NEET Previous Year Questions (2014-2025): Work, Energy & Power

Table of Contents
1. 2025
2. 2024
3. 2023
4. 2022
5. 2021
6. 2020
7. 2019
8. 2018
9. 2017
10. 2016
11. 2015
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The terms 'work', 'energy', and 'power' are frequently used in everyday language. In physics, however, the word 'Work' covers a definite and precise meaning. This chapter of NEET physics covers the fundamental concepts of energy, work, and power, and their applications to various physical systems. Let's have a look at the Previous Year's Questions of NEET Exam of this chapter.

2025

Q1: The kinetic energies of two similar cars A and B are 100 J and 225 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 F_A / F_B is:   [2025]
(a) 1/3
(b) 1/2
(c) 3/2
(d) 2/3
Ans: (d)
From work energy theorm
W . D = ΔKE
⇒ F . S = ΔKE
⇒ (ΔKE)_A / (ΔKE)_B = - F_A S_A / F_B S_B
⇒ (-100) / (225) = - F_A(1000) / F_B(1500)
F_A / F_B = 2 / 3

Q2: 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:    [2025]
(a)
(b)
(c)
(d)
Ans: (b)

At Point P, mg sin θ = m v²/ l ... (1)
By conservation of mechanical energy at point P ∈ Q
(1/2) mv₀² = (1/2) mv² + mg (I + I sin θ)
⇒ v₀² / 2 = v² / 2 + gl(1 + sin θ)
Put gl = v² / l sin θ using (1)
v₀² / 2 = v² / 2 + (v² / sin θ)(1 + sin θ)
v₀² / 2 = (3v²/2) + (v² / sin θ)
⇒v / v₀  = (sin θ /(2 + 3 sin θ))^1/2

2024

where P is the power,  is the force vector, and  is the velocity vector of the particle.
In this question, the force exerted is given as a constant 5N. Since no direction is specified, and the displacement is given in a scalar form, we assume the force acts along the direction of displacement. Therefore, we can treat the vectors as scalars for simplicity.
The displacement of the particle is given by:
$\mathrm{x\; =\; 2t\; -\; 1}$
The velocity, , is the derivative of displacement with respect to time. Deriving the displacement equation with respect to time t gives:

The instantaneous power can now be calculated as:
$Watts$
Thus, the instantaneous power delivered by the force at any time t is 10 Watts$Watts$.
The correct answer is Option A: 10.

Q2: Two bodies A and B of same mass undergo completely inelastic one dimensional collision. The body A moves with velocity v₁ while body B is at rest before collision. The velocity of the system after collision is v₂. The ratio v₁ : v₂ is: [2024]
(a) 1 : 2
(b) 2 : 1
(c) 4 : 1
(d) 1 :  4
Ans: (b)
In a completely inelastic collision, the two bodies stick together and move with a common final velocity. Here, before the collision, body A is moving with a velocity v₁ and body B is at rest. The conservation of momentum must hold true because no external forces are acting on the system.
The momentum before the collision is only due to body A since body B is at rest. Therefore, the total initial momentum p_initial of the system is given by:

where:

• m_A and m_B are the masses of bodies A and B respectively,
• m is the mass of each body,
• v₁ is the velocity of body A,

v₀ = 0 is the velocity of body $B$ because it is initially at rest.
Since they undergo a completely inelastic collision, bodies A and B stick together after the collision and hence move with a common velocity v₂. The total mass of the combined system post-collision is The momentum after the collision is given by:

Applying the conservation of momentum (since no external force implies momentum is conserved), we equate p_initial$initial$ and p_final$final$:

Dividing through by $m$ yields:

Thus, solving for the ratio v₁ / v₂ gives:

Therefore, the ratio of v₁ to v₂ is $\mathrm{2 }:1$, making the correct answer:
Option B: 2 : 1

2023

Q2: A particle moves with a velocity  horizontally under the action of constant force . The instantaneous power supplied to the particle is :     [2023]
(a) 200 W
(b) Zero
(c) 100 W
(d) 140 W
Ans: (d)

2022

Q1: The restoring force of a spring with a block attached to the free end of the spring is represented by             [2022]
(a)

(b)

(c)

(d)
Ans: (a)

Where F is restoring force
x is displacement of block from equilibrium position

Q2: 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 m s ^-2 )           [2022]
(a) 20000 
(b) 34500 
(c) 23500 
(d) 23000 
Ans: b

The power delivered by the motor to the lift can be calculated using the formula:

Power = Force x Velocity

In this case, the force is the sum of the weight of the lift and the passengers and the frictional force opposing the motion:

Force = (Weight of lift + passengers) + Frictional force

Weight of lift + passengers = Mass x gravity

= (2000 kg / 1000) x 10 m/s²

= 20,000 N

Therefore,

Force = (20,000 N) + (3000 N)

= 23,000 N

Now, using the formula for power, we get:

Power = Force x Velocity

= (23,000 N) x (1.5 m/s)

= 34,500 W

Therefore, the minimum power delivered by the motor to the lift in watts is 34,500 W.

Q3: The energy that will be ideally radiated by a 100 kW transmitter in 1 hour is            [2022]
(a) 36 $\times$ 10⁷ J 
(b) 36 $\times$ 10⁴ J 
(c) 36 $\times$ 10⁵ J 
(d) 1 $\times$ 10⁵ J 
Ans: (a)
Energy = Power $\times$ tim
E = 100 $\times$x 10³ $\times$x 3600
= 36 $\times$x 10⁷ J

2021

Q1: A particle is released from height S from the surface of the earth. At a certain height, its kinetic energy is three times its potential energy. The height from the surface of the earth and the speed of the particle at that instant are respectively:          [2021]
(a)
(b)
(c)
(d)

Ans: B
U + KE = E
4U = E = mgS
4mgh = mgS
h = S/4

Q2: 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) [2021]
(a) 12.3 kW
(b) 7.0 kW
(c) 10.2 kW
(d) 8.1 kW
Ans: D

Mass of water falling/second = 15 kg/s h = 60 m, g = 10 m/s², loss = 10 % i.e., 90% is used. Power generated = 15 x 10 x 60 x 0.9 = 8100 W = 8.1 kW.

2020

Q.6. The energy required to break one bond in DNA is 10^-20J. This value in eV is nearly:        [2020]
(a) 0.06 
(b) 0.006 
(c) 6 
(d) 0.6
Ans: a

1.6 × 10^-19 Joule = 1 eV

2019

Q1: A force F = 20 + 10 y acts on a particle in y direction where F is in newton and y in meter. Work done by this force to move the particle from y = 0 to y = 1 m is    [2019]
(a) 30 J
(b) 5 J
(c) 25 J
(d) 20 J
Ans: c

Work done by variable force is

Q2: Body A of mass 4m moving with speed u collides with another body B of mass 2m, at rest. The collision is head on and elastic in nature. After the collision the fraction of energy lost by the colliding body A is:    [2019]
(a) 1/9
(b) 8/9
(c) 4/9
(d) 5/9
Ans: b

Fractional loss of KE of colliding body

Q3: A disc of radius 2 m and mass 100 kg rolls on a horizontal floor. Its centre of mass has speed of 20 cm/s. How much work is needed to stop it?    [2019]
(a) 3 J
(b) 30 kJ
(c) 2 J
(d) 1 J
Ans: a

Work required = change in kinetic energy Final KE = 0
Initial KE

|ΔKE| = 3 J

Q4: The work done to raise a mass m from the surface of the earth to a height h, which is equal to the radius of the earth, is:    [2019]
(a) mgR
(b) 2mgR
(c)

(d)

Ans: c

Initial potential energy at earths surface is

Final potential energy at height h = R

As work done = Change in PE∴ W = U_f - U_i

 (∵ GM = gR²)

2018

Q1: A body initially at rest and sliding along a frictionless track from a height h (as shown in the figure) just completes a vertical circle of diameter AB = D. The height h is equal to    [2018]
(a) 3/2 D
(b) D
(c) 7/5 D
(d) 5/4 D
Ans: d

Q2: 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, then the value of coefficient of restitution (e) will be :-    [2018]
(a) 0.5
(b) 0.25
(c) 0.8
(d) 0.4
Ans: b

Q3: The kinetic energies of a planet in an elliptical orbit about the Sun, at positions A, B and C are KA, KB and KC respectively. AC is the major axis and SB is perpendicular to AC at the position of the Sun S as shown in the figure. Then    [2018]
(a) KA < KB < KC
(b) KA > KB > KC
(c) KB < KA < KC
(d) KB > KA > KC
Ans: b

2017

Q1: Consider a drop of rain water having mass 1 g falling from a height of 1 km. It hits the ground with a speed of 50 m/s. Take 'g' constant with a value 10 m/s². The work done by the [2017]
(i) gravitational force and the
(ii) resistive force of air is :-
(a) (i) 1.25 J (ii) - 8.25 J
(b) (i) 100 J (ii) 8.75 J
(c) (i) 10 J (ii) - 8.75 J
(d) (i) - 10 J (ii) - 8.25 J
Ans: c

Q2: A spring of force constant k is cut into lengths of ratio 1: 2 : 3. They are connected in series and the new force constant is k'. Then they are connected in parallel and force constant is k''. Then k' : k'' is    [2017]
(a) 1 : 6
(b) 1 : 9
(c) 1 : 11
(d) 1 : 14
Ans: c

Force constant of spring

after cutting the spring in the ratio 1:2:3, the force constants will be in the ratio 6:3:2
Force constant in series K' is given by
Force constant in parallel K" : 6k+3k+2k = 11k;hence K'/K" = 1/11

2016

Q1: A particle moves from a point   when a force of  N is applied. How much work has been done by the force ?      [2016]
(a) 8 J
(b) 11 J
(c) 5 J
(d) 2 J
Ans: (c)

Q2: A body of mass 1 kg begins to move under the action of a time dependent force  , whenare unit vectors along x and y axis. What power will be developed by the force at the time t ?    [2016]
(a) (2t³+ 3t⁵)W
(b) (2t² + 3t³)W
(c) (2t² + 4t⁴)W
(d) (2t³ + 3t⁴)W
Ans: a

Q3: A particle of mass 10 g moves along a circle of radius 6.4 cm with a constant tangential acceleration. What is the magnitude of this acceleration if the kinetic energy of the particle becomes equal to 8 × 10^-4 J by the end of the second revoluation after the beginning of the motion ? [2016]
(a) 0.18 m/s²
(b) 0.2 m/s²
(c) 0.1 m/s²
(d) 0.15 m/s²
Ans: (c)

Given: Mass of particle, M = 10g $=$10/1000 Kg
radius of circle R = 6.4 cm
Kinetic energy E of particle = 8 × 10^-4J
acceleration a_t = ? 

Now, using

Q4: What is the minimum velocity with which a body of mass m must enter a vertical loop of radius R so that it can complete the loop?

Q5: Two similar springs P and Q have spring constants K_P and K_Q, such that K_P > K_Q. They are stretched, first by the same amount (case a), then by the same force (case b). The work done by the springs W_P and W_Q are related as, in case (a) and case (b), respectively: [2015]
A: W_P < W_Q; W_Q < W_P
B: W_P = W_Q; W_P > W_Q
C: W_P = W_Q; W_P = W_Q
D: W_P > W_Q; W_Q > W_P

Ans: d

The block of mass M = 10 kg is moving in the x - direction with a speed v = 10 m/s. Its initial kinetic energy is
It is subjected to a retarding force F = 0.1 x J/m during its travel from x = 20 m to 30 m.Work done is given by
Final kinetic energy is, KE_f = KE_i + W = 500 - 25 = 475