Work and Energy

The weightlifter does no work on the weights as he holds them on his shoulders. He does work only when he raises the weights to this height. 'Work is done on an object when a force causes a displacement of the object'.

1.0Introduction

In common usage, the word 'work' means any physical or mental exertion. But in physics, work has a distinctly different meaning. Let us consider the following situations : (1) A student holds a heavy chair at arm's length for several minutes. (2) A student carries a bucket of water along a horizontal path while walking at constant velocity. (3) A student applying force against a wall. (4) A student studying whole day to prepare for examinations.

It might surprise you to know that in all the above situations, no work is done according to the definition of work in physics, even though effort is required in all cases.

2.0The concept of work

In physics, the word 'work' has a definite and precise meaning. Imagine that your car (see figure) has run out of fuel and you have to push it down the road to the gas station. Let your friend push the car with a constant horizontal force. If the car does not move, no work is done. Suppose, he increases the magnitude of this force by pushing the car harder. If the car starts moving, he does a work on the car.

A person exerts a force on the car and displaces it to the left. The work done is done by him on the car.

Work is not done on an object unless the object is moved with the action of a force. The application of a force alone does not constitute work. For example, when a student holds the chair in his hand, he exerts a force to support the chair. But, work is not done on the chair as the chair does not move. Some more examples to understand the concept of work are given below : (1) Push a box lying on a surface. The box moves through a distance. You exerted a force on the box and the box got displaced. In this situation work is done [see figure(a)]

The concept of work (2) A man pushes a trolley and the trolley moves through a distance. A man has exerted a force on the trolley and it is displaced. Therefore, work is done [see figure(b)]. (3) A person lifts a cat through a certain height. To do this he applies a force. The cat rises up. There is a force applied on the cat and the cat has moved. Hence, work is done [see figure(c)]. Two important conditions that must be satisfied for work to be done are: (i) a force should act on an object (ii) the object must be displaced. (iii) direction of force must not be perpendicular to the displacement.

If any one of the above conditions does not exist, work is not done. This is the concept of work that we use in science.

Q. A horse is pulling a cart. The cart moves. Do you think that work is done in this situation?
Explanation: Yes, in this situation a work is done. When a horse pulls a cart, it is applying a force that moves the cart. Since a force is applied on the cart and the cart is displaced, a work is done by the horse on the cart.

Force : A force can be defined as 'a push or a pull exerted on an object that can cause the object to speed up, slow down, or change direction as it moves or it can change its shape and size'.
Displacement : The shortest distance between the initial position and the final position of a moving object in the given interval of time is known as the displacement of the object.

3.0Mathematical definition of work

(1) A constant force is applied in the direction of the displacement of an object: Let a constant force, F acts on an object. Let the object be displaced through a distance, s in the direction of the force (see figure). Let W be the work done.

Work done by a constant force acting in the direction of displacement Here, we define work to be equal to 'the product of the force and displacement'.

Work done = force \times displacement

W = F \times S

(2) A constant force is applied at a certain angle with the direction of the displacement of an object : When the force on an object and the object's displacement are in different directions, the work done on the object is given by,

W = F \times s \times cos θ

where, the angle between the force and the direction of the displacement is

θ

(see figure). Here, we define work to be equal to 'the force multiplied by the displacement multiplied by the cosine of the angle between them'.

Work done by a constant force acting at an angle with the direction of displacement Work is a scalar quantity, it has only magnitude and no direction. SI unit of work: Joule 1 Joule

= 1

newton

\times 1

meter or

1J = 1Nm = 1 kg m^{2} s^{- 2}

C.G.S unit of work : Erg.

1 erg = 1

dyne

cm = 1 g cm^{2} s^{- 2}

1 Joule

= 1 0^{7} ergs

Definition of 1 joule: 1 J is the amount of work done on an object when a force of 1 N displaces it by 1 m along the line of action of the force.

Some important points related to work

(1) If

θ = 0^{\circ}

, then

cos 0^{\circ} = 1

and

W = F \times s

. (2) If

θ = 9 0^{\circ}

, then,

W = 0

because

cos 9 0^{\circ} = 0

. So, no work is done on a bucket being carried by a girl walking horizontally (see figure). The upward force exerted by the girl to support the bucket is perpendicular to the displacement of the bucket, which results in no work done on the bucket.

(3) If,

θ = 18 0^{\circ}

, then

cos 18 0^{\circ} = - 1

and

W = - F \times s

Trigonometry The word trigonometry is derived from the Greek words 'tri' (Meaning three), 'gon' (Meaning sides) and 'metron' (meaning measure). Infact, trigonometry is the study of relationships between the sides and angles of a triangle. Let us take some examples from our surroundings where right triangles can be imagined to be formed. Suppose the students of a school are visiting Statue of unity. Now if a student is looking at the top of the statue, a right-angle triangle can be imagined to be made, as shown in figure.

Trigonometric ratio

Sine and Cosine of an angle: In right $△ A BC$ , the sine of $∠ A$ , which is written " $sin A$ ", is given by $sin A = \frac{Length of leg opposite ∠ A}{Length of hypotenuse} = \frac{a}{c}$ and the cosine of $∠ A$ , which is written "cos A ", is given $by cos A = \frac{Length of legadjacent ∠ A}{Length of hypotenuse} = \frac{b}{c}$

Examples

Indicate the perpendicular, the hypotenuse and the base (in that order) with respect to the angle marked x . Solution: Perpendicular $= a$ Hypotenuse = c Base $= b$
In figure, for $△ ABC$ , find $sin θ$ and $cos θ$ . Solution:

sin θ = \frac{opposite}{hypotenuse} = \frac{6}{10} = 0.6 cos θ = \frac{adjacent}{hypotenuse} = \frac{8}{10} = 0.8

Tangent of an angle

In addition to the sine and cosine ratios, you can use the tangent ratio to find the measures of the sides and angles of a right triangle.

In right $△ ABC$ , the tangent of $∠ A$ , which is written

"tan A " is given by

\frac{length of legopposite to ∠ A}{length of legadjacent to ∠ A} = \frac{a}{b}

Angle (θ)	0°	30°	45°	60°	90°
sin(θ)	0	$1/2$	$1/\sqrt2$	$\sqrt3/2$	1
cos(θ)	1	$\sqrt3/2$	$1/\sqrt2$	$1/2$	0
tan(θ)	0	$1/\sqrt3$	1	$\sqrt3$	∞

4.0Concept of negative and positive work

The work done by a force can be either positive or negative. (1) Whenever angle ( $θ$ ) between the force and the displacement is acute, i.e., $0^{\circ} < θ < 9 0^{\circ}$ , the work done is positive. Also, when angle ( $θ$ ) between the force and displacement is zero, i.e., force and displacement are in same direction, the work done is positive. (2) Whenever angle ( $θ$ ) between the force and the displacement is obtuse, i.e., $9 0^{\circ} < θ < 18 0^{\circ}$ , the work done is negative. Also, when angle ( $θ$ ) between the force and displacement is $18 0^{\circ}$ , i.e., force and displacement are in opposite direction, the work done is negative.

Concept of negative and positive work

Whenever force is in the direction of motion, velocity of the object increases and the work done is positive. Whenever force opposes motion, velocity of the object decreases and the work done is negative.

Work is positive when force and displacement are parallel to each other that is are in the same direction and work is negative when force and displacement are antiparallel to each other.
The following lists the conditions for zero work.

Condition for Zero Work	Calculation of Work Done
The net force should be equal to zero.	W = F × s × cos(θ) = 0 × s × cos(θ) = 0
The net displacement should be equal to zero.	W = F × s × cos(θ) = F × 0 × cos(θ) = 0
The force and displacement should be perpendicular to each other.	W = F × s × cos(θ) = F × s × cos(90°) = 0

Q. A force of 5 N is acting on an object. The object is displaced through 2 m in the direction of the force (see figure). If the force acts on the object all through the displacement, then what is the work done on the object?
Solution: Decode the problem Identify the terms $F = ✓, s = ✓, W = ?$ Apply the formula, $W = Fs$ Given, force, $F = 5 N$ ; displacement, $s = 2 m; W =$ ? Then, the work done $= Fs = 5 \times 2 = 10 J$ .
Q. How much work is done on a vacuum cleaner pulled 3.0 m by a force of 50.0 N at an angle of $3 0^{\circ}$ with the horizontal? Solution: Given, $F = 50.0 N; θ = 3 0^{\circ}$ ; $s = 3.0 m$ ; $W =$ ? Work done, $W = 50 \times 3 \times cos 3 0^{\circ} = 50 \times 3 \times \frac{3}{2}$ or $W = 753 = 75 \times 1.73 = 129.75 J$
Q. An artificial satellite is moving around the Earth in a circular path under the influence of centripetal force provided by the gravitational force between them. What is the work done by this centripetal force? Explanation: Centripetal force ( F ) is always perpendicular to the displacement (s) of the particle moving along a circular path. That is, the angle $(θ)$ between them is.
Now, work done, $W = F$ s $cos θ = F$ s $cos = 0 [∵ cos 9 0^{\circ} = 0]$ Thus, work done by this centripetal force is zero. Work done by the centripetal force is always zero because it is always perpendicular to the displacement. For example, if an electron moves around a nucleus in a circular path due to centripetal force provided by the electric force between them, the work done by this force is zero.

5.0Work done by applied force against gravity

If an object is lifted up to a certain height (see figure), definitely, a work is done by the applied force. The applied force must be equal to the weight ( mg ) of the object. This work done is given by, $W = F \times s = (mg) \times h$ Where, $m =$ mass of object ; $g =$ acceleration due to gravity ; $h =$ height or $W = mgh$ When a body is lifted up, the work done by the applied force is while work done by the gravity is negative. Similarly, when a book is put down from a certain height, the work done by the applied force is negative while work done by the gravity is positive.

Work done against gravity positive

Q. A porter lifts a luggage of 15 kg from the ground and puts it on his head 1.5 m above the ground. Calculate the work done by him on the luggage (Take, $g = 10 m / s^{2}$ ). Solution: Decode the problem Identify the terms $m = ✓, h = ✓, g = ✓, W = ?$ Apply the formula, $W = mgh$ Given, mass of luggage, $m = 15 kg$ ; height, $h = 1.5 m$ ; acceleration due to gravity, $g = 10 m / s^{2}$ ; $W =$ ? Work done, $W = 15 \times 10 \times 1.5 m = 225 J$

6.0Energy

Without light that come to us from the Sun, life on Earth would not exist. With the light energy, plants can grow and the oceans and atmosphere can maintain temperature ranges that support life. Although energy is difficult to define comprehensively, a simple definition is that energy is the capacity to do work. Thus, when you think of energy, think of what work is involved. Let us take some examples: (i) Energy must be supplied to a car's engine in order for the engine to do work in moving the car. In this case, the energy may come from burning petrol. (ii) When a fast moving cricket ball hits a stationary wicket, the wicket is thrown away. Thus, work is done by the energy in the moving ball on the wicket. (iii) An object placed at a certain height has the capability to do work. If it is allowed to fall, it will move downward i.e., a work will be done in this case. (iv) When a raised hammer falls on a nail placed on a piece of wood, it drives the nail into the wood. Thus, energy of hammer does a work on nail. (v) When a balloon is filled with air and we press it we notice a change in its shape. That is, we have done a work on balloon to change its shape. As long as we press it gently, it can come back to its original shape when the force is withdrawn. This means, balloon acquires energy due to which it regains its original shape. (vi) If we press the balloon hard, it can even explode producing a blasting sound. Again, it acquires enough energy that does work to blast the balloon. In all the above examples, the objects acquire the capability of doing work which is called energy. SI unit of energy : Since energy is the capacity to do work, its unit is same as that of work, that is, joule (J). 1 J is the energy required to do 1 joule of work. Sometimes a larger unit of energy called kilo joule ( kJ ) is used, $1 kJ = 1000 J$ .

Q. How does an object with energy do work? Explanation: An object that possesses energy can exert a force on another object. When this happens, energy is transferred from first object to the second object. The second object may move as it receives energy and therefore do some work. Thus, the first object had a capacity to do work. This implies that any object that possesses energy can do work.

7.0Forms of energy

Mechanical waves require a material medium to travel. Sound is also a mechanical wave thus, it requires a material medium like air, water, steel, etc. for its propagation. It cannot travel through vacuum. The world we live in provides energy in many different forms. The various forms include potential energy, kinetic energy, heat energy, chemical energy, electrical energy and light energy.

Mechanical energy

The capacity to do mechanical work is called mechanical energy. Mechanical energy can be of two types : (1) Kinetic energy (2) Potential energy

The sum of the gravitational potential energy and the kinetic energy is called mechanical energy.

Equations of uniformly accelerated motion.

$v = u + at$
$s = ut + \frac{1}{2} at^{2}$
$v^{2} = u^{2} + 2as$ $⋆$ The value of acceleration due to gravity on the surface of earth is $g = 9.8 m / s^{2}$ .

Kinetic energy

Kinetic energy is the energy associated with an object in motion.

Derivation of kinetic energy

Let a constant force F act on a ball of mass m with an initial velocity $u$ . The displacement of the ball be ' $s$ ', time taken to displace it be ' $t$ ', its final velocity be ' $v$ ' and acceleration produced in it be 'a' (see figure).

Now, work done,

W = F \times s

W = (ma) \times s

(

F = m \times a

) Now, from third equation of motion, we have,

v^{2} = u^{2} + 2

as or

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

From (1) and (2), we get,

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

It is clear from equation 3, the work done is equal to the change in the kinetic energy of an object. Equation 3 is called Work-Energy Theorem. If $u = 0$ , the work done will be $W = \frac{1}{2} mv^{2} - \frac{1}{2} m (0)^{2} = \frac{1}{2} mv^{2} s$

Thus, the kinetic energy possessed by an object of mass, $m$ and moving with a uniform velocity, $v$ is $E_{k} = \frac{1}{2} mv^{2}$

If the speed of an object is doubled, its kinetic energy becomes four times the initial value. This is because kinetic energy is directly proportional to the square of the speed of the object. If the mass of an object in motion is doubled, then its kinetic energy is also doubled, as kinetic energy is directly proportional to the mass of the object
Work done by the frictional force is always negative because it decreases the kinetic energy of an object. Also, frictional force is always opposite to the direction of displacement.

When force and displacement are in the same direction, the kinetic energy of the body increases. The increase in K.E. is equal to the work done on the body.

When force and displacement are oppositely directed, the kinetic energy of the body decreases. The decrease in K.E. is equal to the work done by the body against the retarding force.
When a body moves along a circular path with uniform speed, there is no change in its kinetic energy. By W-E. theorem, the work done by the centripetal force is zero.
When K.E. increases, the work done is positive and when K.E. decreases, the work done is negative.

$1 kg = 1000 g$
$xkm / h = x \times \frac{5}{18} m / s$ $⋆ ym / s = y \times \frac{18}{5} km / h$
Q. An object of mass 15 kg is moving with a uniform velocity of $4 ms^{- 1}$ . What is the kinetic energy possessed by the object? Solution: Decode the problem Identify the terms $m = ✓, v = ✓, E_{k} =$ ? Apply the formula, $E_{k} = \frac{1}{2} mv^{2}$ Given, mass of the object, $m = 15 kg$ ; velocity of the object, $v = 4 ms^{- 1}$ ; kinetic energy, $E_{k} =$ ? $E_{k} = \frac{1}{2} \times 15 \times (4)^{2} = 120 J$

You have studied three formulas for the work done of a body.

Use $W = F \times s cos θ$ , when work done by the applied force causes displacement.
Use $W = mgh$ , when work is done by the applied force against the gravity.
Use $W = \frac{1}{2} mv^{2} - \frac{1}{2} m u^{2}$ , when there is a change in the speed/velocity of the moving body.

Q. What is the work to be done to increase the velocity of a car from $30 km h^{- 1}$ to $60 km h^{- 1}$ if the mass of the car is 1500 kg ? Solution: Decode the problem Identify the terms $m = ✓, u = ✓, v = ✓, W = ?$ As there is change in the velocity of the car so we use the formula. $W = \frac{1}{2} mv^{2} - \frac{1}{2} mu^{2}$ Given, mass of the car, $m = 1500 kg$ ; initial velocity of car, $u = 30 km h^{- 1} = 30 \times \frac{5}{18} = \frac{25}{3} m s^{- 1}$ ; final velocity of the car, $v = 60 km h^{- 1} = 60 \times \frac{5}{18} = \frac{50}{3} m s^{- 1}$ ; where, $E_{kf} =$ final kinetic energy and $Eki =$ initial kinetic energy or $W = \frac{1}{2} mv^{2} - \frac{1}{2} mu^{2} = \frac{1}{2} m (v^{2} - u^{2}) = \frac{1}{2} \times 1500 [(\frac{50}{3})^{2} - (\frac{25}{3})^{2}]$ or $W = \frac{1}{2} \times 1500 [\frac{2500}{9} - \frac{625}{9}] = \frac{1}{2} \times 1500 [\frac{1875}{9}] = 156250 J$

Potential energy

The energy possessed by an object due to its position or configuration is called 'potential energy'. Consider the balanced smooth rock shown in figure. As long as the rock remains balanced, it has no kinetic energy. If it becomes unbalanced, it will fall vertically to the ground and will gain kinetic energy as it falls. The origin of this kinetic energy is potential energy present in the rock. Thus, potential energy is stored energy. Potential energy is associated with an object that has the potential to move because of its position or configuration.

Energy is present in the above example, but it is not kinetic energy because there is no motion. It is potential energy.

Gravitational potential energy

The energy associated with an object due to the object's position relative to a gravitational source is called gravitational potential energy.

Gravitational potential energy is energy due to an object's position in a gravitational field. Imagine an egg falling off a table. As it falls, it gains kinetic energy. But, where does the egg's kinetic energy come from? It comes from the gravitational potential energy that is associated with the egg's initial position on the table relative to the floor.

Derivation of potential energy

An object increases its energy when raised through a height. This is because work is done on it against gravity while it is being raised. The gravitational potential energy of an object at a point above the ground is defined as 'the work done in raising it from the ground to that point against gravity'.

Let us consider an object of mass, $m$ which is raised through a height, $h$ from the ground (see figure). A force equal to the weight (mg) of the object is required to do this. The object gains energy equal to the work done on it.

Work done on the object, $W =$ force $\times$ displacement $= mg \times h = mgh$ This work done on the object is the energy gained by the object. This is the potential energy ( $E_{P}$ ) of the object. That is, $E_{p} = W = mgh$ The work done against gravity depends on the difference in

Ground

Deriving formula for potential energy vertical heights of the initial and final positions of the object, and not on the path along which the object is moved. Figure shows a case where a block is raised from position $A$ to $B$ by taking two different paths. Let the height $AB = h$ . In both the situations the work done on the object is mgh.

Work done against gravity is independent of path travelled

The potential energy of an object at a height depends on the ground level or the zero level you choose. An object in a given position can have a certain potential energy with respect to one level and a different value of potential energy with respect to another level.

Q. Find the energy possessed by an object of mass 10 kg when it is at a height of $6 m$ above the ground. Given, $g = 9.8 ms^{- 2}$ . Solution: Decode the problem Identify the terms $m = ✓, s = ✓, g = ✓, W =$ ? Apply the formula, $E_{p} = W = mgh$ Given, mass of the object, $m = 10 kg$ ; displacement (height), $h = 6 m$ ; acceleration due to gravity, $g = 9.8 m s^{- 2}$ . Potential energy $= 10 kg \times 9.8 m s^{- 2} \times 6 m = 588 J$ .

Gravitational potential energy depends on height from an arbitrary zero level

It is important to note that the height, h , is measured from an arbitrary zero level. In the example of the egg, if the floor is the zero level, then $h$ is the height of the table, and mgh is the gravitational potential energy relative to the floor. Alternatively, if the table is the zero level, then h is zero. Thus, the potential energy associated with the egg relative to the table is zero.

Let us take another example : Suppose you drop a volleyball from a second-floor roof and it lands on the first-floor roof of an adjacent building (see figure). If the height is measured from the ground, the gravitational potential energy is not zero because the ball is still above the ground. But if the height is measured from the first-floor roof, the potential energy is zero when the ball lands on the roof.

Gravitational potential energy is measured relative to some zero

If $B$ is the zero level, then all the gravitational potential energy is converted to kinetic energy as the ball falls from A to B. If C is the zero level, then only part of the total gravitational potential energy is converted to kinetic energy during the fall from A to B .

Q. An object of mass 12 kg is at a certain height above the ground. If the potential energy of the object is 480 J , find the height at which the object is with respect to the ground. Given, $g = 10 m s^{- 2}$ . Solution: Decode the problem Identify the terms $m = ✓, Ep_{p} = ✓, h =$ ? Apply the formula, $E_{p} = W = mgh$ Given, mass of the object, $m = 12 kg$ , potential energy, $E_{p} = 480 J$ . Now, $E_{p} = m g h$ or $480 = 12 \times 10 \times h$ or $h = \frac{480}{12 \times 10} = 4 m$ The object is at the height of $4 m$ .

Elastic potential energy

Imagine you are playing with a spring on a tabletop. You push a block into the spring, compressing the spring, and then release the block. The block slides across the tabletop. The kinetic energy of the block came from the stored energy in the compressed spring (see figure). This potential energy is called elastic potential energy.

Elastic potential energy is stored in any compressed or stretched object, such as a spring or the stretched strings of a tennis racket or guitar.

The length of a spring when no external forces are acting on it is called the relaxed length of the spring. When an external force compresses or stretches the spring, elastic potential energy is stored in the spring. The amount of energy depends on the distance the spring is compressed or stretched from its relaxed length.
Understanding elastic potential energy

Any spring that has been stretched or compressed has stored elastic potential energy. This means that the spring is able to do work on another object by exerting a force over some distance as the spring regains its original length. The energy stored in a spring is also called 'strain potential energy'. The energy available for use when a deformed elastic object returns to its original configuration is called 'elastic potential energy'.

8.0The law of conservation of energy

Energy appears in many forms, such as heat, motion, height, pressure, electricity, and chemical bonds between atoms.

Energy transformations

Energy can be converted from one form to another form in different systems, machines or devices. Systems change as energy flows from one part of the system to another. Parts of the system may speed up, slow down, get warmer or colder, etc. Each change transfers energy or transforms energy from one form to another. For example, friction transforms energy of motion to energy of heat. A bow and arrow transform potential energy in a stretched bow into energy of motion (i.e., kinetic energy) of an arrow.

Law of conservation of energy

Energy can never be created or destroyed, just converted from one form into another. This is called the law of conservation of energy.

The law of conservation of energy is one of the most important laws in physics. It applies to all forms of energy.

Energy has to come from somewhere

The law of conservation of energy tells us that energy cannot be created from nothing. If energy increases somewhere, it must decrease somewhere else. The key to understanding how systems change is to trace the flow of energy. Once we know how energy flows and transforms, we have a good understanding of how a system works. For example, when we use energy to drive a car, that energy comes from chemical energy stored in petrol. As we use the energy, the amount left in the form of petrol decreases.

Conservation of mechanical energy

The mechanical energy i.e., the sum of potential and kinetic energies is constant in the absence of any frictional forces. This means that if you calculate the mechanical energy $(Em)$ at any two randomly chosen times, the answers must be equal. Let us take an example of free fall (here, the effect of air resistance on the motion of the object is ignored) : Let us consider an object of mass ' $m$ ' at a certain height ' $h$ ' (see figure). Let it is dropped from this height from point $A$ i.e., $v_{A} = 0$ . Mechanical energy at $A, E_{A} =$ K.E. at $A +$ P.E. at $A = \frac{1}{2} mv_{A}^{2} + mgh_{A}$ or $E_{A} = \frac{1}{2} m (0)^{2} + m g h = m g h$ Let after a certain time ' t ', it reaches point B after covering a distance ' x '. Now, from third equation of motion, we have, $v_{B}^{2} = v_{A}^{2} + 2 gx = (0)^{2} + 2 gx$ or $v_{B}^{2} = 2 gx$ Mechanical energy at B, $E_{B} =$ K.E. at $B +$ P.E. at $B = \frac{1}{2} mv_{B}^{2} + mgh_{B}$

or E_{B} = \frac{1}{2} m (2 gx) + m g (h - x) = m gx + m g h - m gx = m g h

Finally, the ball reaches the ground (at point C) after covering a distance $h$ . Again, from third equation of

Total mechanical energy is conserved in a free fall. motion, we have,

v_{C}^{2} = v_{A}^{2} + 2 gh = (0)^{2} + 2 gh

v_{C}^{2} = 2 gh

Mechanical energy at C,

E_{C} =

K.E. at

C +

P.E. at

C = \frac{1}{2} mv_{C}^{2} + mgh_{C}

E_{C} = \frac{1}{2} m (2 gh) + mg (0) = mgh

From eqs. (1), (3) and (5), we get,

E_{A} = E_{B} = E_{C}

This means total mechanical energy is conserved during the free fall of an object. That is,

\frac{1}{2} mv^{2} + mgh = constant

During the free fall of the object, the decrease in potential energy during a certain time interval in its path, appears as an equal amount of increase in kinetic energy.

According to law of conservation of energy, if potential energy of a body increases, its kinetic energy decreases and vice-versa such that the total mechanical energy always remains constant.
Resistance of air is neglected in verifying the above law.

Q. A $4.0 \times 1 0^{4} kg$ roller coaster starts from rest at point A. Neglecting friction, calculate its potential energy relative to the ground, its kinetic energy, and its speed at point B (see figure). Take $g = 10 m / s^{2}$ . Solution: Decode the problem Identify the terms $m = ✓, u = ✓, g = ✓, h = ✓, E_{A} =$ ? Apply the formula, $EA_{A} = \frac{1}{2} mv_{A}^{2} + mgh_{A}$ Identify the terms $m = ✓, g = ✓, h = ✓$ , $Ep_{p} =$ ? Apply the formula, $Ep_{p} = mgh$ Identify the terms $m = ✓, E_{k} = ✓, v =$ ? Apply the formula, $E_{K} = \frac{1}{2} mv^{2}$ Initial energy (at A), $E_{A} = \frac{1}{2} mv_{A}^{2} + mgh_{A} = \frac{1}{2} m (0)^{2} + mg (54)$
or $E_{A} = 4.0 \times 1 0^{4} \times 10 \times 54 = 2.16 \times 1 0^{7} J$ Potential energy at B, $E_{PB} = m \times g \times h_{B} = 4.0 \times 1 0^{4} \times 10 \times 15 = 6 \times 1 0^{6} J$ Kinetic energy at B, Екв $= E_{A} -$ Ерв

= 2.16 \times 1 0^{7} - 6 \times 1 0^{6} = 15.6 \times 1 0^{6} J

Now, $E_{KB} = \frac{1}{2} mv_{B}^{2}$ or $v_{B} = \frac{2 E _{KB}}{m}$ or $V_{B} = \frac{2 \times 15.6 \times 1 0 ^{6}}{4.0 \times 1 0 ^{4}}$ $= 780 = 27.9 m / s$

9.0Power

The engine in an old school bus could, over a long period of time, do as much work as jet engines do when a jet takes off. However, the school bus engine could not begin to do work fast enough to make a jet lift off. In this and many other applications, the rate at which work is done is more significant than the amount of work done. $Power = \frac{Work done}{Time}$

Power is the rate at which work is done. Power can also be defined as the rate at which energy is transferred. SI unit of power : Watt (W). 1 Watt = 1 joule/second or $1W = 1 J s^{- 1}$ Definition of 1 watt : If 1 joule work is done per second by a device or a machine, then the power of that device or machine is 1 watt.

James Watt did experiments with strong horses and determined that they could lift 550 pounds a distance of one foot in 1 s . He called this amount of power 'one horsepower' (hp). Converting to SI units, $1 hp = 746 W = 0.746 kW$

Power in terms of force ( $F$ ) and velocity ( $v$ )

We know that power,

P = \frac{W}{t} = \frac{F s}{t} = F (\frac{s}{t})

P = F \times v

$1 KW - 1$ kilo watt $= 1 0^{3} W$

$1 MW - 1$ mega watt $= 1 0^{6} W$
1 GW - 1 giga watt $= 1 0^{9} W$
$1 hp - 1$ horsepower $= 746 W$
Q. Two girls, each of weight 400 N climb up a rope through a height of 8 m . We name one of the girls A and the other B. Girl A takes 20 s while B takes 50 s to accomplish this task. What is the power expended by each girl? Solution: Decode the problem Identify the terms $w = mg = ✓, h = ✓, t = ✓, P$ = ? Apply the formula, $P = \frac{Work done}{time taken} = \frac{mgh}{t}$ (i) Power expended by girl A: Weight of the girl, $mg = 400 N$ ; displacement (height), $h = 8 m$ ; time taken, $t = 20 s$ $P = \frac{Work done}{time taken} = \frac{mgh}{t} = \frac{400 \times 8}{20} = 160 W$ (ii) Power expended by girl B : Weight of the girl, $mg = 400 N$ displacement (height), $h = 8 m$ ; time taken, $t = 50 s$ $P = \frac{Work done}{time taken} = \frac{mgh}{t} = \frac{400 \times 8}{50} = 64 W$
Q. A boy of mass 50 kg runs up a staircase of 45 steps in 9 s . If the height of each step is 15 cm , find his power. Take $g = 10 m s^{- 2}$ . Solution: Decode the problem Identify the terms $w = m g = ✓, h = ✓, t = ✓, P$ = ? Apply the formula, $P = \frac{Work done}{time taken} = \frac{mgh}{t}$ Weight of the boy, $W = mg = 50 \times 10 = 500 N$ ; height of the staircase, $h = 45 \times 15 cm = \frac{45 \times 15}{100} = 6.75 m$ ; time taken to climb, $t = 9 s$ $P = \frac{Work done}{time taken} = \frac{mgh}{t} = \frac{500 \times 6.75}{9} = 375 W$

10.0Commercial unit of energy

The unit joule is an extremely small unit. It is inconvenient to express large quantities of energy in terms of joule. We use a bigger unit of energy called kilowatt hour (kWh). It is called commercial unit of energy. Definition of $1 kWh$ : If a machine or a device of power 1 kW or 1000 W is used continuously for one hour, it will consume 1 kWh of energy. Thus, 1 kWh is the energy used in one hour at the rate of 1000 W (or 1 kW ). $1 kWh = 1 kW \times 1 h = 1000 W \times 3600 s = 3600000 J$ or $1 kWh = 3.6 \times 10^{6} J$ .

The energy used in households, industries and commercial establishments is usually expressed in kilowatt hour. For example, electrical energy used during a month is expressed in terms of 'units'. Here, 1 'unit' means 1 kilowatt hour.
Watt hour is a unit in which the amount of electric energy consumed by a device is measured. A meter which measures the electric energy consumed by various electric appliances is called watt hour meter. It is commonly called a electric meter which is installed at every house by the state electricity board.

Q. An electric bulb of 60 W is used for 6 h per day. Calculate the 'units' of energy consumed in one day by the bulb. Solution: Decode the problem Identify the terms $P = ✓, t = ✓$ , energy $=$ ? Apply the formula, Energy $=$ power $\times$ time taken The energy consumed by the bulb is 0.36 'units'. Power of electric bulb $= 60 W = 0.06 kW$ ; time used, $t = 6 h$ Energy $=$ power $\times$ time taken $= 0.06 kW \times 6 h = 0.36 kWh = 0.36 ’units’.$
Q. A light body and a heavy body have same kinetic energy. Which one of the two has greater momentum? Explanation: Firstly, we will find the relationship between kinetic energy and linear momentum. Kinetic energy, $E_{K} = \frac{1}{2} mv^{2} = \frac{1}{2} mv^{2} \times \frac{m}{m} = \frac{( mv ) ^{2}}{2 m}$ or $E_{K} = \frac{p ^{2}}{2m}$ ( $p = mv =$ linear momentum $)$ or $p^{2} = 2 mE$ K or $p = 2 mE_{K}$ This means, if kinetic energy ( $E_{κ}$ ) is constant for both the bodies, then, $p \propto m$ . Thus, heavier body will have greater momentum than the lighter body.

11.0Concept Map

Important Units, Formulae, and Quantity

Quantity	Formula	Unit
Work	F × s, F × s × cos(θ)	Joule
Work against gravity	Mgh	Joule
Energy		Joule, erg
Kinetic energy	(1/2) mv²	Joule
Work-Energy Theorem	W = (1/2) mv² - (1/2) mu²
Momentum and kinetic energy	p = √(2mE)
Potential energy	mgh
Commercial unit of energy		Joule
Power	W/t, F × v	Watt

12.0Some Basic Terms

Horizontal :- Going from side to side, not up and down; flat or level.
Depression :- Depressions are localized pavement surface areas having elevations slightly lower than those of the surrounding pavement.
Penetrate :- To go through or into something.
Gravitational field :- A gravitational field is the force field that exists in the space around every mass or group of masses.
Arbitrary :- Not seeming to be based on any reason or plan and sometimes seeming unfair.
Perpendicular :- at an angle of $9 0^{\circ}$ to another line or surface.
Configuration :- The way in which the parts of something, or a group of things, are arranged.
Identical :- Exactly the same as; similar in every detail.

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