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

Dipanjana Sengupta

Updated On: October 15, 2024 11:51 AM | JEE Main

Important Work Energy and Power JEE Main Questions 2025 are available here and have been prepared based on JEE Main Physics previous year's question papers. Check the Important Practice Questions with PYQs here.
 
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 .

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.


3. At time t = 0 s particle starts moving along the x-axis. If its kinetic energy increases uniformly with time ‘t’, the net force acting on it must be proportional to -

(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

Therefore,
F ∝ 1/√t
Answer: (d) 1/√t

  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


Related Reads

Kinematics Weightage in JEE Mains 2025 Units and Dimensions Weightage in JEE Main 2025
Kinematics JEE Main Questions 2025 Laws of Motion Weightage in JEE Mains 2025
Rotational Motion JEE Main Questions 2025 Gravitation Weightage in JEE Main 2025
Laws of motion JEE Main Questions 2025 JEE Main Question Paper April 2024 with Answer Key Solutions
Work Energy and Power Weightage in JEE Main 2025 Rotational Motion Weightage in JEE Main 2025
JEE Main 2025 Exam JEE Main Preparation Tips
JEE Main Exam Pattern Gravitation JEE Main Questions 2025
Thermodynamics Weightage in JEE Mains 2025 Thermodynamics JEE Main Questions 2025
Kinetic Theory of Gases Weightage in JEE Main 2025 Kinetic Theory of Gases JEE Main Questions 2025
Oscillation Weightage in JEE Mains 2025 Oscillation JEE Main Questions 2025
Electrostatics Weightage in JEE Main 2025 Electrostatics JEE Main Questions 2025
Current Electricity Weightage in JEE Main 2025 Current Electricity JEE Main Questions 2025

For more Work Energy and Power JEE Main Questions 2025, Stay tuned with Collegedekho !

Are you feeling lost and unsure about what career path to take after completing 12th standard?

Say goodbye to confusion and hello to a bright future!

news_cta

FAQs

How much time is required to prepare for the JEE Mains exam?

Although maximum candidates usually require one or two years to prepare for the JEE Main and the JEE Advanced exam, candidates can also ace these exams with a tight study routine of 6 months.

 

Is HC Verma enough for the JEE Mains physics preparation?

HC Verma is considered to be a good book for the JEE Main preparation. It comprises good explanations and many practice problems. However, it is not enough for the preparations and candidates are required to study the NCERT textbooks as well.

 

Is the JEE Mains physics paper easy?

The physics section in the JEE Main exam is typically of moderate level of difficulty. It comprises a mix of numerical and concept-based questions.

Is 75 marks compulsory in the JEE Mains exam?

Candidates must keep in mind that there is no requirement of 75% marks in class for appearing in the JEE Main exam. The JEE Main eligibility criteria of minimum 75% marks in class 12 is required at the time of admission in NITs, IIITs and GFTIs. Candidates can apply and appear in the JEE Main exam irrespective of their class 12 marks.

 

What are the questions asked in the JEE Main exam?

The JEE Main exam has a total of 90 questions from all three subjects. The JEE Main exam has 30 questions each from the subjects of Physics, Chemistry and Mathematics. Out of 30 questions, there are 20 multiple choice questions and 10 are questions with numerical value answers.

 

Is NCERT enough for the JEE Mains preparation?

The official syllabus of JEE Main entrance exam and NCERT books cover almost all topics, and it is preferred by the candidates for their preparation. The syllabus of JEE Main and that of CBSE's Class 11 and 12 board is almost the same. Candidates can consider NCERT books as one of the best reference materials for the JEE Main exam.

 

Can candidates get admission into NIT with 35,000 rank in the JEE Main exam?

Some NITs and IIITs which candidates can get with 35000 rank in the JEE Main entrance exam are Indian Institute of Information Technology Kalyani, National Institute of Technology Mizoram,  Indian Institute of Information Technology Dharwad, National Institute of Technology Nagaland, and National Institute of Technology Manipur.

 

Is it easy to score 99 in the JEE Mains exam?

The exact marks to be obtained for getting the 99 percentile in the JEE Main exam depends on the difficulty level of the paper. If the JEE Main question paper turns out to be easy, then marks secured by most of the candidates are good and the higher percentile is relative and gets pushed for higher marks.

 

 

Is 40 considered to be a good score in the JEE Main exam?

No, 40 is regarded as  a very good score in the JEE Main exam. Since 85 to 95 percentile is considered to be good for admission into NITs, and IITs, securing anything less than that is not a good score for the General, OBC, and EWS candidates. With this score, such candidates can barely make it through a few engineering colleges of India.

 

View More
/articles/work-energy-and-power-jee-main-questions-with-pyqs/
View All Questions

Related Questions

How many marks or percentile do I have to get in jee mains 2025 to getadmission in CSE branch in ISLAMIC UNIVERSITY OF SCIENCE AND TECHNOLOGY

-Basit_showkat2 '::::Updated on October 15, 2024 11:14 AM
  • 1 Answer
Soham Mitra, Content Team

The Islamic University of Science and Technology (IUST) located in Jammu and Kashmir accepts the JEE Main scores of the candidates for providing admission into its B.Tech. courses. The JEE Main (Paper-I) score required for selection into the B. Tech courses offered at IUST ranges between 50% to 60%. For securing admission via lateral entry, candidates must have qualified a minimum of 3 years or 2 years (lateral entry) Diploma examination with at least 45% marks in any branch of Engineering and Technology. Candidates must keep in mind that the IUST cutoff varies every year and is subject to a …

READ MORE...

Btech in robotics and artificial intelligence ke liye kya eligibility hai? Or form kab kab nikalte hai?

-ShanayaUpdated on October 14, 2024 03:31 PM
  • 1 Answer
Rupsa, Content Team

Dear Student,

B.tech in Robotics and Artificial Intelligence aajkal ke aspiring engineering students ke beech ek bahut popular course hai. India me kai colleges, jaise ki DIT University, Dehradun, ye course offer karte hain. B.tech in Robotics and AI ke liye eligibility criteria baaki branch ki tarah hi hai, jisme candidates ko apni 10+2 education kisi recognized school education board se complete karni hoti hai aur unhe Joint Entrance Examination (JEE Main) dena hota hai. Iske liye application forms November ya December me release ho jate hain. Admissions merit list ke basis par diye jaate hain. Candidates DIT …

READ MORE...

parallelogram law of forces

-BasavarajUpdated on October 14, 2024 04:27 PM
  • 1 Answer
Rupsa, Content Team

Dear Student, 

If you are asking about Parallelogram Law of Forces topics in general, then let me tell you that it is indeed a part of the B.Tech in Civil Engineering syllabus. The topic is often asked in entrance exams like JEE Main and JEE Advanced as part of the Physics section. The questions asked from this topic in the JEE papers generally include concepts such as the Parallelogram formula of addition of vectors, formula of resultant vectors etc. The Physics section has a weightage of 33.33% in the JEE Mains examination, covering a significant part of the question …

READ MORE...

Do you have a question? Ask us.

  • Typical response between 24-48 hours

  • Get personalized response

  • Free of Cost

  • Access to community

Recent Articles

Subscribe to CollegeDekho News

By proceeding ahead you expressly agree to the CollegeDekho terms of use and privacy policy

Top 10 Engineering Colleges in India

View All

Tell us your JEE Main score & access the list of colleges you may qualify for!

Predict My College
Top