Work, Energy and Power Class 11 Physics Notes

Work, energy, and power are fundamental concepts in physics that explain how objects move and interact. Work is done when a force is applied to an object, causing it to move. Energy is the capacity to do work and comes in two main forms: Kinetic energy (energy of motion) and potential energy (stored energy due to position). Power refers to the rate at which work is done or energy is transferred. 

1.0Work

• Whenever a force acts on a body, displacing its point of application of force in any direction except perpendicular to it, work is said to be done by the force $\vec{F}and\vec{s}.$$W=\vec{F}\cdot \vec{s}=Fs\cos \theta$
• If a constant force $\vec{F}$ is applied on a body and displaces it through displacement $\vec{s}$ then work done by $\vec{F}$mon the body is given by a scalar product of
• SI unit of work: Joule(J)
• CGS Unit:Erg
• Dimensional Formula $M{L}^2{T}^{-2}$
• Work is a scalar quantity, but it can be positive, negative or zero.

2.0Types of Work

Positive work	Negative work	Zero work
$W=\vec{F}\cdot \vec{s}=Fs\cos \theta$ angle is acute <90°	$W=\vec{F}\cdot \vec{s}=Fs\cos \theta$ angle is obtuse >90°	$W=\vec{F}\cdot \vec{s}=Fs\cos \theta$ W will be zero if F = 0 or s = 0=90°
When a body falls freely under gravity (θ = 0°) , the work done by gravity is positive.	Work done by braking force on the car is negative.	In a pendulum's oscillation, the work done by tension is zero (θ = 90°).

Note:

Dependency on the the Frame of Reference

• Work depends on the choice of reference frame. While the force itself is independent of the frame, displacement varies with it. For example, in a lift moving upward, the man applies a force equal to the weight of the suitcase, but in the lift's frame, the suitcase has zero displacement, so no work is done. However, in the ground frame, the suitcase moves with the lift, and the man’s force does non-zero work.

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3.0Work Done By Force

Cartesian Form

• When the magnitude and direction of the force varies with position, work done by the force for infinitesimal displacement $d\vec{s}isdW=\vec{F}\cdot d\vec{s}$ 

The total work done for the displacement from position A to B is

$W_{AB}={\int }_A^B\vec{F}\cdot d\vec{s}=\int (F\cos \theta )ds$

$\vec{F}=F_x\hat{i}+F_y\hat{j}+F_z\hat{k}=dx\hat{i}+dy\hat{j}+dz\hat{k}$

$W_{AB}={\int }_A^B(F_x\hat{i}+F_y\hat{j}+F_z\hat{k})\cdot (dx\hat{i}+dy\hat{j}+dz\hat{k})$

$W_{AB}={\int }_{x_A}^{x_B}{F}_{x}dx+{\int }_{y_A}^{y_B}{F}_{y}dy+{\int }_{z_A}^{z_B}{F}_{z}dz$

4.0Graphical Meaning of Work

$W_{AB}={\int }_A^B\vec{F}\cdot d\vec{s}$

Area of the shaded portion in the above $\vec{F}-\vec{s}$ graph represents work done between A and B.

5.0Work Done by Spring Force

$dW=\vec{F}\cdot d\vec{x}=-kxdx(\theta =180^{\circ })$

$w_s=\int dW={\int }_{x_i}^{x_f}kxdx=\frac{1}{2}k\left(x_i^2-x_f^2\right)$

$x_{i}and{x}_f$ represent the final and initial deformation of spring.

6.0Energy And Mass Energy Relation

• Energy is defined as the internal capacity to do work. When we say that a body has energy, it means that it can do work.
• Mechanical energy, electrical energy, optical (light) energy, acoustical (sound), molecular, atomic and nuclear energy etc., are various forms of energy. These forms of energy can change from one form to the other.

Mass energy relation

• According to Einstein's mass-energy equivalence principle, mass and energy are inter convertible i.e. they can be changed into each other.

$E=m{c}^2$ (m : mass of the particle ,c : speed of light)

• Energy is a scalar quantity.
• SI Unit : Joule $M{L}^2{T}^{-2}$
• Dimensional Formula

7.0Kinetic Energy

• It is the inherent ability of an object to perform work due to its motion.
• The K.E of a body can be calculated by the amount of work done in stopping the moving body or by the amount of work done in imparting the present velocity to the body from the state of rest.

$K.E=\frac{1}{2}m{v}^2$

8.0Relation Between Kinetic Energy and Momentum

$K.E=\frac{1}{2}m{v}^2$

Momentum p=mv

Hence,

$K=\frac{p^2}{2m}$

$p=\sqrt{2mk}$

$\kappa \propto p^2\Rightarrow \frac{K_1}{K_2}=\frac{p_1^2}{p_2^2}$

For small changes $(<5\%)\text{; }\frac{\Delta K}{K}=2\frac{\Delta p}{p}$

9.0Work-Energy Theorem 

It states that the work done by all the forces (internal and external) on a particle equals the change in its kinetic energy.

$W_{all}=\delta K.E$

$W_{\text{all}}=K_f-K_i=\frac{1}{2}m{v}^2-\frac{1}{2}m{u}^2$

10.0Conservative and Non-Conservative Forces

11.0Potential Energy

Definition: It is the energy due to its position or arrangement within a conservative force field.

Relationship between conservative force field and potential energy

$\vec{F}=-\nabla U=-grad(U)=\frac{\partial u}{\partial x}\hat{i}-\frac{\partial u}{\partial x}\hat{j}-\frac{\partial u}{\partial x}\hat{k}$

If force varies with only one dimension (say along x-axis) then,

$F=-\frac{dU}{dx}=-Fdx$

$\Rightarrow {\int }_{u_1}^{u_2}dU=-{\int }_{x_1}^{x_2}Fdx\Rightarrow \Delta U=-W_c$

12.0Gravitational Potential Energy

• Earth's Gravitational Force is a conservative force hence this work must be stored in the body in the form of Gravitational Potential Energy Ug {\deltaU_g}
• From the concept of potential energy

$\delta U=-W_{Conservative}\to \delta U_g=-W_g\to U_f-U_i=-(-mgh)$

$U_f$ - Gravitational Potential Energy at height h

$U_i$ - Gravitational Potential Energy on ground

• Here we can assume U_i= 0 for our convenience as potential energy is a relative quantity.
• Thus, we have following relation $U_f=mgh$.

13.0Types of Equilibrium

Equilibrium- At equilibrium net force is zero. If the net force acting on a body is zero then the body is said to be in equilibrium and positions where the body achieves equilibrium are called equilibrium positions.

At equilibrium, 

$F=-\frac{dU}{dx}$

14.0Law of Conservation of Mechanical Energy

The Law of Conservation of Mechanical Energy states that in an isolated system, where only conservative forces act, the total mechanical energy (sum of kinetic and potential energy) remains constant. Energy can change between kinetic and potential forms, but the total remains unchanged.

E=K+U=Constant

15.0Definitions of Power, Average Power and Instantaneous Power

Power: It is the ratio of work done(W) to the time taken (t) to do that work. Therefore, it is the rate of energy transfer or work done concerning time, and the symbol (P) denotes it.

$P=\frac{W}{t}$

Average Power: If a force performs a certain amount of work within a given time interval, we can compute the average power generated by the force. It is a scalar quantity.

$P_{\text{avg}}=\frac{W}{\Delta t}$

$P_{\text{avg}}=\frac{\Delta E}{\Delta t}$

Instantaneous Power: Instantaneous power refers to the power consumed or generated at a specific time. It is given by the scalar or dot product of force and velocity.

$P_{=}\frac{dW}{dt}$

$\vec{P}=\vec{F}\cdot \vec{V}$

Instantaneous power=slope of work-time curve =tan

• Power is a scalar quantity with dimensions $M^1{L}^2{T}^{-3}$
• SI Unit-J/s or Watt
• 1 horsepower = 746 watts = 550 ft–lb/s

Area under power–time graph gives the work done $W=\int Pdt$

16.0Efficiency

Efficiency of a machine denotes how much a machine is effective in converting energy into useful work.

$\text{Work Done Energy Input}=\frac{\text{Work Done}}{\text{Energy Input}}$

17.0Sample Questions on Work, Energy and Power

Q-1.Find work done by tension force on a 2kg block during a time interval of 3 sec after release.

Solution:

$\text{Tension }T=\frac{2\times 1\times 2}{1+2}g=\frac{40}{3}\text{N}$

$\text{Acceleration of 2 kg block, }a=\frac{2-1}{2+1}g=\frac{g}{3}$

$\text{Displacement of 2 kg block in 3s, }s=\frac{1}{2}a\left(3^2\right)=\frac{1}{2}\times \frac{g}{3}\times 9=15\text{m}$

$W_T=T.S=T\times S\times cos180^{\circ }=-T\times S$

$\text{Work done by tension force }{W}_T=-\frac{40}{3}\times 15=-200\text{J}$

Q-2.A graph shown below is plotted between force F acting on a block, kept on a horizontal floor v/s position (x) of the particle. Find work done on the particle when the block reaches to x = 10 from x = 0.

Solution:

$\text{Work done}=\text{Area Under }F-x\text{ Curve}$

$W=\frac{1}{2}\times 2\times 20+20\times 2+\frac{1}{2}\times 20\times 2-\frac{1}{2}\times 20\times 20-2\times 20=20J$

Q-3.A position dependent force $F=7-2x+3x^2$ acts on a small body of mass 2 kg and displaces it from x = 0 to x = 5 m . Calculate the work done in joules.

Solution:

$W={\int }_{x_1}^{x_2}Fdx={\int }_0^5(7-2x+3x^2)dx={\left[7x-\frac{2x^2}{2}+\frac{3x^3}{3}\right]}_0^5=135J$

Q-4.The potential energy for a conservative force system is given by $U=a{x}^2-bx$ where a and b are constants. Find out the (a) expression for force (b) potential energy at equilibrium.

Solution:

For Conservative Force 

a. $F=-\frac{dU}{dx}=-(2ax-b)=-2ax+b$

b. $U=a(\frac{b}{2a}{)}^2-b(\frac{b}{2a})=\frac{b^2}{4a}-\frac{b^2}{2a}=-\frac{b^2}{4a}$

Q-5. A chain of mass m and length L is held on a frictionless table in such a way that $\frac{1}{n}th$ part is hanging below the edge of the table. Calculate the work done to pull the hanging part of the chain.

Solution:

Required work done = change in potential energy of the chain

Now, let Potential energy (U) = 0 at the table level

So, the potential energies of the chain initially and finally are respectively

$U_i=-mg(\frac{L}{2n})=-(\frac{M}{L})\frac{L}{n}g(\frac{L}{2n})=-\frac{MgL}{2n^2},U_f=0$

Required Work Done

$U_f-U_i=\frac{MgL}{2n^2}$

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