# What is Energy?

Work and energy are units that are closely associated with one another. In Physics, for 2 objects, the work done is outlined because of the transfer of energy from the primary object to the second object. Also, energy is outlined because of the capability to try to work. Work is believed to be done by a force once an object experiences displacement parallels to the road of action of the force. It’s an activity that has force and movement within the direction of the force. The potential for doing work is what the energy is!

**Work Done**

Work done on an entity is defined as the product of the magnitude of the force acting on the body and the displacement in the direction of the force.

W = F.sHere,

W = work done on an entity

F = Force on the entity

s = Displacement of the entity

**Sign Conventions for Work Done:**

- Positive work is done when both the force and the displacement are in a similar direction.

W = F × s

- The work done is negative when a force acts in a direction opposite to the direction of displacement.

W= − F × s

The angle between force and displacement is 180°.

- Work done will be following if force and displacement are inclined at an angle less than 180°:

W= Fs cosθ

- If force and displacement act at an angle of 90° then work done is zero. The work done is 0, if a force acting on a body causes no displacement. For example, pushing a wall.

**Two circumstances need to be fulfilled with the work to be done:**

- A force should act on the entity
- The entity must be displaced

Work = Force x Displacement

Unit of work done = Joule = Newton x meter. 1 Joule work is said to be done when 1 Newton force is applied on an entity, and it shows the dislocation by 1 meter.

### Energy

The capability to perform work is defined as Energy. Its unit is similar to that of work. SI unit of work or energy = Joule (Nm). Different forms of energy: Light, heat, chemical, electrical or mechanical.

Mechanical energy is the sum of:

**Kinetic energy (K.E)****Potential energy (P.E)**

### Kinetic Energy

Objects in motion have their own energy and are able to do work. This energy is called Kinetic Energy. The K.E. of an object will increase with its speed.

The K.E. of the body moving with a precise rate = work done on that to form it acquire that rate.

To accelerate an object we’ve to use force. To use force, we’d like to try to work. Once work is finished on an object, energy is transferred and also the object moves with a replacement constant speed. The energy that’s transferred is thought of as K.E. Kinetic energy is transferred between objects and maybe remodeled into different varieties of energy. Yo-Yo may be a nice example to explain the transformation of K.E. Whereas getting down to play with it, one starts by holding it rests within the hand, at now, all the energy holds on within the ball in the form of Potential Energy. Once the person drops the plaything the hold on energy is remodeled into K.E., that is that the energy of movement. Once the ball reaches the terribly bottom of the plaything all the energy is regenerated to K.E. Because it moves back to the hand, all the energy is yet again regenerate to P.E. once it reaches the hand.

Units of Kinetic Energy:

- The SI unit of K.E. is Joule which is equal to 1 kg.m
^{2}.s^{-2}. - The CGS unit of K.E. is erg.

**Derivation of the formula:**

Work done, W = F × s …(i)

Due to force the velocity changes to v, and the acceleration produced is a relation between v, u, a and **s = v ^{2}−u^{2 }= 2as**

…(ii)

F = ma …(iii)

Substitute (ii) and (iii) in (i) we get

W = F × s

If u = 0 (object starts at rest)

W

Work done = Change in kinetic energy

The body with an elevated velocity has more K.E. when two alike bodies are in motion.

**Work-energy theorem**

The net work completed by a moving body can be calculated by finding the change in KE. It is the work-energy theorem.

⇒ W_{net} = KE_{final} – KE_{initial}

⇒ W_{net}

**Derivation of the Work-Energy Theorem:**

The work ‘W’ done by the net force on a particle is capable the modification within the particle’s mechanical energy (KE).

Let us take into account a case wherever the resultant force ‘F’ is constant in each direction and magnitude and is parallel to the rate of the particle. The particle is moving with constant acceleration on a line. The link between the acceleration and therefore the internal force is given by the equation “F = ma” (Newton’s second law of motion), and therefore the particle’s displacement ‘d’, is often determined from the equation:

Obtaining,

The work of the net force is calculated with the product of its magnitude (F=ma) and therefore the particle’s displacement. Subbing the higher than equations yields:

W = Fd = = KE_{f} – KE_{i} =

Factors affecting kinetic energy:

- Mass
- Velocity
- Momentum

### Potential Energy

When the work is done on a body, energy gets stored in an entity. For example, stretching a rubber string. The energy that is possessed by a body under its configuration or change in position is known as Potential Energy.

**Types of Potential Energy:**

Gravitational:Energy potential that comes from an object’s height and weight.

Chemical:Energy potential comes from the atoms it contains and the chemical reactions that take place within the entity.

Elastic:Energy potential an object being compressed and stretched.

**At a height the potential energy of an object:**

Work is done against gravity to alter its position when an entity is raised to a certain height. This energy is stored as Potential Energy.

⇒W = Fs

⇒F = ma

In the case of rising the elevation, F = mg

Therefore, W (P.E) = mgh

⇒ ΔPE= mg(h_{final }− h_{initial})

### Law of Conservation of Energy

Energy can neither be created nor destroyed but can be transferred from one form to another. It is the law of conservation of energy. The whole energy before and after the alteration remains constant.

Total energy = KE + PE

**Proof of Law of Conservation of Energy:**

Let from point A, which is at a height h from the ground a body of mass m falls as shown in the following figure:

At point A,

Kinetic energy E_{k} = 0

Potential energy E_{p} = mgh

Total energy, E_{A} = E_{p} + E_{k}

⟹ E_{A} = mgh + 0

⟹ E_{A} = mgh

During the fall, after moving a distance x from A, the body has reached B.

At point B,

Let the velocity at this point be v.

We know, v^{2} = u^{2} + 2as

⟹ v^{2}=0+2×a×x=2×a×x [As, velocity at A, u = 0]

Also, Kinetic energy, E_{k}

⟹ E_{k} =

⟹ E_{k} = mgx

Potential energy, E_{p} = mg(h – x)

So, total energy, E_{B} = E_{p} + E_{k}

⟹ E_{B }= m×g×(h−x)+m×g×x

⟹ E_{B} = mgh – mgx + mgx

⟹ E_{B} = mgh

In the end, the body reaches position C on ground.

At point C,

Potential energy, E_{p} = 0

The velocity of the body is zero here.

So, v^{2 }= u^{2} + 2as

⟹ v^{2} = 0 + 2gh = 2gh

Kinetic energy, E_{k} =

⟹ E_{k} =

Total energy at C

E_{C} = E_{p} + E_{k}

E_{C} = 0 + mgh

E_{C} = mgh

Hence, energy at all points remains same.

### Power

Power is the rate of doing work or the rate of transfer of energy. It is denoted by P

⇒ P = Wt

SI unit is Watt (Js^{−1}).

Average power = Total energy consumed/Total time taken

The commercial unit of power is kWh i.e. energy used in 1 hour at 1000 Joules/second.

1kWh=3.6×10^{6}J

### Sample Questions

**Question 1. How much work is required to stop the car in 30 s if the kinetic energy of car is 6000 J?**

**Solution:**

Work done to stop car = change in Kinetic energy = KE at stop – KE at start= 0 – 6000

=-6000 J (negative sign means work is done against car)

Power = Work done /time == 200 W

**Question 2. Two passengers of mass 40kg each sit in the car then find new kinetic energy of car if mass of car is 700 kg and velocity of car is 18 km/h.**

**Solution:**

K =

K = = 8750J

When two passengers of mass 50kg sit in the car, Total mass becomes =700 + 2 × 40=780 kg

**Question 3. Calculate the work done Find the work done by 10N force acting on object at the angle of 60° which displace the object by 10 m.**

**Solution:**

W= Fs cosθ

) J

= 50J

**Question 4. 245 × 10 ^{2} J of energy were used to raise 50kg boy above the ground, how high would he be raised? And also find the P.E of 20 kg stone which is kept 8m above the ground.**

**Solution:**

The energy posses by a stone of mass 20kg at a height of 8m = PE

PE = mgh

PE = mgh

245 × 10

^{2 }= 50×9.8×hh = 50m

**Question 5. Find the power required to lift 600 kg of water to store in a tank at a height of 1600cm in 10s.**

**Solution:**

Weight of the water = 600 x 10 = 6000 N

Height = 1600 cm = 16 m

T = 10 sec

Power = (workdone)\(time}

=

=

= 9600 Watt

= 9.6 KW

**Question 6. Find the total expenditure if one unit costs 2.75 rupees if four bulbs of 120W and five fans of 130 W each operate for 14 hours daily.**

**Solution:**

Four Bulb energy consumed E

_{b}=0.12×4×14=6.7 KWH5 fans energy consumed

E

_{f}= 0.13×5×14 = 9.1KWHTotal Units consumed =6.7 + 9.1 = 15.8 KWH= 5.8 units

Now if one unit costs Rs 2.75

Total expenditure per day = 2.75 × 15.8 = Rs 43.45