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Last updated date: 25th Mar 2023

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This chapter will teach us about concepts like** work, energy and power**. Work, energy and power are essential concepts in physics. Work is the displacement of an object caused by a force (push or pull). Energy is defined as the ability to perform work. We also get to know about the work done formula and energy formula.

The chapter also includes the following concepts:

Kinetic Energy: It is the energy that a body possesses as a result of its motion.

Potential Energy: It is the energy possessed by a body by virtue of its position or configuration in some field.

Work Energy Theorem: This concept states that the work done by net force in displacing a body is equal to the change in kinetic energy of the body.

Principle of Conservation of Energy: According to this principle, if we account for all forms of energy, the total energy of an isolated system does not change.

Collision: It is an isolated event in which two or more colliding bodies exert relatively strong force on each other for a relatively short time.

Now, let's move onto the important concepts and formulae related to IIT JEE exams along with a few solved examples.

Work done by variable force.

Conservative and non-conservative forces.

Power definition and relation between work and power.

Kinetic energy and linear momentum relation.

Work energy theorem.

Potential energy of a spring.

Mechanical energy and its conservation.

Collision and its types.

1. A particle travels in a straight path, with retardation proportional to displacement. Calculate the kinetic energy loss for every displacement $x$.

Sol:

It is given that retardation($a$) is proportional to displacement ($x$). So in order to solve this problem, we first write retardation in terms of velocity and after separating and integrating the quantities in the given relation, we will be able to reach the answer.

According to the question,

$-a\varpropto x$

$-a=kx$...........(1)

Here, $k$ is the proportionality constant.

Now putting $a=\dfrac{\text{d}v}{\text{d}t}$ in the eq.(1), we get:

$-a=-\dfrac{\text{d}v}{\text{d}t} = kx$

Now after multiplying and dividing on L.H.S by $\text{d}x$, we get:

$\dfrac{\text{d}v}{\text{d}t} \dfrac{\text{d}x}{\text{d}x} =-kx$

As $\dfrac{\text{d}x}{\text{d}t}=v$, therefore, $v\text{d}v=-kx\text{d}x$

$\int_{v}^{0}vdv=-k\int_{x}^{0}xdx$

$[\dfrac{v^2}{2}]^v_u=-k[\dfrac{x^2}{2}]^x_0$

$\dfrac{1}{2}mv^2-\dfrac{1}{2}mu^2=\dfrac{-kx^2m}{2}$

Thus, the loss in kinetic energy is $\Delta K=\dfrac{-kmx^2}{2}$.

Trick: Use the given condition of the question and put the retardation formula in terms of velocity then separate the similar quantities and integrate them.

2. A body falls to the earth from a height of $8,m$ and then bounces to a height of $2,m$. Calculate the ratio of the body's velocities right before and after the contact. Calculate the body's kinetic energy loss as a percentage during the contact with the ground.

Sol:

It is given that heights $h_1=8\,m$ and $h_2=2\,m$. Let $v_1$ be the velocity of the body just before collision with the ground and $v_2$ be the velocity of the body just after collision.

Now, according to the conservation of mechanical energy, we can write:

$\dfrac{1}{2}mv_1^2=mgh_1$ and $\dfrac{1}{2}mv_2^2=mgh_2$

After dividing the above equation, we get:

$\dfrac{v^2_1}{v^2_2}=\dfrac{h_1}{h_2}$

Putting the values of $h_1$ and $h_2$, we get:

$\dfrac{v^2_1}{v^2_2}=\dfrac{8}{2}=4$

$\dfrac{v_1}{v_2}=2$

Therefore, the ratio of the velocities is 2:1.

Now, the percentage loss in kinetic energy is given as:

$\% \text{ age loss in K.E}=(\dfrac{K_1-K_2}{K_1}\times 100$

$\% \text{ age loss in K.E}=\dfrac{mg(h_1-h_2)}{mgh_1}\times 100$

$\% \text{ age loss in K.E}=\dfrac{(8-2)}{8}\times 100$

$\% \text{ age loss in K.E}= 75\%$

Hence, the percentage loss in kinetic energy is 75 %.

Key point: The laws of conservation of mechanical energy is an important concept and so is the loss of kinetic energy expression.

1. A block moving horizontally on a smooth surface with a speed of $40\,m/s$ splits into two equal parts. If one of the parts moves at $60\,m/s$ in the same direction, then the fractional change in the kinetic energy will be $x : 4$ where $x$ = ___________. (JEE Main 2021)

Sol:

Given:

Initial velocity of the block before splitting into two equal parts, $V=40\,m/s$

As the block is split into two equal masses, so if we consider the initial mass of the block $m$ then after splitting, it will be $\dfrac{m}{2}$. Let the velocity of 1st part be $v$ and the velocity of the second part according to the question is $v’=60\,m/s$.

Now, using the momentum conservation principle, we get:

$P_i=P_f$

$mV=\dfrac{m}{2} v+\dfrac{m}{2} v’$

$V=\dfrac{v}{2} +\dfrac{v’}{2}=\dfrac{v}{2}+\dfrac{60}{2}$

$40=\dfrac{v}{2}+30$

$v=10\times 2=20\,m/s$

Now, the initial kinetic energy ($(K.E.)_i$) is

$(K.E.)_i = \dfrac{1}{2}mV^2=\dfrac{1}{2}m(40)^2=800m$

The final kinetic energy ($(K.E.)_f$) is

$(K.E.)_f = \dfrac{1}{2}\dfrac{m}{2}(20)^2+\dfrac{1}{2}\dfrac{m}{2}(60)^2=1000m$

The change in kinetic energy ($\Delta K.E.$) is

$\lvert\Delta K.E.\rvert=\lvert 1000m-800m\rvert=200m$

The fractional change in kinetic energy is

$\text{fractional change in K.E.}=\dfrac{\Delta K.E.}{(K.E.)_i}=\dfrac{200 m}{800m}$

$\text{fractional change in K.E.}= \dfrac{1}{4}=\dfrac{x}{4}$

Hence, the value of $x$ is $1$.

Trick: Just apply the concept of conservation of momentum then fraction change in the kinetic energy.

1. A force of $F = (5y + 20) \widehat{j}\,N$ acts on a particle. The work done by this force when the particle is moved from $y = 0\,m$ to $y = 10\,m$ is ___________ J. (JEE Main 2021)

Sol:

Given:

A variable force expression, $F = (5y + 20) \widehat{j}\,N$

Moved from $y = 0\,m$ to $y = 10\,m$

The work done under the variable force is given by

$W=\int Fdy=\int_{0}^{10} Fdy$

Now, putting the value of $F$ in the above expression and integrating it, we get:

$W=\int_{0}^{10} (5y + 20)dy$

$W=(\dfrac{5y^2}{2}+2y)^{10}_0$

$W=\dfrac{5}{2}\times 100+2\times 10$

$W=250+200=450\,J$

Therefore, the value of work done is 450 J.

Key point: The expression of work done for variable forces is the important key to solve this problem.

1. A 10 kg ball and a 20 kg ball collide at velocities 20 m/s and 10 m/s, respectively. What are their post-collision velocities if the collision is completely elastic?

Ans: -20 m/s; 10 m/s

2. A spring gun has a spring constant of 80 N/cm. A ball of mass 15g compresses the spring by 12 cm. How much is the potential energy of the spring? What is the velocity of the ball if the trigger is pulled?

Ans: 57.6 J; 87.6 m/s

In this article, we have discussed concepts like work done by the variable forces, how kinetic energy is related to linear momentum and the work energy theorem. We have also discussed work and energy definition. The article also includes the concept of **conservation of mechanical energy and collision**.

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JEE Main 2023 January and April Session exam dates and revised schedule have been announced by the NTA. JEE Main 2023 January and April Session will now be conducted on 24-Jan-2023 to 31-Jan-2023 and 6-Apr-2023 to 12-Apr-2023, and the exam registration closes on 12-Jan-2023 and Apr-2023. You can check the complete schedule on our site. Furthermore, you can check JEE Main 2023 dates for application, admit card, exam, answer key, result, counselling, etc along with other relevant information.

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Last updated date: 25th Mar 2023

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NTA has announced the JEE Main 2023 January session application form release date on the official website https://jeemain.nta.nic.in/. JEE Main 2023 January and April session Application Form is available on the official website for online registration. Besides JEE Main 2023 January and April session application form release date, learn about the application process, steps to fill the form, how to submit, exam date sheet etc online. Check our website for more details. April Session's details will be updated soon by NTA.

Last updated date: 25th Mar 2023

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It is crucial for the the engineering aspirants to know and download the JEE Main 2023 syllabus PDF for Maths, Physics and Chemistry. Check JEE Main 2023 syllabus here along with the best books and strategies to prepare for the entrance exam. Download the JEE Main 2023 syllabus consolidated as per the latest NTA guidelines from Vedantu for free.

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JEE Main 2023 Study Materials: Strengthen your fundamentals with exhaustive JEE Main Study Materials. It covers the entire JEE Main syllabus, DPP, PYP with ample objective and subjective solved problems. Free download of JEE Main study material for Physics, Chemistry and Maths are available on our website so that students can gear up their preparation for JEE Main exam 2023 with Vedantu right on time.

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Download JEE Main Question Papers & Answer Keys of 2022, 2021, 2020, 2019, 2018 and 2017 PDFs. JEE Main Question Paper are provided language-wise along with their answer keys. We also offer JEE Main Sample Question Papers with Answer Keys for Physics, Chemistry and Maths solved by our expert teachers on Vedantu. Downloading the JEE Main Sample Question Papers with solutions will help the engineering aspirants to score high marks in the JEE Main examinations.

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Last updated date: 25th Mar 2023

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In order to prepare for JEE Main 2023, candidates should know the list of important books i.e. RD Sharma Solutions, NCERT Solutions, RS Aggarwal Solutions, HC Verma books and RS Aggarwal Solutions. They will find the high quality readymade solutions of these books on Vedantu. These books will help them in order to prepare well for the JEE Main 2023 exam so that they can grab the top rank in the all India entrance exam.

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JEE Main 2023 free online mock test series for exam preparation are available on the Vedantu website for free download. Practising these mock test papers of Physics, Chemistry and Maths prepared by expert teachers at Vedantu will help you to boost your confidence to face the JEE Main 2023 examination without any worries. The JEE Main test series for Physics, Chemistry and Maths that is based on the latest syllabus of JEE Main and also the Previous Year Question Papers.

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Last updated date: 25th Mar 2023

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NTA is responsible for the release of the JEE Main 2023 January and April Session cut off score. The qualifying percentile score might remain the same for different categories. According to the latest trends, the expected cut off mark for JEE Main 2023 January and April Session is 50% for general category candidates, 45% for physically challenged candidates, and 40% for candidates from reserved categories. For the general category, JEE Main qualifying marks for 2021 ranged from 87.8992241 for general-category, while for OBC/SC/ST categories, they ranged from 68.0234447 for OBC, 46.8825338 for SC and 34.6728999 for ST category.

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Last updated date: 25th Mar 2023

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NTA will release the JEE Main 2023 January and April sessions exam dates on the official website, i.e. {official-website}. Candidates can directly check the date sheet on the official website or https://jeemain.nta.nic.in/. JEE Main 2023 January and April sessions is expected to be held in February and May. Visit our website to keep updates of the respective important events of the national entrance exam.

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JEE Main 2023 state rank lists will be released by the state counselling committees for admissions to the 85% state quota and to all seats in IIT colleges. JEE Main 2023 state rank lists are based on the marks obtained in entrance exams. Candidates can check the JEE Main 2023 state rank list on the official website or on our site.

Last updated date: 25th Mar 2023

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Want to know which Engineering colleges in India accept the JEE Main 2023 scores for admission to Engineering? Find the list of Engineering colleges accepting JEE Main scores in India, compiled by Vedantu. There are 1622 Colleges that are accepting JEE Main. Also find more details on Fees, Ranking, Admission, and Placement.

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FAQ

**1. What is the weightage of the Work Energy and Power chapter in JEE?**

From this chapter, approximately two questions are asked every year which leads to a weightage of about 2-3% in the exam.

**2. What is the level of difficulty of the questions from the Work Energy and Power chapter?**

As this chapter deals with daily life applications, the questions asked in this chapter are from easy to moderate levels of difficulty. Thus, it is essential to study this chapter as the level of difficulty is not that hard. Studying this chapter can help you in improving your rank along with a good score.

**3. Is it truly beneficial to review Work Energy and Power chapter questions from previous years for this exam?**

We must practice previous years’ questions in order to score well and become familiar with the exam's difficulty level. It improves our self-esteem while also pointing us to areas where we may improve. Solving previous ten to fifteen years’ question papers will help you better comprehend a concept and also show you how many times a concept or topic has been repeated in the test. It is advisable to practice previous years’ questions in order to prepare for the work power energy notes.

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