JEE

# JEE Important Chapter - Work, Energy and Power

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## Work, Energy and Power

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.

### Important Topics for Work Energy and Power

• 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.

### Work Energy and Power Important Concepts for JEE Main

 Name of the Concept Key Points of the Concepts 1. Work done by variable force Suppose we have to calculate work done in moving a body from a point A($S_A$) to a point B($S_B$) under the action of a variable force. So the work done by this variable force in displacing a body from point $A$ to $B$ is,$W=\int_{S_B}^{S_A}\overrightarrow{F}.\overrightarrow{ds}$ 2. Conservative and non-conservative forces A force is said to be conservative if work done by or against the force in moving a body depends only on the initial and final position of the body, and not on the natural path followed between the initial and final positions.If the work done by or against a force in moving a body from one position to another relies on the path taken between these two places, the force is said to be non-conservative. 3. Power definition and relation between work and power Power is defined as the rate at which work is completed in a given amount of time.The relation between work ($W$) and power ($P$) is expressed as $P=\dfrac{W}{t}$ 4. Kinetic energy and linear momentum relation The energy of a body is defined as the capability or ability of the body to do the work.The relation between kinetic energy ($K.E.$) and linear momentum($P$) shows that a body cannot have kinetic energy without having linear momentum and vice-versa.$K.E.=\dfrac{p^2}{2m}$ 5. Work energy theorem This theorem states that if some work ($W$) is done by the body then the kinetic energy($K.E.$) of the body also increases by the same amount.Work done = increase in K.E. of the body 6. Potential energy of a spring The potential energy ($P.E.$) of the spring is the energy associated with the state of compression or expansion of an elastic spring. It is given as$P.E.=\dfrac{1}{2}kx^2$Here, $k$ = spring constant and $x$ = stretch or compression in the string 7. Mechanical energy and its conservation The total mechanical energy of the system is conserved if the forces doing work on the system are conservative i.e, when net work done by external non-conservative force is zero.The mechanical energy ($E$) of a body is the sum of kinetic energy ($K$) and potential energy ($V$) of the body. Mathematically, we can write it as$E=K+V=\text{constant}$ 8. Collision and its types When two things bump or strike against each other, it is referred to as a collision. There are two types of collision,Elastic collision: A collision in which there is no kinetic energy loss at all.Inelastic collision: A collision that results in some loss of kinetic energy.The basic characteristics of elastic collision are:Linear momentum is conservedTotal energy of the system is conserved.The kinetic energy is conserved.The basic characteristics of inelastic collision are:Linear momentum is conserved.Total energy is conserved.Kinetic energy is not conserved.

### List of Important Work Energy and Power Formulae

 S.No. Name of the Concept Formula Work done by variable force The work done formula by variable force is$W=\int_{S_B}^{S_A}\overrightarrow{F}.\overrightarrow{ds}$The work done formula for non-variable force is$W=\overrightarrow{F}.\overrightarrow{s}$ Kinetic energy and linear momentum relation. The kinetic energy formula ($K.E.$) is$K.E.=\dfrac{1}{2}mv^2$The expression of linear momentum ($p$) is$p=mv$The relationship between kinetic energy and linear momentum is as follows:$K.E.=\dfrac{p^2}{2m}$ Work energy theorem According to this theorem,$W={K_f} - {K_i}$ = increase in $K.E.$ of the bodyHere, ${K_f}$ and ${K_i}$ are the final and initial kinetic energy of the body. Potential energy of a spring The potential energy formula of a spring is$P.E.=\dfrac{1}{2}kx^2$The maximum velocity ($v_m$) of a block of mass $m$ upon maximum displacement ($x_m$) is$v_m=\sqrt{\dfrac{k}{m}}.x_m$ Mechanical energy and its conservation. According to conservation of mechanical energy,$E=K+V=\text{constant}$ Collision and its types The expression of coefficient of restitution ($e$) is$e=\dfrac{v_2-v_1}{u_1-u_2}$Here, $u$ and $v$ are the initial and final velocities of the object.The expressions of final velocities, after elastic collision of two object, are$v_1=\dfrac{(m_1-m_2)u_1}{m_1+m_2}+\dfrac{2m_2u_2}{m_1+m_2}$$v_2=\dfrac{(m_2-m_1)u_2}{m_1+m_2}+\dfrac{2m_1u_1}{m_1+m_2}$The expression of common velocity ($V$) when two object collide inelastically is$V=\dfrac{m_1u_1}{m_1+m_2}$

### Solved Examples

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.

### Previous Years’ Questions from JEE Paper

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.

### Practice Questions

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

### Conclusion

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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English  •   Shift 2
27th July 2022 - Maths, Physics and Chemistry
English  •   Shift 1, 2
27th July 2022 - Maths, Physics and Chemistry
English  •   Shift 1
27th July 2022 - mathematics
English  •   Shift 1
27th July 2022 - physics
English  •   Shift 1
27th July 2022 - chemistry
English  •   Shift 1
27th July 2022 - Maths, Physics and Chemistry
English  •   Shift 2
27th July 2022 - mathematics
English  •   Shift 2
27th July 2022 - physics
English  •   Shift 2
27th July 2022 - chemistry
English  •   Shift 2
28th July 2022 - Maths, Physics and Chemistry
English  •   Shift 1, 2
28th July 2022 - Maths, Physics and Chemistry
English  •   Shift 1
28th July 2022 - mathematics
English  •   Shift 1
28th July 2022 - physics
English  •   Shift 1
28th July 2022 - chemistry
English  •   Shift 1
28th July 2022 - Maths, Physics and Chemistry
English  •   Shift 2
28th July 2022 - mathematics
English  •   Shift 2
28th July 2022 - physics
English  •   Shift 2
28th July 2022 - chemistry
English  •   Shift 2
June, July
mathematics
English  •   Shift 1, 2
mathematics
English  •   Shift 1, 2
mathematics
English  •   Shift 1, 2
mathematics
English  •   Shift 1, 2
mathematics
English  •   Shift 1, 2
chemistry
English  •   Shift 1, 2
mathematics
English  •   Shift 1, 2

View all JEE Main Important Books
In order to prepare for JEE Main 2022, 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 2022 exam so that they can grab the top rank in the all India entrance exam.
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Maths
NCERT Book for Class 12 Maths
Physics
NCERT Book for Class 12 Physics
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## JEE Main Mock Tests

View all mock tests
JEE Main 2022 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 2022 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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JEE MAIN MOCK TEST - 1
3 hr  • 75 questions • OBJECTIVE
JEE MAIN MOCK TEST - 3
3 hr  • 75 questions • OBJECTIVE
JEE MAIN MOCK TEST - 2
3 hr  • 75 questions • OBJECTIVE

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## JEE Main 2022 Cut-Off

JEE Main Cut Off
NTA is responsible for the release of the JEE Main 2022 June and July 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 2022 June and July 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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## JEE Main 2022 Results

JEE Main 2022 June and July Session Result - NTA has announced JEE Main result on their website. To download the Scorecard for JEE Main 2022 June and July Session, visit the official website of JEE Main NTA.
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Rank List
Counselling
Cutoff
JEE Main 2022 state rank lists will be released by the state counselling committees for admissions to the 85% state quota and to all seats in NITs and CFTIs colleges. JEE Main 2022 state rank lists are based on the marks obtained in entrance exams. Candidates can check the JEE Main 2022 state rank list on the official website or on our site.

## JEE Top Colleges

View all JEE Main 2022 Top Colleges
Want to know which Engineering colleges in India accept the JEE Main 2022 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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## Counselling

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## FAQs on JEE Important Chapter - Work, Energy and Power

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.

## JEE Main Upcoming Dates

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