Free practice › Class 11 › Physics › Chapter 5
Work, Energy and Power — NEET & JEE MCQs with answers
Work is done by a force over a displacement ($W = \mathbf{F}\cdot\mathbf{d}$); kinetic energy $K = \tfrac{1}{2}mv^2$ changes by the amount of work done (work-energy theorem). Conservative forces allow the definition of potential energy, and the total mechanical energy $K + V$ is conserved for conservative systems. Power is the rate of doing work ($P = \mathbf{F}\cdot\mathbf{v}$).
Topics in this chapter
- 5.1 Introduction
- 5.2 Notions of Work and Kinetic Energy: The Work-Energy Theorem
- 5.3 Work
- 5.4 Kinetic Energy
- 5.5 Work Done by a Variable Force
- 5.6 The Work-Energy Theorem for a Variable Force
- 5.7 The Concept of Potential Energy
- 5.8 The Conservation of Mechanical Energy
- 5.9 The Potential Energy of a Spring
- 5.10 Power
- 5.11 Collisions
Free practice questions
A force F = 12 N acts on a body at an angle of 120° with its displacement of 5 m. The work done is:
- A. −30 J
- B. +60 J
- C. +30 J
- D. −60 J
Show answer
Answer: A — cos120° = −0.5. W = Fd cosθ = 12 × 5 × (−0.5) = −30 J. The obtuse angle gives negative work.
NCERT Class 11 Physics · Ch 5 · 5.3 Work
Although the scalar product is built from two vectors that each have direction, the resulting dot product:
- A. does not have a direction
- B. has the direction of $\vec{A}+\vec{B}$
- C. has a direction perpendicular to both
- D. has the direction of the larger vector
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Answer: A — The text emphasizes that each vector A and B has a direction, but their scalar product does not have a direction — it is purely a scalar.
NCERT Class 11 Physics · Ch 5 · 5.1 Introduction
The kinetic energy of a 6 kg body moving at 10 m s$^{-1}$ is:
- A. 60 J
- B. 300 J
- C. 150 J
- D. 600 J
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Answer: B — K = ½mv² = ½ × 6 × 10² = ½ × 6 × 100 = 300 J.
NCERT Class 11 Physics · Ch 5 · 5.4 Kinetic Energy
A ball of mass $3\ \text{kg}$ is released from rest from a height of $8\ \text{m}$. Using conversion of gravitational PE to KE, its speed just before hitting the ground is ($g = 10\ \text{m s}^{-2}$)
- A. $8\ \text{m s}^{-1}$
- B. $\approx 12.6\ \text{m s}^{-1}$
- C. $16\ \text{m s}^{-1}$
- D. $160\ \text{m s}^{-1}$
Show answer
Answer: B — $\tfrac{1}{2}mv^2 = mgh \Rightarrow v = \sqrt{2gh} = \sqrt{2 \times 10 \times 8} = \sqrt{160} \approx 12.6\ \text{m s}^{-1}$.
NCERT Class 11 Physics · Ch 5 · 5.7 The Concept of Potential Energy
A force F = 5x − x² (N, x in m) acts on a body from x = 0 to x = 5 m. The work done is:
- A. ≈ 12.5 J
- B. ≈ 20.8 J
- C. ≈ 41.7 J
- D. ≈ 62.5 J
Show answer
Answer: B — W = ∫₀⁵ (5x − x²) dx = [2.5x² − x³/3]₀⁵ = 2.5(25) − 125/3 = 62.5 − 41.667 = 20.83 J ≈ 20.8 J.
NCERT Class 11 Physics · Ch 5 · 5.5 Work Done by a Variable Force
A bob of mass $m$ on a light string of length $L$ is given a horizontal velocity $v_0$ at the lowest point A so that it just completes a vertical circle, the string becoming slack only at the topmost point C. Using NCERT's analysis, the required $v_0$ is:
- A. $\sqrt{gL}$
- B. $\sqrt{2gL}$
- C. $\sqrt{5gL}$
- D. $\sqrt{3gL}$
Show answer
Answer: C — NCERT Example 5.7: combining energy conservation and the condition $mg = mv_C^2/L$ at C gives $v_0 = \sqrt{5gL}$.
NCERT Class 11 Physics · Ch 5 · 5.8 The Conservation of Mechanical Energy (Example 5.7)
A 2 kg body moving at 10 m s$^{-1}$ has 60 J of work done against it by a resistive force. Its final kinetic energy is:
- A. 100 J
- B. 60 J
- C. 160 J
- D. 40 J
Show answer
Answer: D — $K_i = \tfrac{1}{2}(2)(100) = 100$ J. $K_f = K_i + W = 100 + (-60) = 40$ J.
NCERT Class 11 Physics · Ch 5 · 5.2 Work-Energy Theorem
A constant force gives a particle a velocity that varies as v = (4 + 3t) m/s. If the mass is 2 kg, the instantaneous power at t = 2 s is:
- A. 30 W
- B. 24 W
- C. 60 W
- D. 12 W
Show answer
Answer: C — a = dv/dt = 3 m/s². F = ma = 2 × 3 = 6 N. At t = 2 s, v = 4 + 6 = 10 m/s. P = F·v = 6 × 10 = 60 W.
NCERT Class 11 Physics · Ch 5 · 5.10 Power
The work-energy theorem for a variable force (one dimension) is written as $K_f - K_i = \int_{x_i}^{x_f} F\,dx$. If a variable force does $50\,\text{J}$ of work on a particle whose initial kinetic energy is $20\,\text{J}$, what is its final kinetic energy?
- A. $70\,\text{J}$
- B. $30\,\text{J}$
- C. $50\,\text{J}$
- D. $1000\,\text{J}$
Show answer
Answer: A — By the work-energy theorem $K_f - K_i = W$ (Eq. 5.8b). Hence $K_f = K_i + W = 20 + 50 = 70\,\text{J}$.
NCERT Class 11 Physics · Ch 5 · 5.6 Work-Energy Theorem for a Variable Force
A block of mass $4\ \text{kg}$ on a smooth surface attached to a spring ($k = 256\ \text{N m}^{-1}$) is pulled to $0.25\ \text{m}$ and released. Its maximum kinetic energy is
- A. 2.0 J
- B. 4.0 J
- C. 16 J
- D. 8.0 J
Show answer
Answer: D — Maximum KE equals the initial PE: $K_{max} = \tfrac{1}{2}kx_m^2 = \tfrac{1}{2}(256)(0.25)^2 = 128(0.0625) = 8.0\ \text{J}$.
NCERT Class 11 Physics · Ch 5 · 5.9 The Potential Energy of a Spring
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