The Work, Power, and Energy section is a little tricky and takes a lot of time and devotion on your part to understand and master. The questions from this section will test your ability to use equations related to work and power, to calculate the kinetic, potential energy and to use the work-energy relationship to determine the final speed, height etc of an object.

• Candidates should prepare this section properly as most of the questions on mechanics can be solved by using the concepts of Work, Power, and Energy.
• Other than understanding the basics of Work, Potential & Kinetics Energy etc the Work Energy Theorem is the most important topic. Check JEE Main Physics Syllabus

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All the terms like Work, Energy, Force etc have different meanings and explanations in Physics. The definitions are basic and one line but if you forget this one line during the exam then to calculate the answer could become a difficult task for you. Hence we have prepared the quick revision notes that include all the important formulas, previous questions, and some tips and tricks to solve those questions. Candidates can go through the article and can incorporate these notes in their preparation.

Must Read: 

• JEE Main Study Notes on Modern Physics
• JEE Main Study Notes on Laws of Motion

What is Work?

Work done W is explained as the dot product of force F and displacement d.

• Work done by the force is positive in case the angle between force and displacement is acute (0°<θ<90°) as cos θ is positive. This indicates that, when the force and displacement are in the same direction, work done will be positive. This work is said to be done upon the body.
• When the force acts in a direction at the right angle to the direction of displacement (cos 90° = 0), no work is done (zero work).
• Work done by the force is negative in case the angle between force and displacement is obtuse. This means that, when both, force and displacement are in the opposite direction, work done will be negative. This work is said to be done by the bod

Sample Question on Work

Question: Let us consider that the particle is describing a horizontal circle of radius 0.5 m with uniform speed. The centripetal force acting is 10 N. Calculate the work done in describing a semicircle?

1. zero
2. 5J
3. 5πJ
4. 10πJ

Solution: (1)

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What is Work done by a variable force?

The variation of the force and the displacement due to the affluence of force is shown in the graph below:

Work Done by the Variable for is, W = Areas under the curve

W =

• Where dx is the small displacement.
• Units- The unit of work done in represented as S.I is joule (J) and in C.G.S system is erg.
• 1J = 1 N.m , 1 erg = 1 dyn.cm

Sample Question

Question: A bus is stopped by applying a retarding force F in case it is moving at speed v on a level road. The distance covered by it before coming to rest is mentioned as S. If the load of the bus is increased by half because of more passengers, for the same speed and same retarding force, the distance covered by the bus to come to rest shall be

1. 1.5 s 
2.  2 s
3. 1 s 
4. 2.5 s

Solution: (1)

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What is Power and Energy?

Power: The rate at which work is done is known as power and is explained below:

• P = W/t = F.s/v = F.v, where S is the distance and V is the speed.
• Instantaneous power in terms of mechanical energy- P = dE/dt
• Units- The unit of power in the S.I system is J/s (watt) and in C.G.S system is erg/s.

Energy: Energy is the ability of the body to do some work. The unit of energy and work is the same.

• Potential Energy (V) - Potential energy of a body is referred to as V = mgh where m is the overall weight of the body, g is the free fall acceleration and h means the height.

V = mgh

• Kinetic energy - The kinetic energy of a body is the energy possessed by the body by the virtue of its motion. Kinetic Energy (K) is explained as K= ½ mv². Here m means the overall mass of the body while v is the speed of the body.

K= ½ mv² 

What is the relationship between Kinetic Energy (K) and momentum (p)?

K = p²  / 2m  

• In case two bodies with different masses have the same momentum, a body with higher mass shall have lesser kinetic energy.
• In case two bodies of different mass have the same kinetic energy, a body with a greater mass shall have greater momentum.
• For two bodies with the same mass, the body with higher momentum has greater kinetic energy.

Sample Question on Power and Energy

Question: A ball whose kinetic energy is E, is projected at an angle of 45 ° to the horizontal. What will be the kinetic energy when the ball will be at the highest point of its flight?

1. E
2. E/√2
3. E/2
4. zero

Solution: (3)

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Quick Revision Notes on Work Energy Theorem 

The work-energy theorem by a constant force

It states that the change in the kinetic energy of a particle is equal to the work done on it by the net force.

Where Kf is the final kinetic energy and Ki is the initial kinetic energy.

For constant force, W = ½ mv2 – ½ mu2 = Final K.E – Initial K.E

The work-energy theorem by a variable force

Let us assume the variable force works from position xi to xf then according to the work-energy theorem the change in the kinetic energy is equal to the work done on by the variable force.

Sample Question on Work Energy Theorem 

Question: The power supplied to a particle of mass 2 kg varies and the time with which it varies is P = 3t 2 /2 W. If the velocity of the particle at t = 0 is v = 0 then what will be the velocity of the particle at time t = 2 s? Here “t” is in second.

1. 1 ms -1
2. 4 ms -1
3. 2 ms -1
4. 3 ms -1

Solution: (3)

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Potential Energy of a Spring

Let a block is attached with a spring, Initially the spring its natural relaxed condition if we compressed the spring by a constant force F. Then the work done by the force is stored in the spring as the potential energy.

The potential energy of the spring is

Sample Question on Spring Problem

Question: If springs S1 and S2 of the force constants k1 and k2 are stretched by the same force then we can see that work done on spring S1 is more than the work done on spring S2.

Statement 1: If stretched by the same amount work done on S1 will be more than work is done on S2

Statement 2: k1<k2

1. Statement 1 is true, Statement 2 is true and Statement 2 is not the right explanation of Statement 1
2. Statement 1 is false, Statement 2 is true
3. Statement 1 is true, Statement 2 is false
4. Statement 1 is true, Statement 2 is true and Statement 2 is the right explanation of Statement 1

Solution: (2)

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What are the Laws of conservation of energy?

• According to this law of conservation, energy cannot be created nor destroyed. This energy can be converted from one form to another and the sum of total energy will always remain the same.
• The sum of the kinetic and potential energies of an object is being known as mechanical energy.

Represented by K1 + U1 = K2 + U2 = K3 + U3 = Constant

• According to the law of conservation of energy, the total mechanical energy of the system will remain the same.

Represented by mgh + ½ mv2 = constant

• In an isolated system, the total energy total of the system is constant.

Represented by E = U+K = constant or Ui+Ki = Uf+Kf

• Speed of particle v in a central force field- v is equal to √2/m [E-U(x)]

What are conservations of other forms of energy?

• Heat Energy- The total energy of an isolated system remains constant. So, the energy can’t be created and can’t be destroyed.
• Chemical Energy- The energy may change form but total energy in a chemical reaction remains constant.
• Electrical Energy – The total electric charge of a system remains constant. The charge can’t be created or can’t be destroyed.
• Nuclear Energy- In a nuclear reaction total mass of the reactants or starting materials must be equal to the mass of the products. So, the mass can’t be created nor destroyed.

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Conservation of linear momentum

In an isolated system, the total momentum of the system before collision would be equal to the total momentum of the system after the collision. So, pf = pi

Coefficient of restitution (e) - It is explained as the ratio between the magnitude of impulse during a period of restitution to that during a period of deformation.

• e = relative velocity after collision / relative velocity before collision
• = v2 – v1/u1 – u2

Case (i) For perfectly elastic collision, e is equal to 1. Thus, v2 – v1 is equal to u1 – u2. This signifies the relative velocities of two bodies before and after the collision are the same.

Case (ii) For inelastic collision, e<1. Thus, v2 – v1 < u1 – u2. This signifies, the value of e shall depend upon the extent of loss of kinetic energy during the collision.

Case (iii) For perfectly inelastic collision, e = 0. Thus, v2 – v1 =0, or v2 = v1. This signifies the two bodies shall move together with the same velocity. Therefore, there shall be no separation between them.

Sample Question on Momentum 

Question: A 5kg rifle fires a 6g bullet with a Speed of 500m/s. Calculate the ratio of the distance the rifle moves backward whereas the bullet is in the barrel to the distance the bullet moves forward?

1. 1/600
2. 1/666
3. 2/327
4. none of the above

Solution: (2)

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What is an Elastic collision?

In an elastic collision, the momentum and kinetic energy are conserved.

• Post collision, the velocity of both the bodies will be v1 = (m1-m2/ m1+m2)u1 + (2m2/ m1+m2)u2 and v2 = (m2-m1/ m1+m2)u2 + (2m1/ m1+m2)u1

Case: I

• In case both the colliding bodies are of the similar mass, i.e., m1 = m2, in that case, v1 will be equal to u2 and v2 = u1

Case: II

• In case the body B of mass m2 is at rest, means u2 is equal to 0, then, v1 = (m1-m2/ m1+m2)u1 and v2 = (2m1/ m1+m2)u1
• In case m2<
• In case, m2=m1, then, v1 will be zero and v2 will be equal to u1
• In case, m2>>m1, then, v1 will be equal to -u1 and v2 will be very small.

Inelastic collision - In an inelastic collision, only the quantity momentum is conserved but not kinetic energy.

v = (m1u1+m2u2) /(m1+m2) and loss in kinetic energy, E = ½ m1u12+ ½ m2u22 - ½ (m1+ m2)v2

or,

E= ½ (m1u12 + m2u22) – ½ [(m1u1+ m2u2)/( m1+ m2)]2

= m1 m2 (u1-u2)2 / 2( m1 + m2)

Important Points to be Notice:-

• The maximum transfer energy will occur in case both m1 and m2 are equal
• If Ki is the initial kinetic energy and Kf is the final kinetic energy of mass m1, then the fractional down fall in kinetic energy is mentioned as
• Ki – Kf / Ki = 1- v12/u21
• In case both, m2 and nm1 are equal then Ki – Kf / Ki = 4n/(1+n) divided by 2

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Tips to Solve Questions on Work, Power and Energy

1. Question: Q is defined as What is the value of Q in this case?

1. 23361M/8
2. 13361M/8
3. 53361M8
4. none of the above

Tip to Solve the Question: You can reach your answer by using the formula

Solution: (3)

2. Question: At the highest point of a hemisphere of radius R a uniform chain, AB of mass m, and length 1 is placed with one end that is A. Referring to the top of the hemisphere as the datum level what will be the potential energy of the chain?

Tip to Solve the Question: The elemental length is below the datum level. 

Solution: (a)

3. Question: Let us consider that a block is attached to a spring and is also pulled by a constant horizontal force. The block is kept on a smooth surface as shown in the figure. At the initial point, the spring is in the natural length state. Calculate the maximum positive work that the applied force F can do? Also given that string does not break.

Tip to Solve the Question: Apply the work energy theorem on such a question to calculate the answer. 

Solution: (b)

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Tricks to Solve Question on Work, Energy and Power

• Count the number of variables before start solving the questions. You should know all the known and unknown variables. 
• If you have more number of variables and fewer equations then you can start calculating your answer by using Work-Energy Theorem so that you can eliminate the unknown variables.
• Questions from this section are most of the time mixed with other topics so their candidates should be able to identify whether it's a question from work or from some other topic. 
• Pay special attention on spring questions or problems. 

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Some useful Preparation Tips for Students preparing for JEE Main Physics

• The first and most essential thing to be considered before starting the preparation for JEE Examination is to get familiar with the latest syllabus. In case you are not familiar with the latest syllabus, you will not be able to prepare effectively for the same.
• The students are suggested to go through the comprehensive syllabus and focus on the most difficult topics first before moving to the easier topics.
• If you want to develop a deep understanding of the subject and get a good score in JEE Main 2020, going thorough and preparing section-wise notes will help a lot.
• The students must manage the available time in a more productive way as the syllabus of Physics is vast and one needs to study thoroughly.
• Since there are many questions in the Physics section that need to be solved using a pen and paper to solve, so practice more questions in every study session. This way you will be able to strengthen your concepts and broaden your understanding regarding a variety of questions that can be asked in the exam.

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Important Formulas

• Work =
• Work done by Variable force W =
• Power =
• Potential Energy, V = mgh
• Kinetic Energy, K= ½ mv² 
• According to Work Energy Theorem, W = ½ mv² – ½ mu² = Final K.E – Initial K.E
• The potential energy of the spring is