JEE Important Chapter - 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.

JEE Main Physics Chapter-wise Solutions 2022-23 

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

List of Important Work Energy and Power Formulae

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.

Important Related Links for JEE Main 2022-23