Work, Energy and Power NEET PYQs (2005–2026) with Solutions

10 solved Work, Energy and Power PYQs

Real NEET and AIPMT questions from this chapter, newest first. Try each one, then open the answer and the step-by-step explanation.

1. Question 1 (NEET 2026)

   NEET 2026MediumWork done by a variable force

   A particle of mass $M$ moves along a horizontal $x$ axis from $x = 0$ to $x = L$. The coefficient of kinetic friction varies as a function of $x$ as $\mu_k(x) = \mu_0 - \alpha x$, where $\mu_0$, $\alpha$ are constants of appropriate dimensions, so that $\mu_k(L) = 0$. The total work done by the frictional force during the motion is $n\mu_0 MgL$, where $g$ is the acceleration due to gravity. The value of $n$ is:

   1. AOption A: $3$
   2. BOption B: $1$
   3. COption C: $\frac{1}{3}$
   4. DOption D: $\frac{1}{2}$

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   Correct answer: (D) $\frac{1}{2}$

   Explanation

   Since $\mu_k(L) = 0$, $\mu_0 - \alpha L = 0$, so $\alpha = \frac{\mu_0}{L}$ and $\mu_k(x) = \mu_0\left(1 - \frac{x}{L}\right)$. On a horizontal surface the friction force is $f = \mu_k(x) Mg$, opposing the motion. Work done by friction: $$W = -\int_0^L \mu_0 Mg\left(1 - \frac{x}{L}\right)dx = -\mu_0 Mg\left(L - \frac{L}{2}\right) = -\frac{1}{2}\mu_0 MgL$$ The magnitude of the work is $\frac{1}{2}\mu_0 MgL$(the negative sign only shows that friction removes energy), so$n = \frac{1}{2}$, option (D).

2. Question 2 (NEET 2026)

   NEET 2026EasySpring Potential Energy & Power

   The power of a crane, which lifts a mass of 1000 kg to a height of 20 m in 10 s is: (g = 9.8 m/s²)

   1. AOption A: 19.6 W
   2. BOption B: 39.2 W
   3. COption C: 19.6 kW
   4. DOption D: 39.2 kW

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   Correct answer: (C) 19.6 kW

   Explanation

   Power = Work/Time = mgh/t = (1000×9.8×20)/10 = 19600 W = 19.6 kW. Hence option (C) is correct.

3. Question 3 (NEET 2026)

   NEET 2026EasyElastic collision and energy conservation

   Bob $B$ of mass $m$ at rest is hanging vertically from the ceiling via a massless string of length $10\ \mathrm{m}$, as shown in the figure. Point mass $A$ of mass $m$ travelling horizontally with speed $10\ \mathrm{ms^{-1}}$ hits bob $B$ elastically. The bob $B$ rises $h$ meter after the collision. Taking the acceleration due to gravity $g = 10\ \mathrm{ms^{-2}}$ and neglecting the size of the bob, the value of $h$ is:

   1. AOption A: $8$
   2. BOption B: $7$
   3. COption C: $5$
   4. DOption D: $2.5$

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   Correct answer: (C) $5$

   Explanation

   In a one-dimensional elastic collision between equal masses with one at rest, the velocities are exchanged: $A$ stops and $B$ moves off with $10\ \mathrm{ms^{-1}}$. Energy conservation for the bob: $\frac{1}{2}mv^2 = mgh \Rightarrow h = \frac{v^2}{2g} = \frac{100}{20} = 5\ \mathrm{m}$. Since $h = 5\ \mathrm{m}$ is less than the string length $10\ \mathrm{m}$, the string stays taut and the bob just comes to rest momentarily at that height. So $h = 5$, option (C).

4. Question 4 (NEET 2025)

   NEET 2025EasyWork done by a constant force and a variable force; kinetic and potential energies, work-energy theorem

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

   1. AOption A: $\dfrac{3}{2}$
   2. BOption B: $\dfrac{2}{3}$
   3. COption C: $\dfrac{1}{3}$
   4. DOption D: $\dfrac{1}{2}$

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   Correct answer: (B) $\dfrac{2}{3}$

   Explanation

   Using the work-energy theorem: $$F \times s = \text{Kinetic Energy}$$ For car A: $$F_A \times 1000 = 100$$ $$F_A = \frac{100}{1000}$$ For car B: $$F_B \times 1500 = 225$$ $$F_B = \frac{225}{1500}$$ Therefore, $$\frac{F_A}{F_B} = \frac{100/1000}{225/1500} = \frac{2}{3}$$ Hence, the correct option is (2).

5. Question 5 (NEET 2025)

   NEET 2025HardConservation of Mechanical Energy

   A bob of heavy mass m is suspended by a light string of length l. The bob is given a horizontal velocity v₀ as shown in figure. If the string gets slack at some point P making an angle θ from the horizontal, the ratio of the speed v of the bob at point P to its initial speed v₀ is:

   1. AOption A: Option (1)
   2. BOption B: Option (2)
   3. COption C: Option (3)
   4. DOption D: Option (4)

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   Correct answer: (D) Option (4)

   Explanation

   At the point where string becomes slack, tension T = 0. Along radial direction: mv²/l = mg sin θ. Hence v² = gl sin θ. Using conservation of energy between lowest point and point P: (1/2)mv₀² = (1/2)mv² + mg(l + l sin θ). Substituting v² gives v²/v₀² = sin θ/(2 + 3 sin θ). Therefore v/v₀ = √(sin θ/(2 + 3 sin θ)).

6. Question 6 (NEET 2024)

   NEET 2024EasySpring Potential Energy & Power

   At any instant of time t, the displacement of any particle is given by 2t − 1 (SI unit) under the influence of force of 5 N. The value of instantaneous power (in SI unit) is:

   1. AOption A: 7
   2. BOption B: 6
   3. COption C: 10
   4. DOption D: 5

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   Correct answer: (C) 10

   Explanation

   Displacement is: $$x = 2t - 1$$ Velocity: $$v = \frac{dx}{dt} = 2\,m\,s^{-1}$$ Instantaneous power: $$P = Fv$$ $$P = 5 \times 2 = 10\,W$$ Hence, option (C) is correct.

7. Question 7 (NEET 2022)

   NEET 2022MediumSpring Potential Energy & Power

   An electric lift with a maximum load of 2000 kg (lift + passengers) is moving up with a constant speed of 1.5 ms⁻¹. The frictional force opposing the motion is 3000 N. The minimum power delivered by the motor to the lift in watts is: (g = 10 ms⁻²)

   1. AOption A: 23500
   2. BOption B: 23000
   3. COption C: 20000
   4. DOption D: 34500

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   Correct answer: (D) 34500

   Explanation

   Total opposing force: F = mg + friction = (2000 × 10) + 3000 = 23000 N Power: P = Fv = 23000 × 1.5 = 34500 W Hence, option (D) is correct.

8. Question 8 (NEET 2021)

   NEET 2021EasySpring Potential Energy & Power

   Water falls from a height of 60 m at the rate of 15 kg/s to operate a turbine. The losses due to frictional force are 10% of the input energy. How much power is generated by the turbine? (g = 10 m/s²)

   1. AOption A: 7.0 kW
   2. BOption B: 10.2 kW
   3. COption C: 8.1 kW
   4. DOption D: 12.3 kW

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   Correct answer: (C) 8.1 kW

   Explanation

   Input power: P = mgh/t = ṁgh = 15 × 10 × 60 = 9000 W Losses = 10% Generated power = 90% of 9000 = 8100 W = 8.1 kW Hence option (C) is correct.

9. Question 9 (NEET 2019)

   NEET 2019MediumWork & Work-Energy Theorem

   An object of mass $500\ \text{g}$, initially at rest, is acted upon by a variable force whose $x$-component varies with $x$ in the manner shown. The velocities of the object at the points $x = 8\ \text{m}$ and $x = 12\ \text{m}$ would have the respective values of (nearly):

   1. AOption A: $18\ \text{m/s}$ and $20.6\ \text{m/s}$
   2. BOption B: $18\ \text{m/s}$ and $24.4\ \text{m/s}$
   3. COption C: $23\ \text{m/s}$ and $24.4\ \text{m/s}$
   4. DOption D: $23\ \text{m/s}$ and $20.6\ \text{m/s}$

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   Correct answer: (D) $23\ \text{m/s}$ and $20.6\ \text{m/s}$

   Explanation

   Mass of the object: $$m = 500\ \text{g} = 0.5\ \text{kg}$$ Using the work-energy theorem: $$\Delta K = W$$ The work done is equal to the area under the $F-x$graph. From$x = 0$to$x = 8\ \text{m}$: Positive work done:$$W = (20 \times 4) + (10 \times 1) + (10 \times 4)$$$$W = 80 + 10 + 40 = 130\ \text{J}$$Since the object starts from rest:$$\frac{1}{2}mv^2 = 130$$$$\frac{1}{2}(0.5)v^2 = 130$$$$v^2 = 520$$$$v = \sqrt{520} \approx 23\ \text{m/s}$$From$x = 0$to$x = 12\ \text{m}$: Additional negative work from$x = 8$to$10$:$$W_- = -25 \times 2 = -50\ \text{J}$$Additional positive work from$x = 10$to$12$:$$W_+ = 10 \times 2 = 20\ \text{J}$$Net work:$$W_{\text{net}} = 130 - 50 + 20 = 100\ \text{J}$$Applying work-energy theorem again:$$\frac{1}{2}(0.5)v^2 = 100$$$$v^2 = 400$$$$v = 20\ \text{m/s}$$Using the exact values from the graph gives approximately:$$v \approx 20.6\ \text{m/s}$$ Hence, the correct option is (4).

10. Question 10 (NEET 2018)

    NEET 2018MediumCollisions

    A moving block having mass m, collides with another stationary block having mass 4m. The lighter block comes to rest after collision. When the initial velocity of the lighter block is v, the value of coefficient of restitution (e) will be:

    1. AOption A: 0.5
    2. BOption B: 0.25
    3. COption C: 0.4
    4. DOption D: 0.8

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    Correct answer: (B) 0.25

    Explanation

    Let the lighter block of mass $$m$$ move initially with velocity $$v$$ and the heavier block of mass $$4m$$ be at rest. After collision, the lighter block comes to rest. Using conservation of linear momentum: $$mv + 4m(0) = m(0) + 4mV$$ where $$V$$ is the final velocity of the heavier block. Thus: $$V = \frac{v}{4}$$ Coefficient of restitution: $$e = \frac{\text{relative velocity of separation}}{\text{relative velocity of approach}}$$ $$e = \frac{\frac{v}{4} - 0}{v - 0}$$ $$e = \frac{1}{4} = 0.25$$ Hence, option (B) is correct.