Work, Energy and Power NEET PYQs (1996–2026)

Solved Work, Energy and Power PYQs

3 recent questions with answers. The full set of 83 is free to practise.

1. NEET 2026EasyPower
   Q1

   The power of a crane, which lifts a mass of $1000kg$ to a height of $20m$ in $10s$ is: $(g=9.8m/s^2)$

   • A19.6 W
   • B39.2 W
   • C19.6 kW
   • D39.2 kW

   Show answer & solution

   Correct answer: (C)

   Power $=\frac{mgh}{t}=\frac{1000\times 9.8\times 20}{10}=19600$ W
   
   = 19.6 kW.
2. NEET 2025MediumKinetic Energy & Work-Energy Theorem
   Q2

   The kinetic energies of two similar cars $A$ and $B$ are $100\mathbf{J}$ and $225\mathbf{J}$ respectively. On applying breaks, car $A$ stops after 1000 m and car $B$ stops after 1500 m . If $F_A$ and $F_B$ are the forces applied by the breaks on cars $A$ and $B$ respectively, then the ratio of $\frac{F_A}{F_B}$ is

   • A1/3
   • B1/2
   • C3/2
   • D2/3

   Show answer & solution

   Correct answer: (D)

   According to the work-energy theorem, the work done by the braking force is equal to the change in kinetic energy. Thus, for each car, we have:
   
   $$FS=△KE$$
   
   This can be rearranged to:
   
   $$-FS=k_f-k_i\Rightarrow FS=k_i-k_f$$
   
   For both cars, since the final kinetic energy $k_f$ is zero when they stop, the equation simplifies to:
   
   $$FS=k_i$$
   
   For car $A$ and car $B$, the forces $F_A$ and $F_B$ applied by the brakes satisfy:
   
   $$F_A\cdot S_A=k_A\text{ and }{F}_B\cdot S_B=k_B$$
   
   Solving for the ratio of the forces:
   
   $$\frac{F_A}{F_B}=\frac{k_A}{k_B}\times \frac{S_B}{S_A}$$
   
   Substituting the given kinetic energies and stopping distances:
   
   $$\frac{F_A}{F_B}=\frac{100}{225}\times \frac{1500}{1000}$$
   
   This simplifies to:
   
   $$=\frac{150}{225}=\frac{2}{3}$$
   
   Thus, the ratio of the forces $\frac{F_A}{F_B}$ is $\frac{2}{3}$.
3. NEET 2023EasyPotential Energy & Conservation of Energy
   Q3

   The potential energy of a long spring when stretched by 2 cm is U . If the spring is stretched by 8 cm , potential energy stored in it will be
   
   [NEET 2023]

   • A${}^{4}U$
   • B${}^{8U}$
   • C${}^{16U}$
   • D${}^{2U}$

   Show answer & solution

   Correct answer: (C)

   Potential energy stored in spring $U=\frac{1}{2}K{x}^2$
   $U=\frac{1}{2}K(2{)}^2$ where $x=2cm$
   $$U=\frac{1}{2}(K)\cdot (4)U=2K{U}^{\prime }=\frac{1}{2}K(8{)}^2{U}^{\prime }=\frac{1}{2}K\times 64=32K$$On dividing (i) by (ii)
   $$\frac{U}{U}=\frac{2K}{32K}=\frac{1}{16}{U}^{\prime }=16U$$

Topic-wise question count

Where NEET has asked from Work, Energy and Power, most frequent first.

• Collisions24
• Power17
• Potential Energy & Conservation of Energy14
• Work Done13
• Kinetic Energy & Work-Energy Theorem11
• Circular Motion4

Year-wise Work, Energy and Power questions in NEET

Number of questions from this chapter in each year's paper(s).