU/dx = kx - bx³ = 0, so x = 0 is an equilibrium point. The second derivative is d²U/dx² = k - 3bx², which evaluates to +k > 0 at x = 0. Since the second derivative is positive, x = 0 is a point of stable equilibrium, and for small oscillations the restoring force is F ≈ -kx, giving an angular frequency ω = √(k/m).

medium

4. Given below are two statements: Statement I: The potential energy of a spring is zero only when it is at its natural unstretched length (taking unstretched length as reference). Statement II: Spring force is a conservative force because the work done by it over any closed round trip is zero. In the light of the above statements, choose the most appropriate answer from the options given below:

  1. Both Statement I and Statement II are correct
  2. Statement I is correct but Statement II is incorrect
  3. Statement I is incorrect but Statement II is correct
  4. Both Statement I and Statement II are incorrect
Show answer and explanation

Answer: (A) Both Statement I and Statement II are correct

Taking the relaxed state as zero potential energy, U(x) = (1/2)kx² is strictly positive for any non-zero extension or compression (x ≠ 0) and zero only at x = 0. Statement II correctly defines the conservative nature of the ideal spring force, as work done over any closed path is identically zero.

easy

5. A spring of force constant 400 N/m is compressed by 10 cm. The elastic potential energy stored in it is:

  1. 4 J
  2. 20 J
  3. 20000 J
  4. 2 J
Show answer and explanation

Answer: (D) 2 J

U = ½kx² with x = 0.1 m: U = ½ × 400 × (0.1)² = 2 J.

hard

6. A block of mass m is placed on a smooth horizontal floor and connected to a wall by an unstretched spring of force constant k. A constant horizontal pulling force F is suddenly applied to the block directed away from the wall. What is the maximum speed attained by the block?

  1. F / √(mk)
  2. F / √(2mk)
  3. 2F / √(mk)
  4. F / (2√(mk))
Show answer and explanation

Answer: (A) F / √(mk)

The net force on the block at displacement x is F_net = F - kx. Acceleration is zero (and speed is maximum) when F - kx₀ = 0, i.e., x₀ = F/k. Applying work-energy theorem from x = 0 to x = x₀: W_total = ΔK ⇒ F·x₀ - (1/2)kx₀² = (1/2)mv_max². Substituting x₀ = F/k gives F²/k - F²/(2k) = F²/(2k) = (1/2)mv_max². Therefore, v_max = F/√(mk).

medium

7. A block of mass m is attached to one end of a horizontal spring of force constant k and rests on a smooth horizontal table. The other end of the spring is fixed. The block is pulled to a displacement x₀ from the equilibrium position and released from rest. What is the speed of the block when its displacement is x₀/2?

  1. (x₀/2) √(k/m)
  2. (x₀√3/2) √(k/m)
  3. (x₀/√2) √(k/m)
  4. (x₀/4) √(k/m)
Show answer and explanation

Answer: (B) (x₀√3/2) √(k/m)

Total mechanical energy at release is E = (1/2)kx₀². At displacement x = x₀/2, potential energy is U = (1/2)k(x₀/2)² = (1/8)kx₀². By conservation of mechanical energy, K = E - U = (1/2)kx₀² - (1/8)kx₀² = (3/8)kx₀². Since K = (1/2)mv², we have (1/2)mv² = (3/8)kx₀², leading to v = (x₀√3/2)√(k/m).

medium

8. A block of mass m is dropped from a height h above the free end of a light, uncompressed vertical spring of stiffness k fixed to the ground. If the maximum compression produced in the spring is x, the equation describing x is:

  1. (1/2)kx² + mgx - mgh = 0
  2. (1/2)kx² - mgx - mgh = 0
  3. (1/2)kx² - mgh = 0
  4. kx² - 2mgx + 2mgh = 0
Show answer and explanation

Answer: (B) (1/2)kx² - mgx - mgh = 0

From the release point to the point of maximum compression, the total vertical descent of the block is h + x. By conservation of mechanical energy, the loss in gravitational potential energy equals the gain in spring potential energy: mg(h + x) = (1/2)kx². Rearranging gives (1/2)kx² - mgx - mgh = 0.

easy

9. A spring of force constant 900 N/m is compressed by 2 cm. The elastic potential energy stored in it is:

  1. 0.18 J
  2. 9 J
  3. 1800 J
  4. 0.36 J
Show answer and explanation

Answer: (A) 0.18 J

U = ½kx² with x = 0.02 m: U = ½ × 900 × (0.02)² = 0.18 J.

easy

10. A spring of force constant 200 N/m is compressed by 9 cm. The elastic potential energy stored in it is:

  1. 1.62 J
  2. 0.81 J
  3. 9 J
  4. 8100 J
Show answer and explanation

Answer: (B) 0.81 J

U = ½kx² with x = 0.09 m: U = ½ × 200 × (0.09)² = 0.81 J.

easy

11. A spring of force constant 700 N/m is compressed by 8 cm. The elastic potential energy stored in it is:

  1. 2.24 J
  2. 22400 J
  3. 4.48 J
  4. 28 J
Show answer and explanation

Answer: (A) 2.24 J

U = ½kx² with x = 0.08 m: U = ½ × 700 × (0.08)² = 2.24 J.

easy

12. A mass m connected to a light horizontal spring of stiffness k performs oscillations with amplitude A on a smooth floor. At what displacement from the mean position is the kinetic energy of the mass equal to three times the elastic potential energy stored in the spring?

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

Answer: (D) A/2

Total mechanical energy is E = (1/2)kA². Given kinetic energy K = 3U, the total energy is E = K + U = 4U. Thus, (1/2)kA² = 4 × [(1/2)kx²], which simplifies to x² = A²/4 or x = A/2.

Keep going with Work, Energy and Power. All 309 questions with instant answers, your mistakes saved for revision and your accuracy by topic.
Start practising

Other topics in Work, Energy and Power