Please refer to the Work Energy and Power Revision Notes given below. These revision notes have been designed as per the latest NCERT, CBSE and KVS books issued for the current academic year. Students will be able to understand the entire chapter in your class 11th Physics book. We have provided chapter wise Notes for Class 11 Physics as per the latest examination pattern.

**Revision Notes Chapter 6 Work Energy and Power**

Students of Class 11 Physics will be able to revise the entire chapter and also learn all important concepts based on the topic wise notes given below. Our best teachers for Grade 11 have prepared these to help you get better marks in upcoming examinations. These revision notes cover all important topics given in this chapter.

**PHYSICAL DEFINITION**When the point of application of force moves in the direction of the applied force under its effect then work is said to be done.

**MATHEMATICAL DEFINITION OF WORK**

Work is defined as the product of force and displacement in the direction of force

W = F x s

If force and displacement are not parallel to each other rather they are inclined at an angle, then in the evaluation of work component of force (F) in the direction of displacement (s) will be considered.

W = (Fcosθ) x s

or, W = FsCosθ

**VECTOR DEFINITION OF WORK**

Force and displacement both are vector quantities but their product, work is a scalar quantity, hence work must be scalar product or dot product of force and displacement vector.

**WORK DONE BY VARIABLE FORCE**

**Force varying with displacement**

In this condition we consider the force to be constant for any elementary displacement and work done in that elementary displacement is evaluated. Total work is obtained by integrating the elementary work from initial to final limits.

**Force varying with time**In this condition we consider the force to be constant for any elementary displacement and work done in that elementary displacement is evaluated.

**WORK DONE BY VARIABLE FORCE FROM GRAPH**Let force be the function of displacement & its graph be as shown.

find work done from s1 to s2 we consider two points M & N very close on the graph such that magnitude of force (F) is almost same at both the points. If elementary displacement from M to N is ds, then elementary work done from M to N is.

dW = F.ds

dW = (length x breadth)of strip MNds

dW = Area of strip MNds

Thus work done in any part of the graph is equal to area under that part. Hence total work done from s1 to s_{2} will be given by the area enclosed under the graph from s1 to s_{2}.

W = Area (ABS2S1A)

DIFFERENT CASES OF WORK DONE BY CONSTANT FORCE

Case i) Force and displacement are in same direction

θ = 0

Since, W = Fs Cos θ

Therefore W = Fs Cos 0

or, W = Fs

Ex – Coolie pushing a load horizontally

Case ii) Force and displacement are mutually perpendicular to each other

θ = 90

Since, W = Fs Cos θ

Therefore W = Fs Cos 90

or, W = 0

Ex – coolie carrying a load on his head & moving horizontally with constant velocity.

Then he applies force vertically to balance weight of body & its displacement is horizontal.

(3) Force & displacement are in opposite direction

θ = 180

Since, W = Fs Cos θ

Therefore W = Fs Cos 180

or, W = – Fs

Ex – Coolie carrying a load on his head & moving vertically down with constant

velocity. Then he applies force in vertically upward direction to balance the weight of

body & its displacement is in vertically downward direction.

**ENERGY**Capacity of doing work by a body is known as energy.

**Note –** Energy possessed by the body by virtue of any cause is equal to the total work done by the

body when the cause responsible for energy becomes completely extinct.

**TYPES OF ENERGIES**There are many types of energies like mechanical energy, electrical, magnetic, nuclear, solar, chemical etc.

**MECHANICAL ENERGY**Energy possessed by the body by virtue of which it performs some mechanical work is known as mechanical energy.

It is of basically two types-

(i) Kinetic energy

(ii) Potential energy

**KINETIC ENERGY**Energy possessed by body due to virtue of its motion is known as the kinetic energy of the body. Kinetic energy possessed by moving body is equal to total work done by the body just before coming out to rest.

Consider a body of man (m) moving with velocity (vo).After travelling through distance (s) it comes to rest.

**KINETIC ENERGY IN TERMS OF MOMENTUM**K.E. of body moving with velocity v is

**POTENTIAL ENERGY**Energy possessed by the body by virtue of its position or state is known as potential energy. Example:- gravitational potential energy, elastic potential energy, electrostatic potential energy etc.

**GRAVITATIONAL POTENTIAL ENERGY**Energy possessed by a body by virtue of its height above surface of earth is known as gravitational potential energy. It is equal to the work done by the body situated at some height in returning back slowly to the surface of earth. Consider a body of mass m situated at height h above the surface of earth. Force applied by the body in vertically downward direction is \

F = mg

Displacement of the body in coming back slowly to the surface of earth is

s = h

Hence work done by the body is

W = FsCosθ

or, W = FsCos0

or, W = mgh

This work was stored in the body in the form of gravitational potential energy due to its position. Therefore

**G.P.E = mgh**

**ELASTIC POTENTIAL ENERGY **Energy possessed by the spring by virtue of compression or expansion against elastic force in the spring is known as elastic potential energy.

**Spring**

It is a coiled structure made up of elastic material & is capable of applying restoring force & restoring torque when disturbed from its original state. When force (F) is applied at one end of the string, parallel to its length, keeping the other end fixed, then the spring expands (or contracts) & develops a restoring force (FR) which balances the applied force in equilibrium.

On increasing applied force spring further expands in order to increase restoring force for balancing the applied force. Thus restoring force developed within the spring is directed proportional to the extension produced in the spring.

F_{R} ∝ x

or, F_{R} = kx (k is known as spring constant or force constant)

If x = 1, F_{R} = k

Hence force constant of string may be defined as the restoring force developed within spring when its length is changed by unity.

But in equilibrium, restoring force balances applied force.

F = F_{R} = k x

If x = 1, F = 1

Hence force constant of string may also be defined as the force required to change its length by unity in equilibrium.

**Mathematical Expression for Elastic Potential Energy**

Consider a spring of natural length ‘L’ & spring constant ‘k’ its length is increased by xo. Elastic potential energy of stretched spring will be equal to total work done by the spring in regaining its original length.

If in the process of regaining its natural length, at any instant extension in the spring was x then force applied by spring is

F = kx

If spring normalizes its length by elementary distance dx opposite to x under this force then work done by spring is

dW = F. (-dx) . Cos0

(force applied by spring F and displacement –dx taken opposite to extension x are in same direction)

dW = -kxdx

Total work done by the spring in regaining its original length is obtained in integrating

dW from x0 to 0

This work was stored in the body in the form of elastic potential energy.** E.P.E = 1/2 kx _{o}^{2}**

**WORK ENERGY THEOREM**It states that total work done on the body is equal to the change in kinetic energy.(Provided body is confined to move horizontally and no dissipating forces are operating).

Consider a body of man m moving with initial velocity v_{1}. After travelling through

displacement s its final velocity becomes v_{2} under the effect of force F.** **u = v_{1}** **v = v_{2}** **s = s

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

**PRINCIPLE OF CONSERVATION OF ENERGY**

It states that energy can neither be creased neither be destroyed. It can only be converted from one form to another. Consider a body of man m situated at height h & moving with velocity vo. Its energy will be.** **E_{1} = P 1 + K_{1}

or,** ** E_{1} = mgh + ½ mv_{o}^{2}

If the body falls under gravity through distance y, then it acquires velocity v1 and its height becomes (h-y)** **u = v_{o}** **s = y** **a = g** **v = v_{1}

From ** **v

^{2}= u

^{2}+2as

**v**

_{1}

^{2}= v

_{o}

^{2}+ 2gy

Energy of body in second situation

**E**

_{2}= P

_{2}+ K

_{2}

or,

**E**

_{2}= mg (h-y) + ½ mv

^{2}

or,

**E**

_{2}= mg (h-y) + ½ m (v

_{o}

^{2}+ 2gy)

or,

**E**

_{2}= mgh – mgy + ½ mv

_{o}

^{2}+ mgy

or,

**E**

_{2}= mgh + ½ mv

_{o}

^{2}

Now we consider the situation when body reaches ground with velocity v2

**u = vo**

**s = h**

**a = g**

**v = v**

_{2}

From

**v**

^{2}= u

^{2}+2as

**2**

^{2}= v

_{o}

^{2}+ 2gh

Energy of body in third situation

**E**

_{3}= P

_{3}+ K

_{3}

or,

**E**

_{3}= mg0 + ½ mv

_{2}

^{2}

or,

**E**

_{3}= 0 + ½ m (v

_{o}

^{2}+ 2gh)

or,

**E**

_{3}= ½ mv

_{o}

^{2}+ mgh

From above it must be clear that E

_{1}= E

_{2}= E

_{3}. This proves the law of conservation of energy.

**CONSERVATIVE FORCE**

Forces are said to be conservative in nature if work done against the forces gets conversed in the body in form of potential energy. Example:- gravitational forces, elastic forces & all the central forces.

**PROPERTIES OF CONSERVATIVE FORCES**

1. Work done against these forces is conserved & gets stored in the body in the form of P.E.

2. Work done against these forces is never dissipated by being converted into nonusable forms of energy like heat, light, sound etc.

3. Work done against conservative forces is a state function & not path function i.e.

Work done against it, depends only upon initial & final states of body & is independent of the path through which process has been carried out.

4. Work done against conservative forces is zero in a complete cycle.

**TO PROVE WORK DONE AGAINST CONSERVATIVE FORCES IS A STATE FUNCTION**

Consider a body of man m which is required to be lifted up to height h. This can be done in 2 ways.

(i) By directly lifting the body against gravity

(ii) By pushing the body up a smooth inclined plane.

Min force required to lift the body of mass m vertically is** **F = mg

And displacement of body in lifting is** **s = h

Hence work done in lifting is

W_{1} = FsCos0^{o} (since force and displacement are in same direction) mg** **W_{1 }= mgh

Now we consider the same body lifted through height h by pushing it up a smooth inclined plane

Min force required to push the body is

F = mgSinθ

And displacement of body in lifting is s = h/Sinθ

Hence work done in pushing is** **W_{2} = FsCos0

or, ** **W_{2} = mgSinθ . h/Sinθ . 1a

or, ** **W_{2} = mgh

From above W_{1} = W_{2} we can say that in both the cases work done in lifting the body

through height ‘h’ is same.

**To Prove That Work Done Against Conservative Forces Is Zero In A Complete Cycle**

Consider a body of man m which is lifted slowly through height h & then allowed to come back to the ground slowly through height h.

For work done is slowly lifting the body up,

Minimum force required in vertically upward direction is** **W_{2}F = mg

Vertical up displacement of the body is

s = hHence work done is** **W_{2}W = FsCosθ

or, W_{1} = FsCos0 (since force and displacement are in same direction)

or, W_{1} = mgh (since force and displacement are in same direction)

For work done is slowly bringing the body down, Minimum force required in vertically upward direction is** **W_{2}F = mg

Vertical down displacement of the body is** **W_{2}s = h

Hence work done is

or, ** **W_{2}W_{2} = FsCos180

(since force and displacement are in opposite direction)

or, ** **W_{2} = – mgh

Hence total work done against conservative forces in a complete cycle is

** **W_{2}W = W_{1} + W_{2}

or, ** **W_{2}W = (mgh) + (-mgh)

or, ** **W_{2}W = 0

**NON-CONSERVATIVE FORCES**Non conservative forces are the forces, work done against which does not get conserved in the body in the form of potential energy.

**PROPERTIES OF NON-CONSERVATIVE FORCES**1. Work done against these forces does not get conserved in the body in the form of P.E.

2. Work done against these forces is always dissipated by being converted into non usable forms of energy like heat, light, sound etc.

3. Work done against non-conservative force is a path function and not a state function.

4. Work done against non-conservative force in a complete cycle is not zero.

**PROVE THAT WORK DONE AGAINST NON–CONSERVATIVE FORCES IS A PATH FUNCTION**

Consider a body of mass (m) which is required to be lifted to height ‘h’ by pushing it up the rough incline of inclination.

Minimum force required to slide the body up the rough inclined plane having coefficient of kinetic friction μ with the body is** **F = mgSinθ + fk

or, ** **F = mgSinθ + μN

or, ** **F = mgSinθ + μmgCosθ

Displacement of the body over the incline in moving through height h is** **s = h/Sinθ

Hence work done in moving the body up the incline is

W = F.s.Cos0(since force and displacement are in opposite direction)

or, ** **W = (mgSinθ + μmgCosθ). h/Sinθ. 1

or, ** **W = mgh + μmgh/Tanθ

Similarly if we change the angle of inclination from θ to θ1, then work done will be** **W1 = mgh + μmgh/Tanθ1

This clearly shows that work done in both the cases is different & hence work done against non-conservative force in a path function and not a state function i.e. it not only depends upon initial & final states of body but also depends upon the path through which process has been carried out.

**To Prove That Work Done Against Non-conservative Forces In A Complete Cycle Is Not Zero**

Consider a body displaced slowly on a rough horizontal plane through displacement s from A to B.

Minimum force required to move the body is** **F = fk = μN = μmg

Work done by the body in displacement s is

W = F.s.Cos0(since force and displacement are in same direction)** **or, W = μmgs

Now if the same body is returned back from B to A

Minimum force required to move the body is** **F = fk = μN = μmg

Work done by the body in displacement s is

W = F.s.Cos0(since force and displacement are in same direction)

or, ** **W = μmgs

Hence total work done in the complete process** **W = W_{1} + W_{2} = 2μmgs**Note –** When body is returned from B to A friction reverse its direction.

**POWER**

Rate of doing work by a body with respect to time is known as power.

**Average Power**

It is defined as the ratio of total work done by the body to total time taken.

**Instantaneous Power**Power developed within the body at any particular instant of time is known as instantaneous power.

**Or**

Average power evaluated for very short duration of time is known as instantaneous power.

**EFFICIENCY**It is defined as the ratio of power output to power input.

**Or**

It is defined as the ratio of energy output to energy input.

**Or**

I It is defined as the ratio of work output to work input.

**COLLISION**Collision between the two bodies is defined as mutual interaction of the bodies for a short interval of time due to which the energy and the momentum of the interacting bodies change.

**Types of Collision**There are basically three types of collisionsi)

Elastic Collision – That is the collision between perfectly elastic bodies. In this type of collision, since only conservative forces are operating between the interacting bodies, both kinetic energy and momentum of the system remains constant. ii) Inelastic Collision – That is the collision between perfectly inelastic or plastic bodies. After collision bodies stick together and move with some common velocity. In this type of collision only momentum is conserved. Kinetic energy is not conserved due to the presence of non-conservative forces between the interacting bodies. iii) Partially Elastic or Partially Inelastic Collision – That is the collision between the partially elastic bodies. In this type of collision bodies do separate from each other after collision but due to the involvement of non-conservative inelastic forces kinetic energy of the system is not conserved and only momentum is conserved.

**Collision In One Dimension – Analytical Treatment**

Consider two bodies of masses m_{1} and m_{2} with their center of masses moving along the same straight line in same direction with initial velocities u_{1 }and u_{2} with m_{1} after m_{2}. Condition necessary for the collision is u_{1} > u_{2} due to which bodies start approaching towards each other with the velocity of approach u_{1} – u_{2}. Collision starts as soon as the bodies come in contact. Due to its greater velocity and inertia m_{1} continues to push m_{2} in the forward direction whereas m_{2} due to its small velocity and inertia pushes m_{1}in the backward direction. Due to this pushing force involved between the two colliding bodies they get deformed at the point of contact and a part of their kinetic energy gets consumed in the deformation of the bodies. Also m_{1} being pushed opposite to the direction of the motion goes on decreasing its velocity and m_{2} being pushed in the direction of motion continues increasing its velocity. This process continues until both the bodies acquire the same common velocity v. Up to this stage there is maximum deformation in the bodies maximum part of their kinetic energy gets consumed in their deformation.

In case of elastic collision bodies are perfectly elastic. Hence after their maximum deformation they have tendency to regain their original shapes, due to which they start pushing each other. Since m2 is being pushed in the direction of motion its velocity goes on increasing and m1 being pushed opposite to the direction of motion its velocity goes on decreasing. Thus condition necessary for separation i.e. v_{2}>v_{1} is

attained and the bodies get separated with velocity of separation v_{2} – v_{1}.

In such collision the part of kinetic energy of the bodies which has been consumed in the deformation of the bodies is again returned back to the system when the bodies regain their original shapes. Hence in such collision energy conservation can also be applied along with the momentum conservation.

**Applying energy conservation**

** **m_{1}(u_{1 }– v_{1})(u_{1} + v_{1}) = m_{2}(v_{2} – u_{2})(v_{2} + u_{2}) ………(i)**Applying momentum conservation **p

_{i}= p

_{f}

**m**

_{1}u

_{1}+ m

_{2}u

_{2}= m

_{1}v

_{1}+ m

_{2}v

_{2}

**m**

_{1}(u

_{1 }– v

_{1}) = m

_{2}(v

_{2}– u

_{2}) ……….(ii)

Dividing equation (i) by (ii)

**u**

_{1}+ v

_{1}= v

_{2}+ u

_{2}

or,

**v**

_{2}– v

_{1}= u

_{1}– u

_{2}

or, Velocity of separation = Velocity of approach

or,

**v**

_{2}= v

_{1}+ u

_{1}– u

_{2}

Putting this in equation (i)

Hence in perfectly elastic collision between two bodies of same mass, the velocities interchange.ie. If a moving body elastically collides with a similar body at rest. Then the moving body comes at rest and the body at rest starts moving with the velocity of the moving body.

**Case 2-** If a huge body elastically collides with the small body,** **m_{1} >> m_{2}

m_{2} will be neglected in comparison to m_{1}

Hence if a huge body elastically collides with a small body then there is almost no change in the velocity of the huge body but if the small body is initially at rest it gets thrown away with twice the velocity of the huge moving body.eg. collision of truck with a drum.

**Case 3-** If a small body elastically collides with a huge body,

m_{2} >> m_{1}

m_{1} will be neglected in comparison to m_{2}

Hence if a small body elastically collides with a huge body at rest then there is almost no change in the velocity of the huge body but if the huge body is initially at rest small body rebounds back with the same speed.eg. collision of a ball with a wall.

**Inelastic collision**

In case of inelastic collision bodies are perfectly inelastic. Hence after their maximum deformation they have no tendency to regain their original shapes, due to which they continue moving with the same common velocity.

In such collision the part of kinetic energy of the bodies which has been consumed in the deformation of the bodies is permanently consumed in the deformation of the bodies against non-conservative inelastic forces. Hence in such collision energy conservation can-not be applied and only momentum conservation is applied.

Applying momentum conservation** **p_{i} = p_{f}** **m_{1}u_{1} + m_{2}u_{2} = m_{1}v + m_{2}v

or, ** **m_{1}u_{1} + m_{2}u_{2} = (m_{1}+m_{2})v

or, ** **v = m

_{1}u

_{1}+ m

_{2}u

_{2}/(m

_{1 + }u

_{1})

**Partially Elastic or Partially Inelastic Collision**

In this case bodies are partially elastic. Hence after their maximum deformation they have tendency to regain their original shapes but not as much as perfectly elastic bodies. Hence they do separate but their velocity of separation is not as much as in the case of perfectly elastic bodies i.e. their velocity of separation is less than the velocity of approach.

In such collision the part of kinetic energy of the bodies which has been consumed in the deformation of the bodies is only slightly returned back to the system. Hence in such collision energy conservation can-not be applied and only momentum conservation is applied.** **(v_{2} – v_{1}) < (u_{1} – u_{2})

**Collision In Two Dimension – Oblique Collision**

When the centers of mass of two bodies are not along the same straight line, the collision is said to be oblique. In such condition after collision bodies are deflected at some angle with the initial direction. In this type of collision momentum conservation is applied separately along x-axis and y-axis. If the collision is perfectly elastic energy conservation is also applied.

Let initial velocities of the masses m_{1} and m_{2} be u_{1} and u_{2 }respectively along x-axis. After collision they are deflected at angles θ and Ø respectively from x-axis, on its either side of the x axis.

Applying momentum conservation along x-axis** **p_{f} = p_{i}

m_{1}v_{1} Cosθ + m_{2}v_{2} Cos Ø = m_{1}u_{1} + m_{2}u_{2}

Applying momentum conservation along y-axis** **p_{f} = p_{i}** **m_{1}v_{1} Sinθ – m_{2}v_{2} Sin Ø = m10 + m_{2}0

or, ** **m_{1}v_{1} Sinθ – m_{2}v_{2} Sin Ø = 0

or, ** **m_{1}v_{1} Sinθ = m_{2}v_{2} Sin Ø

In case of elastic collision applying energy conservation can also be applied K_{f} = K_{i}

Case-1 For perfectly elastic collision, velocity of separation is equal to velocity of approach, therefore** **e = 1

Case-2 For perfectly inelastic collision, velocity of separation is zero, therefore** **e = 0

Case-3 For partially elastic or partially inelastic collision, velocity of separation is less than velocity of approach, therefore** **e < 1