UP Board Class 9 Science Notes For Chapter 10 Work, Energy And Power

Class 9 Science Notes For Chapter 10  Work, Energy, And Power

Work:  Work is said to be done if by applying a force on an object, it is displaced from its position in the direction of force.

Scientific Conception of Work

From the point of view of science, the following two conditions need to be satisfied for work to be done.

1. A force should act on an object.
2. The object must be displaced.

If any one of the above conditions does not exist, work is not done.

For Example. A girl pulls a trolley and the trolley moves through a distance. In this way, she has exerted a force on the trolley and it is displaced. Hence, work is done.

Work Done by a Constant Force

‘Work done by a force on an object is equal to the magnitude of the force multiplied by the distance moved in the direction of force.’

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Let us assume if a constant force F acts on an object at point A, due to which the object gets displaced through a distance s in the direction of the force and reaches point B, then the work done (W) by force {F) on that object will be equal to the product of the force and displacement.

Work done = Force x Displacement in the direction of force or W = F x s

SI Unit of Work

1. If F = 1 N and s = l m. then the work clone hy the force will he 1 N-m.
2. The SI unit of work is newton-metre (N-m) which is also called joule (J)
3. Thus 1 J is the amount of work done on an object when a three of 1 N displaces it by 1 m along the line of action of the force.
4. 1 joule = 1 newton x 1 metre ⇒ 1J = lN-m
5. Work is a scalar quantity, it has only magnitude and no direction.

Example 1. A force of 10 N is acting on an object. The object is displaced through 5 m in the direction of force. What is the work done in this case?
Answer:

Given, force, F = 10N, displacement, s=5m

Work done, W = f x J = 10 Nx 5m = 50 N-m or 50J

Positive, Negative, and Zero Work

When the force F and displacement s are in the same direction (the angle between the direction of force and displacement is 0°), work done will be positive, i.e. work is done by the force.

For Example. A boy pulls an object towards himself.

W = + F xs

When the force F and displacement s are in opposite directions (the angle between the direction of force and displacement is 180°), work done will be negative, i.e. work is done against the force.

For Example. The frictional force acts in the direction opposite to the direction of displacement, so work done by friction will be negative. W=-Fxs

When the force and displacement are in the perpendicular direction (the angle between the direction of force and displacement is 90°), the work done is zero.

For Example. A coolie carrying a load on his head. In this case, the gravitational force is acting vertically downward (weight of load) and displacement is along the horizontal direction, i.e. force and displacement are perpendicular to each other. So, in this case, the work done by the gravitational force is zero. W= 0

Example 2. A crane lifts” a crate upwards through a height of 20 m. The lilting force provided by the crane is 5 kN. How much work Is done by the force?

Answer: 

Given, force, F = 5 kN = 5000 N Displacement, s = 20 m

Work done, W =?

We know that work done, W = F s

Here, force and displacement are in the same direction.

So, W = Fs ⇒ W =5000Nx 20m = 100000 J

So, the work done by the force is 100000 J or 100 kj

Energy

1. It is the ability to do work It is always essential for performing any mechanical work. An object having the capability to do work is said to possess energy.
2. The object that does the work, loses energy, and the object on which work is done, gains energy.
3. The energy of an object is measured in terms of its capacity to do work.
4. The SI unit of energy is the same as that of work, i.e. joule (J). 1 joule of energy is required to do 1 J of work. A larger unit of energy is kJ.
5. 1 kilo joule (kJ) =10³J
6. Work done against a force is therefore stored as energy.
7. For Example, When a fast-moving cricket ball hits a stationary wicket, the wicket is thrown away.
8. When a raised hammer falls on a nail placed on a piece of wood, it drives the nail into the wood.
9. Sun is the biggest natural source of energy for us. We can also get energy from the nuclei of atoms, the interior of the Earth, and the tides in the ocean.

Forms of Energy

Energy exists in various forms mechanical energy (the sum of potential energy and kinetic energy), heat energy, chemical energy, electrical energy, and light energy.

Kinetic Energy

1. The energy which is possessed by an object due to its motion is called kinetic energy.
2. Its SI unit is joule (J). 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. The kinetic energy of an object increases with its speed.
3. Due to kinetic energy”, a bullet fired from a gun can pierce a target.
4. A moving hammer drives a nail into the wood. Due to its motion, it has kinetic energy or ability to do work.
5. The kinetic energy possessed by an object of mass m, moving with a uniform velocity v is given by,
6. \(\mathrm{KE}\) or \((E_K)=\frac{1}{2} m v^2\)

Calculation of Kinetic Energy

The kinetic energy of an object is measured by the amount of work, it can do before coming to rest. Consider an object of mass m moving with a uniform velocity u.

A force F is applied to it which displaces it through a distance s and it attains a velocity v.

Then, work is done to increase its velocity from u to v. W =Fs

According to the equation of motion,

⇒ \(v^2-u^2 =2 a y\)

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

where a is uniform acceleration, u is initial velocity and v is final velocity,

Also from, F=m a

Substituting the values of F and t in Eq. (1), we have

W = ma \(\cdot \frac{v^2-u^2}{2 a}\) of W=\(\frac{1}{2} m\left(v^2-u^2\right)\)

This is known as the work-energy theorem (i.e. total work is equal to the change in kinetic energy).

If initial velocity,

Then, u = 0

W =\(\frac{1}{2} m v^2\)

This work is equal to the kinetic energy of the object.

⇒ \(\mathrm{KE}\)(or \(E_K)=\frac{1}{2} m v^2\)

Some Important Results can be Derived from the Formula KE=\(\frac{1}{2} m v^2\)

These are given below :

1.  If the mass of an object is doubled, its kinetic energy also gets doubled.
2. If the mass of an object is halved, its kinetic energy also gets halved.
3.  If the speed of an object is doubled, its kinetic energy becomes four times.
4. If the speed of an object is halved, its kinetic energy becomes one-fourth.
5. Heavy objects moving with high speed have more kinetic energy than small objects moving with less speed.

Example 3. A bullet of mass 8 g is fired with a velocity of 80 \(\mathrm{~ms}^{-1}\). Calculate its kinetic energy.
Answer:

Given, mass, m=8 \(\mathrm{~g}=\frac{8}{1000} \mathrm{~kg}, velocity, v=80 \mathrm{~ms}^{-1}\)

KE of the bullet =\(\frac{1}{2} m v^2=\frac{1}{2} \times \frac{8}{1000} \times(80)^2 \)

= \(\frac{1}{2} \times \frac{8}{1000} \times 80 \times 80=25.6 \mathrm{~J}\)

Example 4. If a body of mass 5 kg is moving along a straight line with a velocity of 10 ms^-1 and acceleration of 20 ms^-2. Find its kinetic energy (KE) after 10 s.
Answer:

Given, the mass of the body, m = 5 kg

Initial velocity, u = 10 ms\(\mathrm{~ms}^{-1}\)

Acceleration, a = 20 \(\mathrm{~ms}^{-2}\)

Time, r = 10s Velocity, v = ?; KE = ?

First, we use the equation of meeting, N=N+ at to hinsl p, Then, we see \(KI-\frac{1}{2} m v^7\) to find kinetic energy.

r=\(\Delta+\Delta f\)

r=\((v+a t) m^{-1}\)

As we know, kinetic erg, KE =\(\frac{1}{2} m w^2\)

So. \(\mathrm{KE}=\frac{1}{2} m \times(w+a t)^2\)

KI=\(\frac{1}{2} \times 5 \times(10+20 \times 10)^2\)

∴ \(\mathrm{KE}=\frac{1}{2} \times 5 \times 210 \times 210=110250 \mathrm{j}\)

Example 5. What is the work to be done to increase the velocity of a van from 10 m/s to 20 m/s, if the mass of the is 2000 kg?
Answer:

Given,m=2000 kg, \(v_1=10 ms^{-1}, p_2=20 \mathrm{~ms}^{-1}\)

The initial Kinetic energy of the van

⇒ \(\mathrm{KE}_1.=\frac{1}{2} m \nu_1^2\) \([\mathrm{KE}=\frac{1}{2} m \nu^2]\)

= \(\frac{1}{2} \times 2000 \mathrm{~kg} \times\left(10 m s^{-1}\right)^2=100000 \mathrm{~J}=100 \mathrm{~kJ}\)

The final kinetic energy of the van

⇒ \(\mathrm{KE}_2=\frac{1}{2} m{ }_2^2=\frac{1}{2} \times 2000 \mathrm{~kg} \times\left(20 \mathrm{~ms}^{-1}\right)^2\)

= 400000J = 400 kJ

The work done = Change in kinetic energy = 400 kJ -100 kJ = 300 kJ

So, the kinetic energy of the van increases by 300 kJ when it speeds up from 10 \(m s^{-1}\) to 20 \(m s^{-1}\).

Potential Energy

1. The energy possessed by a body due to its change in position or shape is called potential energy. Its SI unit is joule (J).
2. We can say that the potential energy possessed by a body is the energy present in it by its position or configuration,
3. For Example. a stretched rubber band, spring, string on the bow, etc. Now, we can say that a body possesses energy even when it is not in motion.

Examples of potential energy are

Water stored in a dam has potential energy due to its position at the height.

A stone lying on the roof of the building has potential energy due to its height.

A wound spring of a watch has potential energy due to the change in its shape.

Potential Energy of an Object at a Height

1. When an object is raised through a certain height above the ground, its energy increases. This is because work is done on it against gravity while it is being raised.
2. 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.

Expression for Potential Energy

• Consider an object of mass m, lying at point A on the Earth’s surface. Here, its potential energy is zero and its weight mg acts vertically downwards.
• To lift the object to another position B at a height of h, we have to apply a minimum force that is equal to mg in the upward direction. So, work is done on the body against the force of gravity. Therefore,
• Work done = Force x Displacement or W = F x s
• As, F = mg [weight of the body]
• Here, s = h
• Therefore, W = mg x h = mg i.e. PE = mg
• This work is equal to the gain in energy of the body. This is the potential energy (PE) of the body.
• 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 concerning one level and a different value of potential energy concerning another level.
• The work done by gravity depends on the difference in vertical heights of the initial and final positions of the objects and not on the path along which the object is moved.
• It is clear from the given
• In both the above situations, the work done on the object is much.

Example 6. Suppose you have a body of mass 1 kg in your hand. Tb what height will you raise it, so that it may acquire a gravitational potential energy of 1 J? (Take, g = 10 ms^-2)
Answer:

Given, PE =1J, mass, m = 1 kg,

Acceleration due to gravity, g = 10 ms^-2, h =?

We know that, PE = mg or 1 = 1 x 10 x h

Height, b = \(\frac{1}{1 \times 10}\)=0.1 m = 10 cm

Example 7. A boy weighing 40 kg climbs up a vertical height of 200 m. Calculate the amount of work done by him. How much potential energy does he gain? (Take, g =9.8 ms^-2)
Answer:

Given that, mass, m = 40 kg

Acceleration due to gravity, g = 9.8 ms^-2, height, h = 200 m

Work done by the body = mgt = 40 x 9.8 x 200 = 78400J = 7.84 x 10⁴ J

Gain in PE = Work done = 7.84 x 10⁴J

Example 8. Suppose two bodies A and B having equal masses are kept at heights of h and 3 h, respectively. Find the ratio of their potential energies.
Answer:

Let the mass of each body be m.

PE of body A= mg PE of body B=m g \(\times\) 3 h

Ratio of their potential energies =\(\frac{m g b}{m g \times 3 b}=\frac{1}{3}\)=1: 3

Law of Conservation of Energy

The law of conservation of energy states that energy can neither be created nor be destroyed, it can only be transformed from one form to another. The total energy before and after transformation always remains constant.

Conservation of Energy During the Free Fall of a Body

Consider an object of mass m, lying at position B. It is made to fall freely from a height (h) above the ground

At point B: At the start, the potential energy is mgh and kinetic energy is zero (as its velocity is zero),

i.e. PE = mg

KE = 0

Total energy, TE = PE + KE = mgh

At point A: As it falls, its potential energy will change into kinetic energy. If v is the velocity of the object at a given instant, its kinetic energy would be \(\frac{1}{2} m v^2\).

PE = mg(h – x)

From Newton’s third equation of motion, v

⇒ \(v^2=u^2+2 g x\)

⇒ \(v^2\)=2 g x[u=0]

⇒ \(\mathrm{KE}=\frac{1}{2} m v^2=\frac{1}{2} m \times 2 g\) x=m g x [\(v^2=2 g x]\)

Total energy, TE = mg

At point C: 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.

PE =0

KE =\(\frac{1}{2} m v^2=\frac{1}{2} m\left(2 g^h\right)\)=m g h [ \(v^2=2 g^h\)]

Total energy, TE = mg

Thus, the sum of the potential energy and kinetic energy of the object would be the same at all points,

PE + KE = Constant Or mgh+\(\frac{1}{2} m v^2\)= constant

This verifies the law of conservation of energy.

Example 9. An object of mass 10 kg is dropped from a height of 5 m. Fill in the blanks by computing the potential energy and kinetic energy in each case.(Take, g = 10 ms^-2]
Answer:

Given, mass, m =10 kg Height, h = 5 m

Acceleration due to gravity, g = 10 ms^-2

At height b = 5 m,

KE = 0, as v = 0

PE = mgt =10x 10x 5 = 500J

Total energy (KE + PE) at height 5 m = 500 J

At height h = 4m,

KE =\(\frac{1}{2} m v^2=\frac{1}{2} m \times 2 g y [ v^2=2 g]\)

Distance covered, s = 5-4 = 1 m

KE =\(\frac{1}{2} \times 10 \times 2 \times 10 \times 1=100 \mathrm{~J}\)

PE = mgt =10 \(\times 10 \times 4=400 \mathrm{~J}\)

Total energy \((\mathrm{KE}+\mathrm{PE})\) at height 4 \(\mathrm{~m}=(100+400)=500 \mathrm{~J}\)

At height b=3 m,

s =5-3=2 m

KE =\(\frac{1}{2} m \nu^2=\frac{1}{2} m \times 2 g r=m g s=10 \times 10 \times 2=200 \mathrm{~J}\)

Therefore, PE =\(\mathrm{mgh}=10 \times 10 \times 3=300 \mathrm{~J}\)

Total enetgy (KE + PE) at height 3 m =(200+300)=500 J

At height b=2 m,

s =5-2=3 m

KE=m g y=10 \(\times 10 \times 3=300 \mathrm{~J}\)

PE = mgh =10 \(\times 10 \times 2=200 \mathrm{~J}\)

Tatal encrgy (KE + PE) at height 2 m = 300+200=500 J

At height k=1 m,

s = 5-1 = 4 m

KE = \(\frac{1}{2} m v^2=\frac{1}{2} \times m \times 2 \mathrm{~g}\)

= mgr =10 \(\times 10 \times\) 4=400 J

PE = mgt =10 \(\times 10 \times 1=100 \mathrm{~J}\)

Total energy( KE + PE ) at height 1 m =(400+100)=500 J

At just above the ground, h=0,

s=5-0=5 m

KE=\(\frac{1}{2} m v^2=m_S=10 \times 10 \times 5=500 J\)

PE = m \(\mathrm{mg}^{\prime}=10 \times 10 \times\) 0=0

Total energy (KE + PE) at just above the ground

=(500+0)=500 J

Thus, total mechanical energy remains constant at each height, which proves that energy is always conserved.

Example 10. A man is moving with a high velocity of 30 ms^-1. Determine the total mechanical energy of the man weighing 60 kg, if he is on a height of 50 m at this speed. [Take, g =10 ms^-2)

Answer:

Given, the mass of man, m=60 kg

Velocity, v=30 \(ms^{-1}\), height, b=50 m

The total energy (TE) of the man at a height of 50 m is given by

TE = PE + KE

where, PE = potential energy ( = mg )

and KE = kinetic energy \(\left(=\frac{1}{2} m y^2\right)\)

TE =m g h+\(\frac{1}{2} m v^2\) [from Eq. (1)]

=60 \(\times 10 \times 50+\frac{1}{2} \times 60 \times(30)^2\)

=30000+27000

=57000 J

Transformation of Energy (Are Various Energy Forms Interconvertible?)

1. One form of energy can be converted into another form of energy and this phenomenon is called transformation of energy.
2. When an object is dropped from some height, its potential energy continuously converts into kinetic energy.
3. When an object is thrown upwards, its kinetic energy continuously converts into potential energy.

For Example:

1. Green plants prepare their food (stored in the form of chemical energy) by using solar energy through the process of photosynthesis.
2. When we throw a ball, the muscular energy which is stored in our body gets converted into the kinetic energy of the ball.
3. The wound spring in the toy car possesses potential energy. As the spring is released, its potential energy changes into kinetic energy due to which, the toy car moves.
4. In a stretched bow, potential energy is stored. As it is released, the potential energy of the stretched bow gets converted into the kinetic energy of the arrow which moves in the forward direction with large velocity.

Rate of Doing Work: Power

1. The rate of doing work or the rate at which energy is transferred used or transformed to other forms is called power.
2. If work W is done in time t, then
3. Power, P=\(\frac{\text { Work }}{\text { Time }}\)
4. P=\(\frac{\mathbb{W}}{t}\)
5. The SI unit of power is watt in honor ofJarr.c: Wait having the symbol W. We express a larger rate of energy transfer in kilowatt (kW).
6. 1 W=1 \(\mathrm{Js}^{-1}\) or 1 kW=1000 W=1000 \(\mathrm{Js}^{-1}\)
7. 1 MW = \(10^6\) W, 1 (horsepower) HP =746 W

Average Power

1. Average power is defined as the ratio of total work done to the total time taken. An agent may perform work at different rates at different intervals of time.
2. In such a situation, the average power is considered by dividing the total energy consumed by the total time taken.
3. Average power =\(\frac{\text { Tocal energy consumed }}{\text { Total time raken }}\)

Example 11. A boy does 400 J of work in 20 s and then he does 100 J or work in 2s. Find the ratio of the power delivered by the boy in two cases.
Answer:

Case 1 Work done by the boy. \(W_1\)=400 J

Time taken, \(z_1\)=20 s

power, \(P_1\)=z

Case 2 Work done by the bor, \(W_2\)=100 J

Time taken, \(s_2\)=2 s,

power, \(R_2\)= ?

Power, P=\(\frac{\text { Work dont }(W)}{\text { Time taken }(t)}\)

⇒ \(R_1=\frac{W}{t_1}=\frac{400}{20}=20 W\)

and \(R_2=\frac{W_2}{t_2}=\frac{100}{2}=50 W\)

Time raken,\(s_2=2 s\),

Power, P=\(\frac{\text { Work done }(W)}{\text { Time taken }(t)}\)

and \(P_2=\frac{W_2}{t_2}=\frac{100}{2}=50 W\)

So, \(\frac{B_1}{P_2}=\frac{20}{50}=2: 5\)

Example 12. A boy of mass of 55 kg runs up a staircase of 50 steps in 10 s. If the height of each step is 10 cm, find his power. Take g =10 m/s².
Answer:

Weight of the boy = mg

= 55×10 = 550 N

Height of the stains, h=\(\frac{50 \times 10}{100}\)=5 m

Time taken to dim =10 s

Power, p =\(\frac{\text { Work done }}{\text { Time taken }}\)

=\(\frac{\text { mgt }}{t}=\frac{550 \times 5}{10}=275 W\)

Question 13. The heart does 1.2 J of work in each _g Hp heartbeat. How many times per minute does it beat, if its power is 2 W?
Answer:

Here, work done in each heartbeat =1.2 J

t=1 min =60 s, power, P = 2W = 2 \(Js^{-1}\)

Total work done =P x r = 2 x 60=120 J

Number of times heart beats per minute

= \(\frac{\text { Total work done }}{\text { Work done in each beart beat }}\)

= \(\frac{120}{1.2}=100 \text { times }\)

Example 14. A horse exerts a pull on a cart of 500 N so that the horse cart system moves with a uniform velocity of 36 km^-1. What is the power developed by the horse in watts as well as in horsepower?
Answer:

Given, F=500 N

v=\(36 \mathrm{kmh}^{-1}=\frac{36 \times 1000}{3600} \mathrm{~ms}^{-1}=10 \mathrm{~ms}^{-1}\)

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

\(p^2 =\frac{W}{t}=\frac{F_s}{t}\) [ W=\(F_s\) and v = t/t]

=F v=500 x 10=5000 W

In horsepower.

P=\(\frac{5000}{746}\)=6.70 HP

Activity Zone

Activity 1

Objective: To understand why we use the term work differently in science.

Procedure

We come across several activities which are normally considered as work in our day-to-day life.

For each of these activities, the teacher should ask the following questions

1.  What is the work being done?
2. What is happening to the object?
3. Who (what) is doing the work?

Conclusion

1.  When an object moves a distance by applying force on it, work is said to be done.
2. The object moves under the influence of the applied force.
3. The agency that exerts the force is doing work.

Question 1. Work done by the force depends on.
Answer:

Work done by the force depends on the path or displacement of the body.

Question 2. Net work during the motion of the Earth around the Sun Is zero. Why?
Answer:

Net work during the motion of the Earth around the Sun is zero because the angle between force and displacement is 90°.

Question 3. What Is the SI unit of work?
Answer:

The SI unit of work is N-m or joule (J).

Question 4. Give mathematical expression for work done.
Answer:

The mathematical expression for work done (W) is as follows: W = Fs cos θ

where F is force, s is displacement, and θ is the angle between F and s.

Question 5. During a circular motion, what will be the work done?
Answer:

During a circular motion, force is perpendicular to the displacement, so the net work done is zero.

Activity 2

Objective: To understand how work is done.

Procedure:

1. Think of some situations from your daily life involving work.
2.  List them.
3.  Discuss with your friends whether work is being done in each situation or not.
4. Try to reason out your response by taking an example of an engine pulling a train.

Conclusion

The engine exerts a force on the train. Work is being done on the train due to which the train begins to move.

Question 1. 5 N of force Is applied to an object, but the object does not move. Then, how much work Is being done?
Answer:

If there is no displacement due to the application of force, then the net work done will be zero.

Question 2. What are the two conditions that need to be satisfied during work?
Answer:

The two conditions that need to be satisfied during work are

1. A force should act on an object.
2. There must be displacement due to applied force.

Question 3. When an engine pulls a train, who exerts force on It?
Answer:

When an engine pulls a train, the engine exerts a force on it

Question 4. 2 N of force displaces a body by 2 m, the work done will be.
Answer:

Work done by the given force = Force x Displacement =2 x2 = 4 J

Question 5. N-m Is the SI unit of which physical quantity?
Answer:

Ans N-m is the SI unit of work done.

Activity 3

Objective: To understand how work is not done.

Procedure:

1. Think of some situations when the object is not displaced despite a force acting on it Also, think of situations when an object gets displaced in the absence of a force acting on it
2. List all the situations that you can think of for each.
3. Discuss with your friends whether work is done in these situations. Suppose, we work hard to push a huge rock.
4. The rock does not move despite all the effort and we get completely exhausted. However, we have not done any work on the rock as there is no displacement of the rock.
5. Suppose a car is moving at a constant speed. The car moves without any external force acting on it However, there is no work done as there is no external force acting on the car.

Conclusion

1. When a force is applied to an object and it is not displaced from its position, the work done will be zero.
2. When an object gets displaced in the absence of a force acting on it, the work done will be zero.

Question 1. When a bullock pulls a cart, who does work?
Answer:

When a bullock pulls a cart, work is being done by the bullock.

Question 2. We pick a suitcase from the Earth to a height of h. Net work done by us will be (where, mass of suitcase =m, acceleration due to gravity =g).
Answer:

Net work done during the pickup of a suitcase to a height h is W = Fxs = mg

Question 3. Net work done during motion in a circular path is
Answer:

Net work done during motion in a circular path is zero as the angle between force and displacement is 90°.

Question 4. If a body does not move by the application of force, the net work done will be
Answer:

If a body does not move by the application of force, then the work done will be zero because of zero displacement.

Question 5. A waiter is carrying a tray to the table, the net work done by the waiter will be
Answer:

A waiter is carrying a tray to the table, the net work done by him will be zero as the weight of the tray is acting downwards and he is moving horizontally. The angle between force and displacement is 90°.

Hence, work done = 0.

Activity 4

Objective: To understand that the work done by a force can be either positive or negative.

Procedure

1. Lift an object. Work is done by the force exerted by you on the object. The object moves upwards.
2. The force exerted by you is in the direction of displacement. However/there is the force of gravity acting on the object

Conclusion

1. The work done by the force applied by you on the object is positive because it acts in the direction of displacement of the object
2. As the object moves up the force of gravity acts in the downward direction, i.e. in the opposite direction of the displacement of the object.
3. Hence, the work done by the force of gravity on the object is negative.

Question 1. What is the direction of force and displacement, if net work done is zero?
Answer:

If the net work done is zero, it means the direction of force is perpendicular to the direction of displacement.

Question 2. Give the work done in the given diagram.

Answer:

Work done when force and displacement are in opposite direction will be negative, i.e. W = -Fs

Question 3. Give the work done in case of a given diagram

Answer:

Work done in the given diagram is (W) = Fs, as the angle between F and s is 0°.

Question 4. What is the angle between force and displacement for maximum work?
Answer:

Work done will be maximum, if

cos θ = 1 or 0 = 0°, i.e. W = Fs cosθ°=1

Question 5. What is the work done by frictional force when the body is dragged along a rough surface?
Answer:

Work done is negative because the displacement of the body and frictional force is in the opposite direction

Activity 5

Objective: To understand that a falling body possesses kinetic energy due to its motion.

Procedure

1. Take a heavy ball. Drop it on a thick bed of sand. A wet bed of sand would be a better way than any other material.
2. Drop the ball on the sand bed from a height of about 25 cm. The ball creates a depression.
3. Repeat this activity from heights of 50 cm, lm, and 1.5 m.
4. Ensure that all the depressions are distinctly visible.
5. Mark the depressions to indicate the height from which the ball was dropped. Compare their depths.

Observation

1. The depression is deepest when the ball is dropped from 1.5 m height.
2. The depression is shallowest when the ball is dropped from 25 cm height, as it is the minimum height from which, the ball is dropped.

Conclusion

The larger the height from which the ball is dropped, the larger the kinetic energy gained by the ball on reaching the ground and more will be its capability to do work.

Question 1. A stone Is dropped from some height. What kind of energy is present in stone on reaching the ground?
Answer:

On reaching the ground, the stone has kinetic energy.

Question 2. A moving object has velocity due to which it has which type of energy?
Answer:

A moving object has velocity due to which it has kinetic energy.

Question 3. If two balls A and 8 are dropped from the same height. Which one has maximum kinetic energy?
Answer:

If two balls are dropped from the same height, then the heavier ball has greater kinetic energy w.r.t. the lighter one.

Question 4. If two bodies A and 8 of the same mass fall from height hJ and h, respectively, on the sand, where h₂ >hv Which body has more energy?
Answer:

Body B falling from height h₂ has more energy.

Question 5. A ball Is allowed to fall freely from a tower. Which energy Is gained at every point during the fall?
Answer:

Kinetic energy.

Activity 6

Objective: To understand the conversion of energy from one form to another.

Procedure

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

Observation and Conclusion

1. A list of such activities and gadgets is as follows:
2. Electric cell → Chemical energy changes into electrical energy.
3. Electric motor  → Electrical energy changes into mechanical energy.
4. Electric heater → Electrical energy changes into heat energy.
5. Dynamo → Mechanical energy changes into electrical energy.
6. Headphone → Electrical energy changes into sound energy.
7. Hydroelectric power station → Mechanical energy changes into electrical energy.
8. Microphone → Sound energy changes into electrical energy.
9. Steam engine → Heat energy changes into mechanical energy.
10. Photoelectric cell → Light energy changes into electrical energy.

Question 1. In a photocell, which type of energy conversion takes place?
Answer:

In photocell, light energy is converted into electrical energy.

Question 2. In solar cells, light energy Is converted to what?
Answer:

In solar cells, Sun energy is converted into electrical energy.

Question 3. Give the sequence of energy changes during the production of electricity In the dam.
Answer:

In the production of electricity in the dam, firstly stored potential energy of water is converted into mechanical energy to move the turbine and this mechanical energy gets converted into electricity or electrical energy by a dynamo.

Question 4. Give the energy change In the electric motor.
Answer:

In an electric motor, electrical energy is converted into mechanical energy (to move the body).

Question 5. In the headphones, which type of energy gets converted into sound energy?
Answer:

In the headphones, electrical energy gets converted into sound energy.

Activity 7

Objective: To understand that the sum of kinetic energy and potential energy of an object is its total mechanical energy.

Procedure

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 ms^-2.

Calculation:

At height h=4 m ,

⇒ \(E_P\)=m g h=20 x 10 x 4=800 J

V = 0

⇒ \(E_K=\frac{1}{2} m v^2\)=0

Total energy =\(E_p+E_K\)=800+0=800 J

At height h=3 m,

⇒ \(E_P=m g t=20 \times 10 \times 3\)=600 J

Distance covered, s=4-3=1 m

From \(v^2-w^2\)=2 g s,

⇒ \(v^2 =u^2+2 g s \Rightarrow v^2\)=0+2 g s=2 g s

⇒ \(E_K =\frac{1}{2} m v^2=\frac{1}{2} m \times\) 2 g s=m g s

=20 \(\times 10 \times 1=200 \mathrm{~J}\)

Total energy =\(E_P+E_K\)=600+200=800 J

At height h=2 m,

⇒ \(E_P =m g h=20 \times 10 \times 2\)=400 J

s =4-2=2 m

⇒ \(E_E =m g r=20 \times 10 \times\) 2=400 J

Total energy =\(E_P+E_K\)=400+400=800 J

At height h =1 m,

⇒ \(E_P =m g h=20 \times 10 \times 1\)=200 J

s =4-1=3 m

⇒ \(E_K =m g s=20 \times 10 \times\) 3=600 J

Total energy =\(E_{\Gamma}+E_K\)=200+600=800 J

At just above the ground h=0,

s =4-0=4 m

⇒ \(E_r\)=m g h=0

⇒ \(E_K\) =m g s=20 x 10 x 4=800 J

Total energy =\(E_R+E_K\)

=0+800=800 J

Therefore, the table will be as follows:

Conclusion:

The total mechanical energy remains constant at each height, hence energy is conserved.

Question 1. When a body falls from a certain height, which energy transformation takes place?
Answer:

When a body falls suddenly, then its potential energy gradually gets converted into kinetic energy. On just reaching the ground, the whole potential energy of the body gets converted into kinetic energy.

Question 2. A car accelerates up a hill. What happens to its kinetic energy and its potential energy?
Answer:

The speed of the car is increasing so its kinetic energy increases. Going up the hill implies a gain in the vertical height so its potential energy increases.

Question 3. Give the variation of KE with height (h) graphically.
Answer:

Question 4. Give the variation of PE with height (h) graphically.
Answer:

Question 5. A ball Is thrown upwards from point A. It reaches up to the highest point 8 and then returns what Is Its PE at 8?
Answer:

From the law of conservation of energy, PE at point B = KE at point A.

Activity 8

Objective: To understand the agents that transfer energy to do work at different rates.

Procedure

1. Consider two children, say A and B. Let us say, they have the same weight. Both start climbing up a rope separately and both reach a height of 8 m.
2. For this A takes 15 s while B takes 20 s to accomplish the task.
3. Work done by each child is mgh, where mg is the weight of each child.

Observation

The same amount of work is being done by both but A takes less time (15 s) while B takes more time (20 s) to do the same work.

Conclusion

1. So, A has done more work in 1 s.
2. Hence, A has more power than B.

Question 1. Give the relation between work and power.
Answer:

The relation between work and power is as follows:

Power, P=\(\frac{\text { Work }}{\text { Time }}=\frac{W}{t}
Time t\)

Question 2. A body takes 2 s to do 10 J of work. What is its power?
Answer:

Power =\(\frac{\text { Work done }}{\text { Time taken }}=\frac{10}{2}=5 \mathrm{~W}\)

Question 3., Boy A does 400 J of work In 10 min her boy 8 does 500 J of work in 20 min. Who expends more power A or 8?
Answer:

Power expended by A=\(\frac{W}{t}=\frac{400}{10 \times 60}=0.67 W[ 1 \mathrm{~min}=60 \mathrm{~s}]\)

and power expended by B=\(\frac{W}{t}=\frac{500}{20 \times 60}\)=0.42 W

A expends more power.

Question 4. If a force F acts on the body and moves it with constant velocity v. Then, the power of the body will be
Answer: 

The power of a body moving with velocity v by a force F is given by

Power, P = Force x Velocity ⇒ P = Fv

Question 5. The unit kWh Is used for which physical quantity?
Answer:

kWh is the unit of energy. It is used for commercial purposes.

Work, Energy, And Power Question And Answers

Question 1. A force of 7 N acts on an object. The displacement is say 8 m, in the direction of the force. Let us take it that the force acts on the object through the displacement. What is the work done in this case?

Answer:

As work done, W = Fs[Work done is positive because force and displacement are in the same direction] W=7×8=56J

Question 2. When do we say that work is done?

Answer:

If a force acting on a body causes some displacement, then we can say that work is being done by the force on the body that is displaced.

Question 3. Write an expression for the work done when a force is acting on an object in the direction of its displacement.
Answer:

Expression for the work done is given by

Work done, W = + Fs [F and s are in the same direction] where, F = force, s = displacement,

So, W = Fs

Question 4. Define 1 J of work.

Answer:

1 J of work is defined as 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.

Question 5. A pair of bullocks exerts a force of 140 N on a plow. The field being ploughed is 15 m long. How much work is done in plowing the length of the field?

Answer:

Given, force, F =140 N, displacement, s =15 m

Work done, W = + F . s [Fand s are in the same direction]

So, W =F-s = 140×15 =2100J

Question 6. What is the kinetic energy of an object? kinetic energy at the surface of the earth 

Answer:

The kinetic energy of an object is defined as the energy due to its motion.

Question 7. Write an expression for the kinetic energy of an object.

Answer:

The expression for kinetic energy for an object is given by KE=\(\frac{1}{2} m v^2\)

where, KE = kinetic energy, w = mass of the body, v – velocity of the body.

Question 8. The kinetic energy of an object of mass m moving with a velocity of 5 ms^-1 is 25 J. What will be its kinetic energy when its velocity is increased three times?

Answer:

As for the question,

Kinetic energy \(\left(\mathrm{KE}_i\right)\) initially is given by \(\frac{1}{2} m v_i^2\) where, m= mass of the body, v_i= initial velocity

⇒ \(\mathrm{KE}_l \left.=\frac{1}{2} m v_i^2\left[\text { given, } \mathrm{KE}_i=25\right], w_i=5 \mathrm{~ms}^{-1}\right]\)

25 =\(\frac{1}{2} m\left(5^2\right)\)

m =\(\frac{25 \times 2}{5 \times 5}\)

m =2 kg

Now, as from the question, final velocity \(\left(v_f\right)\) becomes 3 times its initial velocity, i.e.

⇒ \(v_f=3 v_i \Rightarrow v_f=3 \times 5=15 \mathrm{~ms}^{-1}\)

Now, kinetic energy, KE =\(\frac{1}{2} m v_f^2=\frac{1}{2} \times 2 \times 15 \times 15\)

=225

Question 9. What is power?

Answer:

Power is defined as the rate of doing work. If the work done by an object in time t is W Then, power, P = \(\frac{W}{t}\)

Its unit is Js^-1 or watt.

Question 10. Define 1 watt of power.

Answer:

Power, P=\(\frac{W}{t}\), If W =1 J, t=1 s

Then, P=\(\frac{1 \mathrm{~J}}{1 \mathrm{~s}}=1 \mathrm{~W}\)

The power of an object is said to be 1 watt if it does 1 J of work in 1 s.

Question 11. A lamp consumes 1000 J of electrical energy in 10 seconds. What is the power?

Answer:

Given energy = 1000

i.e. wark done, W=1000

Time, r=10 s

Power of lamp, P=\(\frac{\mathrm{W}}{t}=\frac{1000}{10}\)=100 W

Question 12. Define average power.
Answer:

Average power is defined as the ratio of total work done to the total time taken.

Exercises

Question 1. Look at the activities listed below. Reason out whether or not work is done in the light of your understanding of the term ‘work.’

1. Seema is swimming in a pond.
2. A donkey is carrying a load on its back.
3. A windmill is lifting water from a well.
4. A green plant is carrying out photosynthesis.
5. An engine is pulling a train.
6. Food grains are getting dried in the Sun.
7. A sailboat is moving due to wind energy.

Answer:

1. Work is being done by Seema because she displaces the water by applying force.
2. No work is being done by the gravitational force because the direction of force, i.e. load is vertically downward and displacement is horizontally. If displacement and force are perpendicular, then no work is done.
3. Work is done because the mill is lifting the water, i.e. it is changing the position of water.
4. No work is done because there is no force and displacement.
5. Work is done because the engine is changing the position of the train.
6. No work is done because there is no force and no displacement.
7. Work is done because of the force acting on the boat, it starts moving.

Question 2. An object thrown at a certain angle to the ground moves in a curved path and falls back to the ground. The initial and the final points of the path of the object lie on the same horizontal line. What is the work done by the force of gravity on the object?

Answer:

As we know work done is the product of force and displacement and here in this case, displacement in the direction of gravitational force (change in height) is zero, so work done by the force of gravity on the object is zero.

Question 3. A battery lights a bulb. Describe the energy changes involved in the process.

Answer:

A battery contains chemicals and supplies electrical energy. So, a battery converts chemical energy into electrical energy.

In an electrical bulb, the electrical energy is first converted into heat energy. This heat energy causes the filament of the bulb to become white-hot and produce light energy.

Thus, the energy changes are Chemical energy →  Electrical energy → Heat energy → Light energy

Question 4. Certain force acting on a mass 20 kg mass changes its velocity from 5 ms^-1 to 2 ms^-1. Calculate the work done by the force.

Answer:

Given, mass, m = 20 kg Initial velocity, u – 5 ms 1 Final velocity, v = 2 ms^-1

Work done by the force = Change in kinetic energy = Final kinetic energy – Initial kinetic energy

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

= \(\frac{1}{2} m\left(y^2-v^2\right)=\frac{1}{2} \times 20\left[(2)^2-(5)^2\right]\)

= 10 (4-25)=10 x (- 2,1) = – 210 J

Question 5. A mass of 10 kg is at point A on the table. It is moved to a point. If the line joining A and 6 is horizontal, what is the work done on the object by the gravitational force? Explain your Solwer.

Answer:

Here, both the initial and final positions are on the same horizontal line. So, there is no difference in height,

i.e. h =0. where, h = difference in the heights of initial and final positions of the object.

We know that work done by gravitational force, W = mg
Work done, W = mg x 0 = 0

Question 6. The potential energy of a freely falling object decreases progressively. Does this violate the law of conservation of energy? Why?

Answer:

Indeed, the potential energy of a freely falling object decreases progressively. But as the object falls Ground^ down, its speed increases, i.e. the kinetic energy of the object increases progressively (kinetic energy will increase with the increase in speed).

Now, we can say that the law of conservation of energy is not violated, because the decrease in potential energy results in the increase of kinetic energy.

Question 7. What are the various energy transformations that occur when you are riding a bicycle?

Answer:

In the case of riding a bicycle, the muscular energy is converted into the kinetic energy of the bicycle.

Question 8. Does the transfer of energy take place when you push a huge rock with all your might and fail to move it? Where is the energy you spend going?

Answer:

When we push a huge rock, then the rock also exerts a large force on us (according to Newton’s third law of motion). The muscular energy spent by us in the process is used to oppose the huge force acting on us due to the rock.

Question 9. An object of mass 40 kg is raised to a height of 5 m above the ground. What is its potential energy, if the object is allowed to fall? Find its kinetic energy when it is halfway down.

Answer:

Given, mass, m = 40 kg, height, h = 5 m As potential energy is given by PE = mgh

So, PE= 40 x 10 x 5 =2000J [v =10 ms^-2]

When it is allowed to fall, its PE gets converted into kinetic energy KE. So, when it reaches to half-way, half of its PE gets converted to KE.

So, mg \(\frac{b}{2}=\frac{1}{2} m v^2\) [where, v= velocity at the bottom]

So,KE =m g \(\frac{b}{2}\)

=40 \(\times 10 \times \frac{5}{2}\)=1000 J

Question 10. What is the work done by the force of gravity on a satellite moving around the Earth? Justify your answer.

Answer:

Work done by the force of gravity on a satellite moving around the Earth is zero. Because of the angle between force (centripetal) and displacement in the case of circular motion.

So, work done, W = 0

Question 11. Can there be displacement of an object in the absence of any force acting on it? Think, and discuss this question with your friends and teacher.

Answer:

Yes, if an object moves with a constant velocity, i.e. there is no acceleration, then no force acts on it. As the object is moving, so it gets displaced from one position to another position.

Question 12. A person holds a bundle of hay over his head for 30 min and gets tired. Has he done some work or not? Justify your answer.

Answer:

On holding a bundle of hay over the head, the work done by the person is zero because there is no displacement.

Question 13. Illustrate the law of s conservation of energy by discussing the energy changes which occur when we draw pendulum bob to one side and allow it to oscillate. Why does the bob eventually come to rest? What happens to its energy eventually? Is it a violation of the law of conservation of energy?

Answer:

1. Let a simple pendulum be suspended from a rigid support S and OS be the equilibrium position of the pendulum. Let the pendulum be displaced to a position P, where it is at rest.
2. At position P, the pendulum has potential energy {mgh). When the pendulum is released from position P, it begins to move towards position O.
3. The speed of the pendulum increases and its height decreases which means the potential energy is converted into kinetic energy.
4. At position O, the whole of the potential energy of the pendulum is converted into its kinetic energy.
5. Then, the pendulum swings to the other side due to the inertia of motion. As the pendulum begins to move towards position Q, the speed of the pendulum decreases and height increases which means kinetic energy is converted into potential energy.
6. At point Q, the whole of the kinetic energy is converted into potential energy. Thus, we find that the potential energy is converted into kinetic energy and vice-versa during the motion of the pendulum. But the total energy remains constant.
7. When the pendulum oscillates in air, the air friction opposes its motion. So, some part of the kinetic energy of the pendulum is used to overcome this friction.
8. With time, the energy of the pendudecreasessing and finally becomes zero.
9. The energy of the pendulum is transferred to the atmosphere. So, energy is being transferred, i.e. is converted from one form to another. So, no violation of the law of conservation of energy takes place.

Question 14. An object of mass m is moving with a constant velocity How much work should be done on the object to bring the object to rest?

Answer:

Concept Change in kinetic energy (KE) =Work done

Given, mass = m, initial velocity, u = v

Final velocity, v = 0

So, W=\(\frac{1}{2} m v^2-\frac{1}{2} m s^2\)

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

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

Hence, the work that should be done to bring the object to rest is \(\frac{1}{2} m v^2\).

Question 15. Calculate the work required to be done to stop a car of 1500 kg moving at a velocity of 60 km-1.

Answer:

Concept Change in kinetic energy is equal to the work done W.

Given, initial velocity, u=60 \(\mathrm{kmh}^{-1}\)

=60 \(\times \frac{5}{18}=\frac{50}{3} \mathrm{~ms}^{-1} {\left[1 \mathrm{kmh}^{-1}=5 / 18 \mathrm{~ms}^{-1}\right] }\)

Final velocity v=0

So, the magnitude of change in kinetic energy =W

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

W =\(\frac{1}{2} m\left(v^2-s^2\right)\)

=\(\frac{1}{2} \times 1500 \times\left(\frac{-50 \times 50}{9}\right)\)

=-\(\frac{1}{2} \times \frac{1500 \times 50 \times 50}{9}\)

W\(\left.=-\frac{625000}{3}=-208333.3\right)\)

Hence, the work required to be done to stop a car is 208333.3 J.

Question 16. In each of the following, a force F is acting on an object of mass m. The direction of displacement is from West to East shown by the longer arrow. Observe the figure and state whether the work done by the force is negative, positive, or zero.

Answer:

In given. (1), the angle between F and s is 90°, so the work done is zero.

In given. (2), the angle between F and s is 0°, so the work done is positive.

In given. (3), the angle between F and s is 180°, so the work done is negative.

Question 17. Soni says that the acceleration in an object could be zero even when several forces are acting on it. Do you agree with her? Why?

Answer:

Yes, I agree with Soni, the acceleration of an object can be zero even when several forces are actonng onacting the resultant of all the forces acting on an object is zero.

Question 18. A freely falling object eventually stops on reaching the ground. What happens to its kinetic energy?

Answer:

When a freely falling object strikes the Earth, some sound and heat are produced. So, the kinetic energy of the object converts into sound energy and heat energy.

Summary

Work is said to bo done in a physical activity involving n form and movement in the direction of force and the process an equal amount of energy is used up.

Two conditions need to be satisfied for work to be done.

1. A force should act on the object.
2. The object must be displaced.

Work done by a force on an object is equal to the magnitude of the force multiplied by the distance moved in the direction of force, i.e. U’ = F xs.

• The SI unit of work is joule (J).
• 1 J is the amount of work done on an object when a force of 1 X displaces it by 1 m along the line of action of force.
• Energy is the ability to do work.
• The SI unit of energy is the same as that of work, i.e. joule.
• A larger unit of energy is kilojoule (kJ).
• Relation between Joule and Kilojoule
• 1 kilojoule = 1000 joule.
• The SI unit of work and energy is named after British Physicist, James Prescott Joule.
• Then energy that is possessed by an object due to its motion is called kinetic energy.
• The SI unit of kinetic energy is joule (J).
• The kinetic energy possessed by an object of mass m, moving
  noth a uniform velocity v is KE = \(\frac{1}{2} m v^2\)
• The work-energy theorem is given asW = \(\frac{1}{2} m\left(v^2-u^2\right)\). If initial velocity u = 0, then W = \(\frac{1}{2} m v^2\)
• The energy possessed by a body due to its position or change in its configuration (i.e. shape or position) is known as potential energy.
• The SI unit of potential energy is the joule (J).
• Tho potential energy of an object can be given ns PE = mg.
• According to the law of conservation of energy, energy can neither be created nor be destroyed but it can be transformed from one form to another.
• During the free fall of a body, its energy is always conserved. The sum of the potential energy and kinetic energy of an object is the same at all points.
• One form of energy can be converted into other forms of energy, this phenomenon is known as transformation of energy.

Some Energy Transformations

1. Electric Motor Electrical energy into mechanical energy.
2. Electric Generator Mechanical energy into electrical energy.
3. Steam Engine Heat energy into kinetic energy.
4. Electric Bulb Electrical energy into light energy.
5. Dry Cell Chemical energy into electrical energy.
6. Solar Cell Light energy into electrical energy.
7. The rate of doing work or the rate at which the energy is
   transformed is known as power (P).
8. Power =\(\frac{\text { Work done }}{\text { Time }}\)
9. P =\(\frac{W}{t}\).
10. The SI unit of power is watt (W).
11. The relation between watt and horsepower can be given as 1 (horsepower) HP = 746 W.
12. Average power is defined as the ratio of total work done to total time taken.