Class 9 ENERGY

Topic Covered

`color{red} ♦` ENERGY

ENERGY

Life is impossible without energy. The demand for energy is ever increasing. Where do we get energy from? The Sun is the biggest natural source of energy to us.

Many of our energy sources are derived from the Sun. We can also get energy from the nuclei of atoms, the interior of the earth, and the tides. Can you think of other sources of energy?

Activity _____________`11.5`

♦ A few sources of energy are listed above. There are many other sources of energy. List them.
♦ Discuss in small groups how certain sources of energy are due to the Sun
♦ Are there sources of energy which are not due to the Sun?



The word energy is very often used in our daily life, but in science we give it a definite and precise meaning.

Let us consider the following examples: when a fast moving cricket ball hits a stationary wicket, the wicket is thrown away. Similarly, an object when raised to a certain height gets the capability to do work.

You must have seen that when a raised hammer falls on a nail placed on a piece of wood, it drives the nail into the wood. We have also observed children winding a toy (such as a toy car) and when the toy is placed on the floor, it starts moving.

When a balloon is filled with air and we press it we notice a change in its shape. As long as we press it gently, it can come back to its original shape when the force is withdrawn.

However, if we press the balloon hard, it can even explode producing a blasting sound. In all these examples, the objects acquire, through different means, the capability of doing work.

An object having a capability to do work is said to possess energy. The object which does the work loses energy and the object on which the work is done gains energy.


How does an object with energy do work? An object that possesses energy can exert a force on another object. When this happens, energy is transferred from the former to the latter.

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.

The energy possessed by an object is thus measured in terms of its capacity of doing work. The unit of energy is, therefore, the 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 equals 1000 J.
Q 3265201165

An object of mass `15 kg` is moving with a uniform velocity of `4 m s^(–1)`. What is the kinetic energy possessed by the object?


Class 9 Chapter 11 Example 3
Solution:

Mass of the object, `m = 15 kg`, velocity
of the object, `v = 4 m s^(–1)`.
From Eq. (11.5),

` E_k = 1/ 2 mv^2`

` = 1/2 xx 15 kg xx 4 m s^(–1) xx 4 m s^(–1)`

`= 120 J`
The kinetic energy of the object is `120 J`.
Q 3275201166

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`?


Class 9 Chapter 11 Example 4
Solution:

Mass of the car, `m =1500 kg`,

initial velocity of car, `u = 30 km h^(–1)`

` = ( 30 xx 1000 m)/( 60 xx 60 s)`

` = 8.33 m s^(–1)`.
Similarly, the final velocity of the car,
`v = 60 km h^(–1)`
`= 16.67 m s^(–1)`.
Therefore, the initial kinetic energy of the car,

` E_(kt) = 1/2 m u^2`

` = 1/2 xx 1500 kg xx (8.33 m s^(–1) )^2`

`= 52041.68 J`.
The final kinetic energy of the car,

` E_(kt) = 1/2 xx 1500 kg xx (16.67 m s^(–1) )^2`
`= 208416.68 J.`
Thus, the work done = Change in
kinetic energy
`= E_(kf) – E_(kt)
`= 156375 J.`

FORMS OF ENERGY

Luckily 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.

`bbul" KINETIC ENERGY"`


Activity _____________`11.6`

♦ Take a heavy ball. Drop it on a thick bed of sand. A wet bed of sand would be better. Drop the ball on the sand bed from height of about 25 cm. The ball creates a depression.
♦ Repeat this activity from heights of 50 cm, 1 m and 1.5 m.
♦ Ensure that all the depressions are distinctly visible
♦ Mark the depressions to indicate the height from which the ball was dropped.
♦ Compare their depths.
♦ Which one of them is deepest?
♦ Which one is shallowest? Why?
♦ What has caused the ball to make a deeper dent?
♦ Discuss and analyse.




Activity _____________`11.7`

♦ Set up the apparatus as shown in Fig. 11.5.
♦ Place a wooden block of known mass in front of the trolley at a convenient fixed distance
♦ Place a known mass on the pan so that the trolley starts moving.

♦ The trolley moves forward and hits the wooden block.
♦ Fix a stop on the table in such a manner that the trolley stops after hitting the block. The block gets displaced.
♦Note down the displacement of the block. This means work is done on the block by the trolley as the block has gained energy.
♦ From where does this energy come?
♦ Repeat this activity by increasing the mass on the pan. In which case is the displacement more?
♦ In which case is the work done more?
♦ In this activity, the moving trolley does work and hence it possesses energy.



A moving object can do work. An object moving faster can do more work than an identical object moving relatively slow. A moving bullet, blowing wind, a rotating wheel, a speeding stone can do work.

How does a bullet pierce the target? How does the wind move the blades of a windmill? Objects in motion possess energy. We call this energy kinetic energy.

A falling coconut, a speeding car, a rolling stone, a flying aircraft, flowing water, blowing wind, a running athlete etc. possess kinetic energy. In short, kinetic energy is the energy possessed by an object due to its motion.

The kinetic energy of an object increases with its speed.

How much energy is possessed by a moving body by virtue of its motion? By definition, we say that the kinetic energy of a body moving with a certain velocity is equal to the work done on it to make it acquire that velocity.

Let us now express the kinetic energy of an object in the form of an equation. Consider an object of mass, m moving with a uniform velocity, u.

Let it now be displaced through a distance s when a constant force, F acts on it in the direction of its displacement. From Eq. (11.1), the work done, W is F s.

The work done on the object will cause a change in its velocity. Let its velocity change from u to v. Let a be the acceleration produced.

We studied three equations of motion. The relation connecting the initial velocity (u) and final velocity (v) of an object moving with a uniform acceleration a, and the displacement, s is

` v^2 – u^2 = 2 a s` .........(8.7)

This gives

` s = (v^2 - u^2)/(2a)` ..........(11.2)

From section 9.4, we know `F = m a`. Thus, using (Eq. 11.2) in Eq. (11.1), we can write the work done by the force, F as

` W = m a xx ( (v^2 - u^2)/(2a))`

or

` W = 1/2 m ( v^2 - u^2)` ............(11.3)

If the object is starting from its stationary position, that is, `u = 0`, then

` W = 1/2 m v^2` ........(11.4)

It is clear that the work done is equal to the change in the kinetic energy of an object. If `u = 0`, the work done will be ` 1/2 mv^2`

Thus, the kinetic energy possessed by an object of mass, m and moving with a uniform velocity, `v` is

` E_k = 1/2 m v^2` ........(11.5)

POTENTIAL ENERGY

Activity _____________`11.8`

♦ Take a rubber band.
♦ Hold it at one end and pull from the other. The band stretches.
♦ Release the band at one of the ends.
♦ What happens?
♦ The band will tend to regain its original length. Obviously the band had acquired energy in its stretched position.
♦ How did it acquire energy when stretched?



Activity _____________`11.9`

♦ Take a slinky as shown below.
♦ Ask a friend to hold one of its ends. You hold the other end and move away from your friend. Now you release the slinky.



♦ What happened?
♦ How did the slinky acquire energy when stretched?
♦ Would the slinky acquire energy when it is compressed?




Activity ____________`11.10`
♦ Take a toy car. Wind it using its key.
♦ Place the car on the ground.
♦ Did it move?
♦ From where did it acquire energy?
♦ Does the energy acquired depend on the number of windings?
♦ How can you test this?




Activity ____________`11.11`

♦Lift an object through a certain height. The object can now do work. It begins to fall when released.
♦ This implies that it has acquired some energy.
If raised to a greater height it can do more work and hence possesses more energy.
♦ From where did it get the energy? Think and discuss.



In the above situations, the energy gets stored due to the work done on the object. The energy transferred to an object is stored as potential energy if it is not used to cause a change in the velocity or speed of the object.

You transfer energy when you stretch a rubber band. The energy transferred to the band is its potential energy. You do work while winding the key of a toy car.

The energy transferred to the spring inside is stored as potential energy. The potential energy possessed by the object is the energy present in it by virtue of its position or configuration.

Activity ____________`11.12`

♦ Take a bamboo stick and make a bow as shown in Fig. 11.6.
♦ Place an arrow made of a light stick on it with one end supported by the stretched string.
♦ Now stretch the string and release the arrow.
♦ Notice the arrow flying off the bow. Notice the change in the shape of the bow.
♦ The potential energy stored in the bow due to the change of shape is thus used in the form of kinetic energy in throwing off the
arrow.




`bbul"POTENTIAL ENERGY OF AN OBJECT AT A HEIGHT"`

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 energy present in such an object is the gravitational potential energy.

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.

It is easy to arrive at an expression for the gravitational potential energy of an object at a height.



Consider an object of mass, m. Let it be raised through a height, h from the ground. A force is required to do this. The minimum force required to raise the object is equal to the weight of the object, mg.

The object gains energy equal to the work done on it. Let the work done on the object against gravity be W. That is,

work done, `W =` force `xx` displacement
`= mg xx h`
`= m g h`

Since work done on the object is equal to mgh, an energy equal to mgh units is gained by the object. This is the potential energy (EP) of the object.

`E_p = m g h` ........(11.7)

It is useful to note that the work done by gravity depends on the difference in vertical heights of the initial and final positions of the object and not on the path along which the object is moved.

Fig. 11.8 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`.




`ulbb"ARE VARIOUS ENERGY FORMS INTERCONVERTIBLE?"`

Can we convert energy from one form to another? We find in nature a number of instances of conversion of energy from one form to another.

Activity ____________`11.13`
♦ Sit in small groups.
♦ Discuss the various ways of energy conversion in nature.
♦ Discuss following questions in your group :
(a) How do green plants produce food?
(b) Where do they get their energy from?
(c) Why does the air move from place to place?
(d) How are fuels, such as coal and petroleum formed?
(e) What kinds of energy conversions sustain the water cycle?



Activity ____________`11.14`

♦ Many of the human activities and the gadgets we use involve conversion of energy from one form to another.
♦ Make a list of such activities and gadgets.
♦ Identify in each activity/gadget the kind of energy conversion that takes place.

Q 3285201167

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 m s^(–2)`.
Class 9 Chapter 11 Example 5
Solution:

Mass of the object, `m = 10 kg`,
displacement (height), `h = 6 m`, and
acceleration due to gravity, `g = 9.8 m s^(–2)`.
From Eq. (11.6),
Potential energy `= m g h`
`= 10 kg xx 9.8 m s^(–2) xx 6 m`
`= 588 J`
The potential energy is `588 J`.
Q 3205201168

An object of mass `12 k g` 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)`.


Class 9 Chapter 11 Example 6
Solution:

Mass of the object, `m = 12 kg`,
potential energy, `E_p = 480 J`.
`E_p = m g h`
`480 J = 12 kg xx 10 m s^(–2) xx h`

` h = (480 J)/(120 kg m s ^(-2)) = 4m`

The object is at the height of `4 m.`

LAW OF CONSERVATION OF ENERGY

We learnt that the form of energy can be changed from one form to another. What happens to the total energy of a system during or after the process? Whenever energy gets transformed, the total energy remains unchanged.

This is the law of conservation of energy. According to this law, energy can only be converted from one form to another; it can neither be created or destroyed. The total energy before and after the transformation remains the same.

The law of conservation of energy is valid in all situations and for all kinds of transformations.

Consider a simple example. Let an object of mass, m be made to fall freely from a height, h. At the start, the potential energy is mgh and kinetic energy is zero. Why is the kinetic energy zero? It is zero because its velocity is zero.

The total energy of the object is thus mgh. As it falls, its potential energy will change into kinetic energy. If v is the velocity of the object at a given instant, the kinetic energy would be ½mv2.

As the fall of the object continues, the potential energy would decrease while the kinetic energy would increase. When the object is about to reach the ground, h = 0 and v will be the highest.

Therefore, the kinetic energy would be the largest and potential energy the least. However, the sum of the potential energy and kinetic energy of the object would be the same at all points. That is,

potential energy + kinetic energy = constant

or

` m g h + 1/2 m v^2 =` constant ..........(11.7)

The sum of kinetic energy and potential energy of an object is its total mechanical energy.

We find that during the free fall of the object, the decrease in potential energy, at any point in its path, appears as an equal amount of increase in kinetic energy. (Here the effect of air resistance on the motion of the object has been ignored.)

There is thus a continual transformation of gravitational potential energy into kinetic energy.

Activity ____________`11.15`

♦ An object of mass `20 kg` is dropped from a height of `4 m`. Fill in the blanks in the following table by computing the potential energy and kinetic energy in each case.



♦ For simplifying the calculations, take the value of `g` as `10 m s^(–2)`.


 
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