Work, Energy and Power

definition

Work

Work is defined as the force in the direction of displacement times displacement. It is a scalar quantity having S.I unit Joule.

W = F s

where

F

is the component of force in the direction of displacement.

definition

Work done by a constant force

When the force is constant, the work done is defined as the product of the force and distance moved in the direction of force.
Example: Suppose a body is kept on the frictionless surface and a force of constant magnitude of

10 N

is acting on it, due to the action of forces the body will completed a distance of

5

meter, then the work done will be given as

W = F s = 10 \times 5 = 50 J o u l e

definition

Work Done by Gravity

Gravitational force is the force acting on a body due to gravity.
Example: For a freely falling particle, the particle moves in the direction of gravity. Hence work done by gravity,

W = m g h

Note: For a particle falling under gravity, work done by gravity is dependent only on the difference in height and independent of any vertical displacement.

definition

Energy

Energy is defined as the capacity of a system to perform work. Suppose a body having mass

m

kg moving with a linear velocity of

v

meter/sec so its energy is is in the form of kinetic energy which is equal to

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

SI unit of energy is Joule.
CGS unit of energy is erg.

definition

Forms of Energy

Some forms of energy are as follows:

Kinetic energy(due to motion).
Potential energy(due to position).
Mechanical Energy(sum of kinetic and potential energy).
Heat energy(due to temperature).
Electrical energy(due to electrical current).
Chemical energy(stored in a material).

definition

Condition for Positive, Negative & Zero work done by a force

We know that:

W = \vec{F} . \vec{d} = F d c o s θ

Sign of work done depends on the angle between force and displacement vector.
If

θ = 0^{0}

- Work done is positive.
If

θ = 90^{0}

- Work done is zero.
If

θ = 180^{0}

- Work done is negative.

example

Work Done

Example: A force

\vec{F} = 6 \hat{i} - 8 \hat{j} N

, acts on a particle and
displaces it over

4

m along the positive X-axis and then displaces it over

6

m along the positive Y-axis. What is the total work done over the two displacements?

Solution:

W . D = \int \vec{F} . \vec{d s}

= (6 \hat{i} - 8 \hat{j}) . (4 \hat{i}) + (6 \hat{i} - 8 \hat{j}) . (6 \hat{j})

(6 \times 4) + (- 8 \times 6) = 24 - 48 = - 24 J

example

Work Done by a Variable Force

Example: A bicycle chain of length

1.6

m and of mass

1

kg is lying on a horizontal floor. What is the work done in lifting it with one end touching the floor and the other end

1.6

m above the floor? (Take

g = 10 m s^{- 2}

)
Solution:
Consider chain to be concentrated on its center of mass.
Therefore, change in potential energy = work done by external force.
Work done

= m g \frac{h}{2} = 8

example

Work Done by ideal springs

Example: A spring obeying the linear law

F = - K x

is first compressed by

10

cm and the work done is

W_{1}

. Next it is compressed by
another

10

cm, the work done now is

W_{2}

, then what is the value of

W_{1} : W_{2}

?
Solution:

W_{1} = Δ (P . E)

= \frac{1}{2} K (x {_{2}}^{2} - x {_{1}}^{2})

= \frac{1}{2} K (10^{2} - 0^{2})

Similarly,

W_{2} = \frac{1}{2} K (20^{2} - 10^{2})

So,

W_{1} : W_{2} = 100 : 300

= 1 : 3

example

Work done by frictional force

Example: A

5.0 k g

box rests on a horizontal surface. The coefficient of
kinetic friction between the box and the surface is

0.5

. A horizontal force pulls the box at a constant velocity for

10 c m

. What will be the work done by the applied horizontal force and the frictional force ? (take

g = 10 m / s^{2}

)

Work done by friction force

= \int \vec{F} . d \vec{s}

= - (μ_{k} N) \times 10^{- 1}

= - 0.5 \times 5 \times 10 \times 10^{- 1}

= - 2.5 J

Using work energy theorem
Work done by force + Work done by friction

= Δ K . E

Work done by force

- 2.5

= 0

\Rightarrow

Work Done by force

= 2.5 J

example

Work Done as area under force Vs displacement graph

Example: A force acts on a body and displaces it in it's direction. The graph
shows the relation between the force and displacement. The work done by the force is:
Work done by force =

\int F . d s

=

Area under F.s graph

= A r e a o f △

= \frac{1}{2} \times b a s e \times h e i g h t

= \frac{1}{2} \times (14 - 2) \times 60

= 360 J

definition

Kinetic Energy and Momentum

K . E . = p^{2} / 2 m

K . E . : Kinetic energy, p : momentum, m : mass

definition

Calculation of Kinetic Energy

Example: Calculate the kinetic energy and potential energy of the ball half way up, when a ball of mass

0.1 k g

is thrown vertically upwards with an initial speed of

20 m s^{- 1}

.
Solution:
Total energy at the time of projection

= \frac{1}{2} m v^{2} = \frac{1}{2} \times 0.1 (20)^{2} = 20 J

Half way up, P.E. becomes half the P.E. at the top i.e.

P . E . = \frac{20}{2} = 10 J

∴ K . E . = 20 - 10 = 10 J

example

Kinetic Energy of a System of Particles

The mass of a simple pendulum bob is

100 g

.The length of the pendulum is

1 m

. The bob is drawn aside from the equilibrium position so that the string makes an angle of

60^{0}

with the vertical and let go. The kinetic energy of the bob while crossing its equilibrium position will be:

K . E_{A} = Δ

P . E .

b/w A & B

= m g (L - L \cos 60^{0})

= 10^{- 1} \times 9.8 \times (1 - \frac{1}{2})

= 0.49 J

example

Application of work-energy theorem in problems involving gravity

Example:
A particle of mass

100

g is thrown vertically upwards with a speed of

5

m/s. Find the work done by the force of gravity during the time the particle goes up.
Solution:
Change in kinetic energy is

Δ K E = 0 - \frac{1}{2} \times 0.1 \times 25 = - 1.25 J

Using work-energy theorem, work done is given by,

W = - 1.25 J

example

Application of work-energy theorem in problems involving springs

Example:
A body of mass

2

kg is connected to a spring. Under influence of a constant external force, maximum elongation in the spring is

0.5

m. Now the external force is removed, find the maximum speed of the mass. Given spring constant =

1 N / m

.
Solution:
Spring force on the body is given by,

F = - k x

Work done by spring force is given by,

W = \int F d x = \frac{1}{2} k (x_{i}^{2} - x_{f}^{2})

W = \frac{1}{2} \times 1 ({0.5}^{2} - x_{f}^{2})

By work-energy theorem, this equals the change in kinetic energy.

Δ K E = W

K E_{f} - 0 = \frac{1}{2} ({0.5}^{2} - x_{f}^{2})

For maximum speed,

K E_{f}

is maximum, i.e.

x_{f} = 0

\frac{1}{2} m v^{2} = 0.125

v = 0.353 m / s

example

Friction and Spring Force acting on a body

Question: A

2

kg body is dropped from height of

1

m onto a spring of spring constant

800

N/m as shown in the figure. A frictional force equivalent to

0.4

kgwt acts on the body. What will be the speed of the body just before striking the spring? (Take

g = 10 m / s^{2}

)

Solution: We know that the change of kinetic energy is equal to work done by the system.
Change of Kinetic energy

= \frac{1}{2} m v^{2} - 0 = \frac{1}{2} m v^{2}

Work done

= (m g - f) h

, where

f = 0.4 k g w t = 0. 4 \times 10 = 4 N

, friction force.
Now,

\frac{1}{2} m v^{2} = (m g - f) h

\frac{1}{2} 2 v^{2} = (20 - 4) 1

v^{2} = 16

v = 4 m / s

example

Use of Work Energy Theorem in Body Under Frictional Force

Example: A

2

kg block slides on a horizontal floor with a speed of

4

m/s. It strikes a uncompressed spring, and compresses it till the block is motionless. The kinetic friction force is

15

N and spring constant is

10, 000

N/m. What is the spring compression?

Solution: The total kinetic energy possessed by the block goes into the potential energy of the spring and the work done against friction.Let

x

be the compression of the spring.Thus

\frac{1}{2} m v^{2} = f_{r} x + \frac{1}{2} k x^{2}

⟹ 5000 x^{2} + 15 x - 16 = 0

⟹ x = 0.055 m = 5.5 c m

example

Use of Work Energy Theorem in Pulley Mass System

Question. A light inextensible string that goes over a smooth fixed pulley as shown in the figure connects two blocks of masses

0.36 k g

and

0.72 k g

. Taking

g = 10

m/s

^{2}

, find the work done (in joules) by the string on the block of mass

0.36

kg during the first second after the system is released from rest.

Solution:

2 m g - T = (2 m) a

T - m g = m a

\Rightarrow a = g / 3

T = \frac{4 m g}{3}

w = T \times s = T \times \frac{1}{2} a t^{2} = \frac{4 m g}{3} \times \frac{1}{2} \times \frac{g}{3} \times 1^{2} = \frac{200}{9} \times 0.36 = 8 J o u l e s

example

Application of work-energy theorem for a body under constant external force

Example:
A

500

gm ball moving at

15

m/s slows down uniformly until it stops. If the ball travels

15

m, what was the average net force applied while it was coming to a stop?
Solution:
Initial kinetic energy,

K E_{1} = \frac{1}{2} m u^{2} = \frac{1}{2} \times 0.5 \times 15^{2} = 56.25 J

Final kinetic energy,

K E_{2} = \frac{1}{2} m v^{2} = 0 J

Change in kinetic energy,

Δ K E = - 56.25 J

By work energy theorem, work done equals change in kinetic energy.

W = Δ K E = - 56.25 J

For a constant force, work done is

W = F . s

F \times 15 = - 56.25

F = - 3.75 N

definition

Work done on a body by a variable external force

According to work-energy theorem,

W = Δ K E

This is very useful in finding work done by variable forces when the initial and final velocity of a body is given.
Example:
Under the action of a force, a

2

kg body moves such that its position

x

as a function of time is given by

x = \frac{t^{3}}{3}

, where

x

is in metre and

t

in second. Find the work done by the force in the first two seconds.Solution:

v = \frac{d x}{d t} = t^{2}

W = \frac{1}{2} m v^{2} = \frac{1}{2} m t^{4}

= \frac{1}{2} \times 2 \times (2)^{4} = 16 J

example

Work Energy theorem with multiple external forces

Question: A body is moving up on inclined plane of angle

θ

with an initial kinetic energy

E

. The coefficient of friction between the plane and the body is

μ

. What is the work done against friction before the body comes to rest?

Work energy theorem

\frac{1}{2} m u^{2} - \frac{1}{2} m v^{2} = W =

S

a = - (g \sin θ + μ_{k} g \cos θ)

v^{2} - u^{2} = 2 a s

s = \frac{E}{m g (s i n θ + μ_{k} c o s θ)}

f = μ_{k} m g c o s θ

=

F . S

= \frac{μ E \cos θ}{μ \cos θ + \sin θ}

example

Change in energy of finite length objects

Question: A uniform chain is held on a frictionless table with half of its length hanging over the edge. If the chain has a mass

m

and length

L

,how much work is required to pull the hanging part back onto the table?
(Given acceleration due to gravity

= g

)

Solution:
Work is required to pull against the gravity.Let

ρ

be linear density

ρ = \frac{m}{L}

. Let x=0 at the bottom most point of the hanging chainThe displacement of each differential element of mass dm at a distance x from the bottom =

\frac{L}{2} - x

Work done W=

\int_{0}^{L / 2} F d x = \int_{0}^{L / 2} - d m \times g \times (\frac{L}{2} - x) \int_{0}^{L / 2} - ρ g (\frac{L}{2} - x) d x = {- ρ g (\frac{L x}{2} - \frac{x^{2}}{2})}_{0}^{L / 2} = \frac{- ρ g}{8} = \frac{m g L}{8}

example

Work done by an impulsive force

Example:
A batsman hits a ball of mass m moving with an initial velocity of u. After the impact, the direction of motion of the ball reverses and velocity becomes v. Find the work done by the batsman on the ball.
Solution:
Change in kinetic energy ,

Δ K = \frac{1}{2} m v^{2} - \frac{1}{2} m u^{2}

By work-energy theorem, work done by batsman is,

W = Δ K = \frac{1}{2} m v^{2} - \frac{1}{2} m u^{2}

example

Application of work-energy theorem in non-inertial frame of reference

Example:
A body of mass

1 k g

is in rest inside a vehicle on a friction less surface. When the vehicle starts moving with an acceleration of

1 m / s^{2}

, the body moves a distance of

1 m

in the backward direction. Find the speed of the body at this point.
Solution:
Force on the body in the vehicle frame of reference,

F = m a = 1 N

Work done by this force,

W = F s = 1 J

This equals change in kinetic energy by work-energy theorem.

K E_{f} - K E_{i} = W

\frac{1}{2} \times 1 \times v^{2} - 0 = 1

v = \sqrt{2} m / s

definition

Examples of Conservative and Non-Conservative Forces

Force due to gravity is conservative force as work done from taking an object from height

h

to ground is

+ m g h

whereas it's

- m g h

on the other way around. However, friction is an example of non-conservative force. Work done by friction in dragging an object to distance

r

- f r

and in retracing the same path work done would be

- f r

so total work done will be

- 2 f r

definition

Properties of Conservative Forces

In this particular case in order to calculate work done by gravity in the closed path direct formula

m g h

can be applied owing to conservative nature of gravitational force.

definition

Forms of Potential Energy

The different forms of potential energy are:

Gravitational potential energy-The rock hanging above the ground has a form of stored energy called gravitational potential energy.
Elastic potential energy-Elastic potential energy is the energy stored when an object is squeezed or stretched. This stored energy then can cause the rubber band to fly across the room when you let it go.
Chemical potential energy-Chemical potential energy is the energy stored in bonds between the atoms that make up matter.

definition

Potential energy with respect to a reference line

Potential energy is always defined with respect to a reference line in space. Usually infinity is taken as the reference line and potential energy defined is zero at infinity. In gravitation, for objects close to surface of earth, ground surface is taken as reference line and energy at ground surface is zero.
Note:
Choice of reference line is not fixed and can be redefined. This helps in solving of problems.

definition

Calculation of Potential Energy

For an object having mass

m

having its center of gravity at height

h

and

g

be the acceleration due to gravity.

P . E . = m g h

definition

Potential energy of a spring

A spring stores potential energy due to extension. Since an unextended spring does not store potential energy, it is used as the point of zero energy.
For a spring, potential energy is defined as

U = \frac{1}{2} k x^{2}

where

x

is the extension of the spring.

result

Relationship between force and potential energy

Force is equal to the gradient of potential energy.

\vec{F} = - \frac{δ U}{δ x} \hat{i} - \frac{δ U}{δ y} \hat{j} - \frac{δ U}{δ z} \hat{k}

Example:
A particle of mass

1

kg is moving in the

x - y

plane and its potential energy

U

in joule obeys the law

U = 6 x + 8 y

, where (

x, y

) are the coordinates of the particle in meter. If the particle starts from rest at (9,3) at time

t = 0

, then find the force on the body.Solution:

\vec{F} = - \frac{δ U}{δ x} \hat{i} - \frac{δ U}{δ y} \hat{j} - \frac{δ U}{δ z} \hat{k}

\vec{F} = - 6 \hat{i} - 8 \hat{j}

diagram

Plot of potential energy v/s position

F = - \frac{d U}{d x}

Force is defined as negative of the slope of potential energy v/s position plot.
Example:
Consider the given plot defined by the equation:

U = \frac{a X^{2}}{2} - \frac{b X^{3}}{3}

Between A and B, slope is positive, so force is negative.
Between B and C, slope is negative, so force is positive.

F = - d U / d x = - a x + b x^{2}

diagram

Plots of potential energy, kinetic energy and total energy v/s position

Total energy of an isolated system is constant.
For mechanical energy,

E = K + U

where

E, K, U

are the total, kinetic and potential energies.
Variation of energies with position for a spring is shown in the attached plot.
Note:
Kinetic energy is always greater than or equal to zero. This property must be satisfied in energy v/s position plot.

diagram

Stable and unstable equilibrium points from potential energy plot

Stable equilibrium means that, with small deviations of the body from this state, forces or moments of forces emerge which tend to return the body to the state of equilibrium. A convex minima (

\frac{d^{2} U}{d x^{2}} < 0

) in potential energy v/s position plot refers to a point in unstable equilibrium.
Unstable equilibrium means that, with small deviations of the body from this state, forces or moments of forces emerge which tend to return the body away from the state of equilibrium. A concave maxima (

\frac{d^{2} U}{d x^{2}} > 0

) in potential energy v/s position plot refers to a point in unstable equilibrium.

definition

Equilibrium point using potential energy plot

F = - \frac{d U}{d x}

If slope of the potential energy v/s position plot is zero, then

F = 0

.
For example, in the given plot, B corresponds to the point of equilibrium.

definition

Mechanical Energy

Mechanical energy is the energy of an object due to its position and motion. It is equal to the sum of kinetic energy and potential energy. Example: A freely falling body is comprised of mechanical energy.

definition

Conservation of total mechanical energy

Mechanical energy is the sum of the potential and kinetic energies in a system. The principle of the conservation of mechanical energy states that the total mechanical energy in a system (i.e., the sum of the potential plus kinetic energies) remains constant as long as the only forces acting are conservative forces.
Example: Consider a person on a sled sliding down a

100 m

long hill on a

30

^{0}

incline. The mass is

20 k g

, and the person has a velocity of

2 m / s

down the hill when they're at the top. How fast is the person traveling at the bottom of the hill?

Solution:At the top:

P . E = m g h = (20) (9.8) (100 s i n 30

) = 9800 J

K E = 1 / 2 m v^{2} = 1 / 2 \times (20) \times (2)^{2} = 40 J

Total mechanical energy at the top

= 9800 + 40 = 9840 J

At the bottom:

P E = 0, K . E . = 1 / 2 m v^{2}

Total mechanical energy at the bottom =

1 / 2 m v^{2}

If we conserve mechanical energy, then the mechanical energy at the top must equal what we have at the bottom. This gives:

1 / 2 m v^{2} = 9840, v = 31.3 m / s .

definition

Application of conservation of energy in spring

Example:
A mass of

0.5 k g

moving with a speed of

1.5 m / s

on a horizontal smooth surface, collides with a nearly weightless spring of force constant

k = 50 N / m

. Find the maximum compression of the spring.
Solution:

\frac{1}{2} m v^{2} = \frac{1}{2} k {x_{m}}^{2}

∴ x_{m} = \sqrt{\frac{m}{k}} . v

= \sqrt{\frac{0.5}{50}} \times 1.5

= 0.15 m

example

Springs in series

When springs are connected in series, force is same in both the springs. Equivalent spring constant is given by:

\frac{1}{k_{e q}} = \frac{1}{k_{1}} + \frac{1}{k_{2}}

Elongation and energy in the springs are divided in the ratio,

\frac{x_{1}}{x_{2}} = \frac{E_{1}}{E_{2}} = \frac{k_{2}}{k_{1}}

Example:
In the given diagram, the maximum displacement on application of a force is A. Find the maximum speed of the mass attached after the external force is removed.
Total energy before removal of force,

E = U = \frac{1}{2} k_{e q} A^{2}

At zero extension,

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

Using conservation of energy,

v = \sqrt{\frac{k_{e q} A^{2}}{m}}

example

Springs in parallel

When springs are connected in parallel, extension is same in both the springs. Equivalent spring constant is given by:

k_{e q} = k_{1} + k_{2}

Force and energy in the springs are divided in the ratio,

\frac{F_{1}}{F_{2}} = \frac{E_{1}}{E_{2}} = \frac{k_{1}}{k_{2}}

E = U = \frac{1}{2} k_{e q} A^{2}

At zero extension,

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

Using conservation of energy,

v = \sqrt{\frac{k_{e q} A^{2}}{m}}

definition

Change in mechanical energy and work done by external forces

Case 1: Potential energy is defined for the external force (valid only when the applied force is conservative)
By work-energy theorem,

Δ K = W

By conservation of energy,

Δ E = Δ U + Δ K = 0

Δ U = - W

Case 2: Potential energy is not defined for the external force
Here, the force doing work is not part of the system and hence conservation of mechanical energy of the system is not valid. However, total energy of the system and the external force is constant.

Δ E + W = 0

Δ E = - W

The change in mechanical energy of the system may occur as change in kinetic energy or potential energy.

example

Application of conservation of energy for spring in horizontal plane

Example:
A block of mass

2 k g

is on a smooth horizontal surface. A light spring of force constant

800 N / m

has one end rigidly attached to a vertical wall and lying on that horizontal surface. Now the block is moved towards the wall compressing the spring over a distance of

5 c m

and then suddenly released. By the time the spring regains its natural length and breaks contact with the block, the velocity acquired by the block will be?
Solution:Using conservation of energy
Initial

M . E =

Final

M . E

i.e

(P . E)

spring

=

(K . E)

block

\frac{1}{2} k x^{2} = \frac{1}{2} m v^{2}

800 \times 25 \times 10^{- 4} = 2 \times v^{2}

v = 1 m / s

example

Application of conservation of energy for spring in vertical plane

Example:
A toy gun consists of a spring and a rubber dart of mass

25

gm. When the spring is compressed by

4

cm and the dart is fired vertically, it projects the dart to a height of

2

m. If the spring is compressed by

8

cm and the same dart is projected vertically, the dart will rise to a height of?
Solution:

\frac{1}{2} K x^{2} = m g h

hence ,

x^{2} \propto h

hence ,

\frac{h}{2} = \frac{8^{2}}{4^{2}}

∴ h = 8 m

example

Conservation of energy when one of the forces is non conservative

Example:
A coconut of mass

m

falls from the tree through a vertical distance of

s

and could reach ground with a velocity of

v

^{- 1}

due to air resistance. Find the work done by air resistance.
Solution:
Mechanical energy before falling = Mechanical energy after falling + Energy loss due to air resistanceEnergy loss due to air resistance ,

E_{a} = m g s - \frac{1}{2} m v^{2}

Work done by air resistance is,

W = - E_{a} = - m g s + \frac{1}{2} m v^{2}

definition

Power

Power is defined as the rate of doing work. It also equals the amount of energy consumed in unit time. It is a scalar quantity.

∴ P = \frac{W}{t}

Power can also be found as the product of force and velocity in the direction of force.

∴ P = F v

result

Units of power

Different units of power are:

SI unit: $1 W = 1 J / s = 1 k g m s^{- 3}$

1 h p (horsepower) = 746 W

1 e r g / s = 10^{- 7} J / s

definition

Efficiency

Efficiency of a machine is defined as the ratio of actual power output to the ideal power output with no loss. Its value lies in the range

0 < η < 1

.
Example:
A crane is used to lift 1000 kg of coal from a mine 100 m deep. If the time taken by the crane is 1 hour, then find the power of the crane assuming its efficiency to be 80%.Solution:
Given that

m = 1000 K g

h = 100 m

t = 1 h o u r = 60 \times 60 s = 3600 s

η = 80

%
We have

P = \frac{m g h}{t}

P = \frac{(1000) (10) (100)}{3600}

P = 0.02778 \times 10^{4} = 277.8 W

Now power of crane is

P^{'} = P \times \frac{1}{η}

P^{'} = 277.8 \times \frac{100}{80} = 346.8 W

definition

Mathematical Expression of Power

Power is expressed as:
Power =

\vec{F} . \vec{v}

definition

Average Power

It is the average amount of work done or energy converted per unit of time. The average power is often simply called "power" when the context makes it clear. The instantaneous power is then the limiting value of the average power as the time interval t approaches zero.

Average Power =

\frac{δ W}{δ t}

example

Using power in problems of conservation of energy

Example:
A crane is used to lift

1000 k g

of coal from a mine

100 m

deep. The time taken by the crane is

1

hour. The efficiency of the crane is

80

%. If

g = 10 m s^{- 2}

, then find the power of the crane.Solution:

Power supplied = \frac{m g h}{t}

Power used by crane = \frac{m g h}{t} \times \frac{100}{80}

= \frac{1000 \times 10 \times 100}{3600} \times \frac{100}{80} = \frac{10^{5}}{36 \times 8} W

example

Power in variable mass system

Example:
A pump having efficiency 75% lifts 800 kg water per minute from a 14 m deep well and throws at a speed of 18 ms

^{- 1}

. Find the power of the pump.
Solution:Power used to lift the water,

P = \frac{d m}{d t} g h = \frac{800}{60} (10) (14) = \frac{5600}{3} W

Efficiency

= \frac{Power Used}{Actual Power}

\Rightarrow 0.75 = \frac{5600 / 3}{P_{a}}

\Rightarrow P_{a} = \frac{5600 / 3}{0.75} = 2488.88 W

Monday, July 31, 2017