Physics Class 12 Work, Energy, and Power - Constant and Variable Force, Kinetic and Potential Energy

Samia Miraj

Updated On: October 25, 2022 10:00 am IST

The science students must prepare for the Kinetic and Potential Energy for the Physics exam as soon as possible. Given below, we are sharing the details of Kinetic and Potential Energy, Kinetic energy, Potential energy, and much more given below.

Work, Energy, and Power

Work has a very simple concept because it basically means the movement which is done to the object when a force is applied to it. This basically results in the displacement of the object either by a pushing force or a pulling force. However, your capacity to do a particular work will be called your energy. Power can be referred to as the work that is done for a unit of time so for example the amount of work that you are able to perform in 1 minute is called your power. Check out more specifications on the topic of work, energy, and power given below:

Work, Energy, and Power- Definition of Work

Work done can be identified as the product of your force in the direction of the displacement of the object. You will have to calculate work by multiplying the force and the distance in the direction of the force. The SI unit of work is Joule (J).

Work, Energy, and Power- Definition of Energy 

Your capacity to do any work or displacement of any object is known as energy. Energy equals the position and the height of a particular object. The SI unit of energy is also Joule (J).

Work, Energy, and Power- Definition of Power 

Power is the rate at which the work is done by a particular person. The formula for identifying power is W/t. The SI unit of power is Watt (W).

Work Done By A Constant Force and A Variable Force

There are two types of forces in nature however the more often force is the variable force. 

Constant Force

The work done by a constant force of magnitude F, as we know, that displaces an object by Δx can be given as:

W = F.Δx

Variable Force

In the case of a variable force, work is calculated with the help of integration. For example, in the case of a spring, the force acting upon any object attached to a horizontal spring can be given as:

Fs = -kx

Where,

k is the spring constant

x is the displacement of the object attached

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Work, Energy, and Power - Kinetic Energy

Kinetic energy comes into play when an object is in motion, basically, kinetic energy is the energy that is gained by the result of the motion of the object. The motion can be rotation, vibration, or any other type of motion. The SI unit of kinetic energy is Joules which is equal to kg-m2s-2.

Formula For Kinetic Energy

The formula for kinetic energy is-

Work Energy Theorem 

Work energy theorem means the net work done on an object equals the object's change in kinetic energy. The equation for the formula is-

W net = K−K0 =ΔK

Where W net is net work, K is the final kinetic energy, K0 is the initial kinetic energy and the result is the change in kinetic energy.

Work, Energy, and Power - Notion of Potential Energy

Potential energy basically refers to the energy which is stored in a particular object. There are different types of potential energies and the SI unit of potential energy is the joule (J).

Formula of Potential Energy

The formula for potential energy is-

W = m×g×h = mgh

Where,

m is the mass in kilograms

g is the acceleration due to gravity

h is the height in meters

Types of Potential Energy

There are two types of main potential energies mentioned below:

  • Gravitational potential energy is defined as the energy which is in an object when the object is raised to a certain height against gravity.
  • Elastic potential energy is an object when it is compressed or stretched. The main examples of elastic potential energy come from rubber bands, trampolines, and bungee cords.

Potential Energy of a Spring 

When energy is stored in a compressible or stretchable object then it is called elastic potential energy. One of the very basic examples of elastic potential energy is Spring because the movement will occur in the spring when we alter the usual position. 

Formula of Potential Energy of a Spring

String potential energy = force × distance of displacement.

Also, the force is equal to the spring constant × displacement.

The Potential energy of a spring is:

P.E=1/2k×x2

P.E= The potential energy of a spring

k= Spring constant

x= Spring displacement

Work, Energy, and Power - Conservative Forces 

A conservative force is a force done in moving a particle from one point to another, such that the force is independent of the path taken by the particle. It depends only on the initial and final position of the particle. There are two types of conservative forces such as gravitational force and elastic spring force. The following are the properties of a conservative force-

  • When the force depends only on the initial and final position irrespective of the path taken.
  • In any closed path, the work done by a conservative force is zero.
  • The work done by a conservative is reversible.

Work, Energy, and Power - Non-Conservative Forces 

The Non-Conservative force refers to the force which is occurring when the work done depends upon the path that is taken. Friction is one of the best examples of this type of force. If the force results in the change of mechanical energy then it is a particular example of a non-conservative force. The following are the properties of a non-conservative force-

  • It is path-dependent therefore it also depends on the initial and final position.
  • In any closed path, the total work done by a non-conservative force is not zero.
  • The work done by a non-conservative is irreversible.

Work, Energy, and Power - Motion in a Vertical Circle 

Circular motion is the motion in which the distance of the body remains constant from a fixed plane and it can be classified into non-uniform and uniform motion. 

Vertical Circular Motion Derivation

Vertical circular motion derivation is used to find out the tension at the high and the low points as well as the velocity of the object in motion. It is an expression of the difference in tensions at the highest and lowest points for a particle performing the vertical circular motion.

Suppose a body of mass ‘m’ performs vertical circular motion on a circle of radius r.

Let, 

TL = Tension at the lowest point

TH = Tension at the highest point

VL = Velocity at the lowest point

VH = Velocity at the highest point

At the lowest point L, 

TL = mv L2/r + mg —(1)

At the highest point H,

TH = mvH2/r – mg —-(2)

Subtracting (1) by (2)

TL – TH = mvL2/r + mg – ( mv2H/r – mg)

= m/r ( vL2 – vH2) + 2mg

Therefore, TL – TH = m/r ( vL2 – vH2) + 2mg -–(3)

By the law of conservation of energy,

(P.E + K.E) at L = (P.E + K.E) at H 

0 + 1/2mvL2 = mg.2r + ½ mvH2

½ m(vL2 – vH2) = mg.2r

vL2 – vH2 = 4gr —(4)

From equations (3) and (4),

TL – TH = m/r(4gr) + 2mg = 4mg + 2mg

TL – TH = 6mg

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Elastic And Inelastic Collisions in One and Two Dimensions

A collision can be defined as a transfer of momentum from one object to another and there are mainly two types of collisions such as elastic and inelastic collisions. 

Elastic Collision

If the kinetic energy of the system is constant then the collision will be called an elastic collision.

Elastic Collision in One Dimension 

Assume that two particles with masses m1 and m2 collide with velocities u1 and u2, and their velocities after the collision are v1 and v2, respectively. The collision of these two particles is a head-on elastic collision. As per the law of conservation of momentum equation,

m1u1 + m2u2 = m1v1 + m2v2 (1)

We also have the law of kinetic energy conservation for elastic collisions

12m1u21+12m2u22=12m1v21+12m2v22 (2)

We get, 

m1u21+m2u22=m1v21+m2v22(Factoringout12) (3)

On rearranging the equation, we get, 

m1u21−m1v21=m2v22−m2u22 (4) 

Now, 

m1(u21−v21)=m2(v22−u22) (5) 

If we elaborate the above equation, it will now become, 

m1(u1 + v1 ) (u1 − v1 ) = m2 (v2 + u2 ) (v2 − u2 ) (6) 

Applying the conservation of momentum equation, 

m1u1 + m2u2 = m1v1+ m2v2

Regroup the equation with the same masses, 

m1u1 − m1v1 = m2v2 − m2u2 (7) 

Therefore, m1 (u1 − v1 ) = m2(v2 − u2 ) (8)

On dividing equation (6) by (8), we get, 

m1(u1 + v1 ) (u1 − v1 ) =m2(v2+u2)(v2−u2)m1(u1−v1) = m2(v2−u2) (9)

u1 + v1 = v2 + u2 (10)

Now, 

v1= v2 + u2 − u1 (11)

Using the values of v1 from equation (11) in the law of conservation of momentum equation, we get, 

v2 = [2 m1u1 + u2(m2 − m1 ) ] / (m1+ m2 ) (12)

Now, use the value of v2 in equation (11) v1= v2 + u2 − u1

v1=[2m1u1+u2(m2−m1)](m1+m2)+u2−u1 (13)

v1=[2m1u1+u2(m2−m1)+u2(m1+m2−u1(m1+m2)](m1+m2) (14)

As a result, we get, 

v1=[2m2u2+u1(m1−m2)](m1+m2) (15)

When the masses of both bodies are equal, their velocities are generally exchanged after the collision.

v1= u2 and v2= u1

This means that if two objects of the same mass collide, if the second mass is at rest and the first mass collides with it, the first mass comes to rest, and the second mass moves at the same speed as the first mass.

In this case, v1= 0 and v2= u1

And, If m1 < m2 

Then, v1= − u1 and v2= 0

This means that the lighter body will use its own velocity to bombard the heavier mass, while the heavier mass will remain static.

But if, m1 > m2 

Then, v1= u1 and v2= 2u1

Elastic Collision in Two Dimension

Assume that m1 and m2 are two mass particles in a laboratory frame of reference and that m1 collides with m2 which is initially at rest. Let the velocity of mass m1 before the collision be u1, and after the collision, it moves with a velocity v1 and is deflected by an angle θ1 with its incident direction, while m2 moves with a velocity v2 and is deflected by an angle θ2 with its incident direction.

According to the law of conservation of linear momentum, for components along the x-axis

m1u1=m1v1cosθ1 +m2v2cosθ2 ---(1)

For components along the y-axis,

0=m1v1sinθ1 − m2v2sinθ2 --(2)

As per the law of conservation of Kinetic energy, 

12m1u21+12m2u22=12m1v21+12m2v22 ---(3)

Analyzing the above equations reveals that finding values for four unknown quantities v1, v2, θ1, and θ2 using the above three equations is not possible. As a result, it cannot predict the variable because there are four of them. However, if we measure any one variable, we can uniquely determine the other variable using the above equation.

Inelastic Collision 

If the kinetic energy gets converted into heat or deformation of shape or any other type of reaction then the type of collision will be called an inelastic collision.

Inelastic Collision in One Dimension

Consider the masses m1 and m2 of two particles. Let u1 and u2 represent the respective velocities before the collision. Assume that both particles stick together after colliding and move at the same velocity v. As per the law of linear momentum conservation,

m1u1 + m2u2=(m1+ m2) v or V=m1u1+m2u2m1+m2 (1)

On considering the second particle to be stationary or at rest, u2=0 then

m1u1 = v(m1+m2)

or

V=m1u1m1+m2

Hence, |v| < |u1|.

The Kinetic energy before the collision is as follows:

KE1= 12

m1u21

The Kinetic Energy of the system after the collision is as follows:

KE2= 12

(m1+m2) v2

KE2=12(m1+m2)(m1u1m1+m2)2=12m21u21(m1+m2)

Hence,

K.E.2K.E.1

=12m21u21(m1+m2)12m1u21

 = m1m1+m2

< 1 (4) 

As a result, K2 < K1; hence, energy loss would be there after the particles collide.

The Fraction of Kinetic Energy Loss can be Calculated by Using the Following Formula:

K.E1−K.E2K.E1=[1−(m1m1+m2)]K.E1K.E1=m2m1+m2

Inelastic Collision in Two Dimension

In a laboratory frame of reference, let m1 and m2 be the two mass particles, and m1 collides with m2, which is initially at rest. Let the velocity of mass m1 before the collision be u1, and after the collision, it moves with a velocity v1 and is deflected by an angle θ1 with its incident direction, while m2 moves with a velocity v2 and is deflected by an angle θ2 with its incident direction. According to the law of conservation of linear momentum, for components along the x-axis,

m1u1=m1v1cosθ1 +m2v2cosθ2 ---(1)

And, for components along the y-axis,

0=m1v1sinθ1 − m2v2sinθ2 --(2)

Analyzing the above equations reveals that we must find the values of four unknown quantities v1, v2, θ1, and θ2 with the help of the above three equations, which is impossible because we need at least four equations to find the values of four unknown quantities. As a result, we cannot predict the variables because there are four of them. However, if we measure any two variables, then the other variable can be uniquely determined using the above equation.
Also Read: 

It is important to complete the unit of work, energy, and power because it is one of the most anticipated topics included in physics. Check out the specifications related to the topics from the notes mentioned above:

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