Work, Energy and Power Page 1

Chapter No. 9

Work, Power and Energy

1) Work:

Work is the product of force and distance and the distance should be in the direction of the

force i.e. work is the product of force and displacement.

Consider a force ‘P’ is acting on the body which moves a distance of ‘S’ in the direction of the

force as shown in Figure (a) below.

Figure (a)

∴ Work Done = Force x Distance

= P x s

But if the force acting on the body and the distance moved are not in the same direction as shown

in Figure (b) below, then the work done on the body is given by,

∴Work done = (Component of force in the direction of Motion) x (Distance)

= P cos θ x s

Figure (b)

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Unit = W = F x s

W = Nm or J (joule)

Work done by a force of 1 N on a body which moves a distance of 1 m, is called as 1 Nm. This is

also known as one Joule (J). Hence one joule is the work done by the force of 1 N on a body

whose displacement is 1 m.

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Questions:

1) Find the work done in drawing the body:

a) Weighing 1000 N through a distance 10 m along the horizontal surface by a

horizontal force of 400 N.

b) Weighing 1000 N through a distance 10 m along the horizontal surface by a

horizontal force of 400 N whose line of action makes an angle of 300 with the

horizontal.

2) A body of weight 2000 N moves on a level horizontal rough road for a distance of 200 m.

The resistance of the road is 10 N per 1000 N weight of the body. Find the work done by

the resistance on the body.

3) A block of wood of weight 1200 N is placed on a smooth inclined plane which makes an

angle of 300 with the horizontal. Find the work done in pulling the block up the length of

8 m.

4) If in Q.3) the inclined plane is rough and μ = 0.3, then find the work done in pulling up

for the length of 8 m.

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Power:

The rate of doing work is known as power

Power = Work done per second

= Force x Distance

Time

Work, Energy and Power Page 3

= Force x Distance

Time

∴ Power = Force x Velocity

Note: (force and velocity should be in the same direction)

Unit: Power = Force x Distance

Time

= N m

sec or Watt (W)

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Questions:

1) A train of weight 4000 N is pulled by an engine on a level track at a constant speed of 54

kmph. The resistance due to friction is 5 N per kN weight of the train. Find the power of

the engine.

2) If in the above question, the train is to move with an acceleration of 0.5 m/s2 on a level

track at a constant speed of 54 kmph. Then find the power of the engine?

3) A train of weight 2000 kN is ascending a slope 1 in 100 with an uniform velocity of 36

kmph. Find the power exerted by the engine. If the road resistance is 8 kN per kN weight

of the train.

Work, Energy and Power Page 4

4) A train of weight 2000 kN moves down a slope of 1 in 50 at 18 kmph and engine

develops a power of 35 kN. If the train is pulled up at the same speed, what power will be

required to pull the train?

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Energy:

The capacity to do work is known as energy. It is the product of power and time

The energy which appears in various forms gets converted from one form into the other.

During this conversion work is done by it.

Unit: Joule (J)

Forms of energy:

i) Mechanical Energy

ii) Thermal (heat) energy

iii) Chemical energy

iv) Light energy

v) Sound energy

vi) Nuclear energy

vii) Electrical energy

(In applied mechanics we are dealing with Mechanical Energy)

Mechanical Energy:

A) Potential Energy

B) Kinetic Energy

A) Potential Energy

(Also known as position energy or datum energy)

Definition: Potential Energy is the energy possessed by the body due to its position with respect

to certain reference level.

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Potential Energy due to Gravity:

If a body is raised to a height above a certain reference level, then its P.E. with respect to that

level is given by

P.E. = m x g x h ---- (joule)

Where,

m = mass of body in kg

g = acceleration due to gravity = 9.81 m/s2

h = height above reference level in m

B) Kinetic Energy:

The energy possessed by the body due to its motion or velocity

a) K.E. for linear Motion:

K.E. = 𝟏

𝟐 mv2 ---- (joule)

Where,

m = mass of body in kg

v = velocity in m/s

b) K.E. for circular or Rotational Motion:

K.E. = 𝟏

𝟐 Iω2 ---- (joule)

Proof: Consider a body of mass ‘m’ starting from rest. Let it is subjected to an accelerating force

‘F’ and after covering a distance‘s’, its velocity becomes ‘v’

(Initial velocity) u = 0

(Work done) W = F x s

But F = m x a

W = m x (a x s)

We have,

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V2 = u2 + 2as

V2 = 2as

∴ a s = v2

2

∴ W = mv2

2

Work done = K.E. = 1

2 mv2

Work energy Principle (Work energy Equation):

Statement:

It states that, when a particle moves under the action of a force, the work done by the

force is equal to the change in kinetic energy of the particle i.e.

Work Done = Change in K.E.

Proof:

We know that,

F = m x a

F = m x dv

dt

F = m x dv

ds x

ds

dt

F = m x dv

ds x v

F ds = m v dv

Integrating on both sides,

F ∫ dss2

s1 = m ∫ v dv

v2

v1

F (s2 – s1) = m [v2

2]

v1

v2

F (s2 – s1) = 1

2 mv2 -

1

2 mu2

∴ Work Done = (Final K.E.) - (Initial K.E.)

= Change in K.E.

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Problems:

1) A car travelling at a speed of 75 kmph. Suddenly applies breaks and halts after skidding

50 m. Determine:

a) Time required stopping the car.

b) The coefficient of friction between tyres and roads.

Dec. (2005) (12 MKS)

2) A car moving on a straight level road is skidding for total distance of 60 m after breaks

were applied. Determine the speed of the car just before breaks were if co-efficient of

friction μ = 0.4. (May 2002 8 MKS)

3) A body weighing 300 N is pushed up a 300 plane by a 400 N force acting parallel to the

plane. If the initial velocity of the body is 1.5 m/s and co-efficient of kinetic

friction μ = 0.2. What velocity will the body have after moving 6m?

4) A car is traveling with a speed of 50 kmph is brought to rest in 5 sec by applying

breaking force. Find the magnitude of breaking force If the mass of car is 8 tones. Use the

work energy principle.

5) An object weighing 5 N falls through a height of 40 m and burries itself 1.5 m deep in the

sand. Find the average resistance of the penetration using work energy principle.

6) A stone of mass 50gm falls from a height of 35 m and burries itself 0.22 m deep into the

sand. Find the average resistance of sand and time of penetration. (May 2005 12 MKS)

Note:

A) When a body of some weight is moving along horizontal plane then, its weight

will not contribute into the force.

B) When a body of same weight is falling freely then, its weight will contribute into

the force.

7) A block weighing 2500 N rests on a level horizontal plane for which μ = 0.2. This block

is pulled by a force of 1000 N acting at an angle of 300 to the horizontal. Find the velocity

of the block after it moves 30 m starting from rest. If the force of 1000 N is removed,

how much further will it moves? Use work energy principle method.

8) A small block starts from rest at a point A and slides down the inclined plane as shown in

Figure below. What distance along the horizontal plane will it travel before coming to the

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rest? μ = 0.3. Assume that the initial velocity with which it starts to move along BC is of

the same magnitude as that gained in sliding from A to B.

9) A bullet of mass 81 gm and moving with a velocity of 300 m/s is fired in to a block of

woo and it penetrates to a depth of 10 cm. If the bullet moving with the same velocity

were fired into a similar piece of wood 5 cm thick, with what velocity would it emerge?

Find also the force of resistance, assuming it to be uniform.

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Law of conservation of energy:

Statement:

The energy can neither be created nor destroyed though it can be transformed from one form to

another form. The total energy possessed by a body remains constant provided no energy is

added to or taken from it.

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Proof:

Following Figure shows a body resting on the top of the tower.

Let,

W = weight of the body

m = mass of the body = W

g

h = height of tower.

Now, for the top most position of the body,

Potential energy of the body with respect to ground = W h

Kinetic energy of the body = K.E. = 1

2 mv2 = 0 (since v = 0)

∴ Total energy at the top of the tower = P.E. + K.E. = W h ------------- (1)

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Now,

The body falls down by a height ‘h1’ to position 1 as shown in Figure above. Let ‘v1’ be

the velocity at position 1

Using,

V2 –u2 = 2gh

V12 – 0 = 2gh1

∴ v1 = √2gh1

Kinetic energy of the body at position 1 = K.E. = 1

2 mv2 =

1

2 mv1

2 = 1

2 m2gh1

And the P.E. of the body at position 1 = W (h-h1) ---- (with respect to ground)

∴ Total energy at the top of the tower = P.E. + K.E. = W (h-h1) + 1

2 m2gh1

= W h – Wh1 + mgh1

= W h – Wh1 + Wh1

= W h ----------- (2)

And it is the same energy, the body possessing at the top of the tower.

Now at the ground level,

P.E. = 0

And K.E. = 1

2 mv2

V2 –u2 = 2gh

V = √2gh ----------------- (with respect to ground)

K.E. = 1

2 m (2gh)

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= m g h

= W h

∴ Total energy at the ground level = W h

Which is same, the body possessing at top of the tower and position 1 ----------- (proved)

Impact:

Impact means the collision of two bodies which occurs in a very small interval of time

and during which the two bodies exert very large force on each other. It is defined as the

collision between two bodies which takes place for a very short interval of time and

during which each body exerts a relatively large force on each other.

Figure (a)

Momentum:

It is the product of mass and velocity of the body in motion.

a) For linear motion, linear momentum = mV

b) For circular motion, angular momentum = I ω

Impulse:

It is the product of the force and the time during which it is applied.

∴ Impulse = F x t

For a variable force, Impulse of force = ∫ F dt

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Impulse is a vector quantity and it acts for a very small time.

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Types of impact:

Line of impact:

It is the common normal to the surface in contact during the impact.

Figure (b)

a) Central Impact:

The impact in which the mass centres or (centres of mass)of the two bodies lie on the line

of impact as shown in Figure (b) above.

b) Eccentric Impact:

The impact in which the mass centres of the two bodies does not lie on the line of impact

as shown in Figure (c) below.

c) Direct Impact:

If the velocities of the two bodies are directed along the line of impact, then the impact is

known as direct impact as shown in Figure (b) above.

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Figure (c)

d) Oblique impact:

If the velocities of either or both bodies are not directed along the line of impact, then the

impact is known as oblique impact as shown in Figure (d) below.

Figure (d)

1) Direct central Impact:

If the velocities u1 and u2 of two bodies are directed along the line of impact and if the

centres c1 and c2 of those bodies lie on the line of impact as shown in Figure (b)

above. The impact is known as the direct central impact.

2) Oblique central Impact:

If the velocities u1 and u2 of two bodies are not directed along the line of impact and

if the centres c1 and c2 of those bodies lie on the line of impact as shown in Figure (d)

above. The impact is known as the oblique central impact.

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Restitution (e): (for direct central impact)

The two bodies try to regain their original size and shape during a short interval of

time after deformation. It is known as restitution.

Co-efficient of restitution:

It is defined as the ratio of impulse during restitution to the impulse during

deformation.

∴ e = 𝐢𝐦𝐩𝐮𝐥𝐬𝐞 𝐝𝐮𝐫𝐢𝐧𝐠 𝐫𝐞𝐬𝐭𝐢𝐭𝐮𝐭𝐢𝐨𝐧

𝐢𝐦𝐩𝐮𝐥𝐬𝐞 𝐝𝐮𝐫𝐢𝐧𝐠 𝐝𝐞𝐟𝐨𝐫𝐦𝐚𝐭𝐢𝐨𝐧

or

e = 𝐯𝐞𝐥𝐨𝐜𝐢𝐭𝐲 𝐨𝐟 𝐬𝐞𝐩𝐚𝐫𝐚𝐭𝐢𝐨𝐧

𝐯𝐞𝐥𝐨𝐜𝐢𝐭𝐲 𝐨𝐟 𝐚𝐩𝐩𝐫𝐨𝐚𝐜𝐡

∴ e = 𝐫𝐞𝐥𝐚𝐭𝐢𝐯𝐞 𝐯𝐞𝐥𝐨𝐜𝐢𝐭𝐲 𝐚𝐟𝐭𝐞𝐫 𝐢𝐦𝐩𝐚𝐜𝐭

𝐫𝐞𝐥𝐚𝐭𝐢𝐯𝐞 𝐯𝐞𝐥𝐨𝐜𝐢𝐭𝐲 𝐛𝐞𝐟𝐨𝐫𝐞 𝐢𝐦𝐩𝐚𝐜𝐭

∴ e = v2 − v1

u1 − u2

Where,

u1 and u2 = initial velocities of m1 and m2

v1 and v2 = final velocities of m1 and m2

When e = 0 plastic impact

e = 1 perfectly elastic impact

0 < e < 1 partially elastic impact

Classification of Impacts:

1) Perfectly elastic impact:

In this impact, momentum is conserved

i.e. Momentum before impact = Momentum after impact

∴ m1 u1 + m2 u2 = m1 v1 + m2 v2

Kinetic energy is also conserved in this impact

1

2 m1u1

2 + 1

2 m2u2

2 = 1

2 m1v1

2 + 1

2 m2v2

2

So there is no loss of kinetic energy in this impact

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Velocities of bodies get interchanged i.e. u1 = v2 and u2 = v1

Co-efficient of restitution = e = 1

2) Partially elastic impact (semi – plastic impact)

Only momentum is conserved in this impact

∴ m1 u1 + m2 u2 = m1 v1 + m2 v2

In this impact kinetic energy is not conserved. So there is loss of kinetic energy in this

impact.

Loss in K.E. = K.E. before impact – K.E. after impact

Co-efficient of restitution = 0 < e < 1

After this impact the distance between two bodies goes on increasing

3) Partially plastic impact:

If the two bodies move together with common velocity after impact then it is said to be

plastic impact.

Only momentum is conserved in this impact

∴ m1 u1 + m2 u2 = m1 v1 + m2 v2

But v1 = v2 = v = common velocity after impact

m1 u1 + m2 u2 = (m1 + m2) v

K.E. is not conserved in this impact

Loss in K.E. = K.E. before impact – K.E. after impact

Co-efficient of restitution = e = 0

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Newton’s Law of collision:

It states that,” The relative velocity of separation, of two the moving bodies collide with

each other, bears a constant ratio of their velocity of approach”. The constant of proportionality

is known as co-efficient of restitution.

Law of Conservation of Momentum:

It states that. “If the resultant of the external forces acting on the system is zero, the

momentum of the system remains constant”.

This means that the total momentum of the system before collision is equal to the total

momentum of the system after collision.