Work Energy and Power JEE Main Questions 2025: Important Practice Questions with PYQs

Work Energy and Power JEE Main Questions 2025 - Work Energy and Power from JEE Main Physics is one of the most important topics of the exam. The topic holds at least 2 - 4 questions every year which stands to approximately 8 to 12 marks in the JEE Main Physics exam. However, the expected Work Energy and Power weightage for JEE Main 2025 Physics is 5 to 7% of the entire JEE Main 2025 syllabus.

Latest update: NTA has declared the JEE Main 2025 exam dates! JEE Main 2025 exam will be conducted between January 22 to 31, 2025.

Quick Link: JEE Main Application Form 2025

In this article, we have discussed the Important Practice Questions of Work Energy and Power JEE Main 2025 along with the solutions. 

Also Read: JEE Main Previous Year Question Paper

JEE Main 2025 Work Energy and PowerImportant Questions

Check the Important Practice Questions with PYQs related to Work Energy and Power from JEE Main Physics.

1.  Two masses of 1 g and 4g are moving with equal kinetic energy. The ratio of the magnitudes of their momenta is

(a) 4 : 1

(b) √2 : 1

(c) 1: 2

(d) 1 : 16

Answer - 

Where p denotes momentum, E denotes kinetic energy and m denotes mass.

Therefore the answer is (c) 1:2

1. If a machine is lubricated with oil -

(A) The Mechanical Advantage Of The Machine Increases

(B) Its Efficiency Increases, But Its Mechanical Advantage Decreases.

(C) Both Its Mechanical Advantage And Efficiency Increase

(D) The Mechanical Efficiency Of The Machine Increases 

The answer is (D) When a machine is lubricated with oil friction decreases. Hence the mechanical efficiency of the machine increases.

(a) √t

(b) t

(c) constant

(d) 1/√t

Solution

Given that kinetic energy increases uniformly with time K.E ∝ t

Kinetic energy can be written as KE= ½ mv2 so we can write ½ mv2 ∝ t Therefore, v ∝ √t 

F ∝ 1/v

1. This question has Statement 1 and Statement 2. Of the four choices given after the statements, choose the one that best describes the two statements.

If two springs S1 and S2 of force constants k1 and k2, respectively, are stretched by the same force, it is found that more work is done on spring S1 than on spring S2.

• Statement-1: If stretched by the same amount, work done on S1, will be more than that on S2
• Statement-2 :k1 < k2

(A) Statement-1 is False, Statement-2 is true

(B) Statement-1 is True, Statement-2 is true and Statement-2 is not the correct explanation of Statement-1.

(C Statement-1 is True, Statement-2 is false

(D) Statement-1 is True, Statement-2 is true and Statement-2 is the correct explanation of statement-1.

Solution

Given same force F = k1x1 = k2x2

k1/k2 = x1/x2

W1 = ½ k1x12 and W2=½ k2x22

As W1/W2 ＞1 so [½ k1x12/ ½ k2x22]＞1

Fx1/Fx2＞1 ⇒ k2//k1＞1

Therefore k2 ＞ k1 Statement-2 is true

Or 

if x1 = x2 = x

W1/W2 = ½ k1x2 / ½ k2x2

W1/W2 = k1/ k2 ＜1

W1 ＜W2 , Statement-1 is False

Answer: (A) Statement-1 is False, Statement-2 is true

1. A stone tied to a string of length L is whirled in a vertical circle with the other end of the string at the centre. At a certain instant of time, the stone is at its lowest position and has a speed u. The magnitude of the change in its velocity as it reaches a position where the string is horizontal is - 

Answer - 

At lowest position B, Potential Energy = 0

Kinetic Energy = ½ mu2

Total Energy = ½ mu2

At C, when the string is horizontal,

Potential Energy = mgL

Kinetic Energy = ½ mv2

Total Energy = ½ mv2 + mgL

Since energy is conserved,

½ mv2 + mgL = ½ mu2

v2 = u2 -2gL

Since v is in the vertical direction and u is in the horizontal direction, they are mutually perpendicular to each other.

Change in velocity

The answer is 

6. When a rubber band is stretched by a distance x, it exerts a restoring force of magnitude F = ax + bx2 where a and b are constants. The work done in stretching the unstretched rubber band by L is

(a) aL²/2 +bL³/3

(b) ½ (aL²/2 +bL³/3)

(c ) aL² +bL³

(d) ½ (aL² +bL³)

Solution

Work done by a variable force

7. A person trying to lose weight by burning fat lifts a mass of 10 kg to a height of 1 m 1000 times. Assume that the potential energy lost each time he lowers the mass is dissipated. How much fat will he use up considering the work done only when the weight is lifted? Fat supplies 3.8 X 10⁷ J of energy per kg which is converted to mechanical energy with a 20% efficiency rate. 

Take g = 9.8 m/s²

(a) 12.89 X 10-³ kg

(b) 2.45 X 10-³ kg

(c) 6.45 X 10-³ kg

(d) 9.89 X 10-³ kg

Solution

Work done against gravity = (mgh) 1000 in lifting 1000 times

= 10 x 9.8 X 10³

= 9.8 x 10⁴ Joule

20% of the efficiency is used to convert fat into energy

(20% of 3.8 X 10⁷ J ) x m = 9.8 x 10⁴

Where m is the mass

m = 12.89 x 10-³ Kg

8. A time-dependent force F = 6t acts on a particle of mass 1 kg. If the particle starts from rest, the work is done by the force during the first 1 sec. will be :

(a) 9 J

(b) 18 J

(c) 4.5 J

(d) 22 J

Solution

F= 6t, m = 1Kg, u =0

Now, F=ma =m(dv/dt)=1x(dv/dt)

dv/dt = 6t

v= 3(12– 0) = 3m/s

From work-energy theorem

W =Δ KE = ½ m(v2 – u2)

W = ½ x 1(32 – 0) = 4.5 J

Answer: (c) 4.5 J

9. If a machine is lubricated with oil [1980 - 2 marks]

A)The Mechanical Advantage Of The Machine Increases

B)Its Efficiency Increases, But Its Mechanical Advantage Decreases

C)Both Its Mechanical Advantage And Efficiency Increase

D)The Mechanical Efficiency Of The Machine Increases

Ans. (D) When a machine is lubricated with oil friction decreases. Hence the mechanical efficiency of the machine increases.

10. If W1, W2 and W3 represent the work done in moving a particle from A to B along three different, paths 1,2 and 3 respectively (as shown) in the gravitational field of a point mass m, find the correct relation between W1, W2 and W3 [2003]

a)W1>W2>W3

b)W1 = W2 = W3

c)W1 < W2 < W3

d)W2 >W1 >W3

Ans. (b) The gravitational field is conservative. In a conservative field, the work done by W does not depend on the path (from A to B). It depends on initial and final points W1=W2 = W3

11. If the heart pushes 1 cc of blood in 1 s under pressure 20000 Nm-², the power of the heart is

1) 0.02 W

2) 400 W

3) 5 × 10-10 W

4) 0.2 W

Answer: 1) 0.02 W

Given,

Pressure, P = 20000 Nm-²

Volume per second = 1cc

Power of heart = Pressure x volume per second

= 20000 Nm-² x 10-⁶ = 0.02 w

12. Potential energy of a particle is related to x coordinate by equation x2 – 2x will be equilibrium at

1) x = 0.5

2) x = 1

3) x = 2

4) x = 4

Answer: 2) x = 1

Solution:

For stable equilibrium, Fnet = 0

F = – dU/dx

F = -d(x2 – 2x)/dx

= -(2x -2)

-2x + 2 = 0

⇒ x = 2/2 = 1

⇒ x = 1

13. A bomb of mass 9 kg explodes into two parts. One part of mass 3 kg moves with a velocity 16 m/s, then the kinetic energy of the other part is -

1) 162 J

2) 150 J

3) 192 J

4) 200 J

Answer: 3) 192 J

Solution:

Given,

Mass of the bomb = 9 Kg

Mass of smaller part = 3 kg

Mass of the bigger part = 6 Kg

Velocity, v = 16 m/s

Momentum Before Explosion = Momentum After Explosion

M x 0 = 3 x 16 + 6 x v

6v = – (3 x 16)

v = – 8 m/s

Kinetic energy of mass 6 Kg = (1/2) x 6 x (8)2

= 192 J

14. The kinetic energy of a body becomes four times its initial value. The new momentum will be

a) Same As The Initial Value

b) Twice The Initial Value

c) Thrice The Initial Value

d) Half Of Its Initial Value

Answer: b) Twice The Initial Value

E = p2/2m

E ∝ p2

(E1/E2) = p,2/p22

⇒ 1/4 = p,2/p22

⇒ 1/2 = p1/p2

p2 = 2p1

15. A particle of mass 100 g is thrown vertically upwards with a speed of 5 m/s. The work done by the force of gravity during the time, the particle goes up is

1) –0.5 J

2) –1.25 J

3) 1.25 J

4) 0.5 J

Answer: 2) –1.25 J

Solution:

Mass = 100 g

Speed, u = 5 m/s

Work done by the force of gravity is the kinetic energy while moving up

Wg = (1/2)mvf2 – (1/2)mvi2

= 0 – (1/2) x (0.1 kg) x 25

= -1.25 J article 

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