 # MSBSHSE Class 9 Science Chapter 2 Work and Energy Solutions

MSBSHSE Class 9 Science Chapter 2 Work And Energy Solutions are important from the viewpoint of your MSBSHSE Class 9 Science examination, as the study material helps the students to score well in exams. The answers, with detailed step -by-step solutions to the questions provided here, is the best resource for the students to grasp the basic concepts of the chapter.

Work and Energy is a significant topic of Science and it is discussed in detail in this chapter covering concepts such as work, energy, mechanical energy, law of conservation of energy, freefall and so on. Students can learn thoroughly about the subject by revising these MSBSHSE Class 9 Solutions of Science Chapter 2 Work and Energy. These solutions are prepared following proper research and will help students to gain a thorough conceptual understanding. The well structured content makes it easier for students to learn and comprehend even the most complex topics from the chapter. The content is created on the basis of the latest MSBSHSE Syllabus for Class 9. The solutions provided here are designed to meet the multiple criteria required to do well in the exams. Practising these questions and answering with the highly relevant answers is the best way to retain the performance through various classes.

## Maharashtra Board Class 9 Science Chapter 2- BYJU’S Important Questions & Answers

1. Explain unit of work as per SI and CGS system.

Answer: Given that, Work = Force × Displacement, the unit of force in SI system is newton (N), while the unit of displacement is metre (m). Hence, the unit of force is newton-metre, also called joule. If a force of 1 newton displaces an object through 1 metre in the direction of the force, the amount of work done on the object is 1 joule.

1 joule = 1 newton × 1 metre. Therefore 1J = 1 N × 1m.

Alternately, the unit of force In CGS system is dyne, while that of displacement is a centimetre (cm). So, the unit of work done is dyne-centimetre, also called an erg.

Now, If a force of 1 dyne displaces an object through 1 centimetre in the direction of the force, the amount of work done is 1 erg. That is, 1 erg = 1 dyne × 1 cm.

2. Define the relationship between joule and erg using the equation.

Answer: It is given that 1 newton (1N) = 105 dyne, while 1m = 102 cm. So, if

Work= forces × displacement, then

1 joule = 1 newton × 1 m

1 joule = 105 dyne × 102 cm

= 107 dyne cm.

Hence, 1 joule = 107 erg.

3. With the force and the displacement specified, define the work done by the force.

Answer: Work done by the force in the various scenarios are explained here:

1. The work done by the force is positive when the force and the displacement are in the same direction (θ=00)

2. Work done by the force is negative when the force and the displacement are in opposite directions (θ=1800)

3. Work done by the force is zero when the applied force does not cause any displacement or when the force and the displacement are perpendicular to each other (θ=900)

4. Give an example of where the work done by the gravitational force is zero.

Answer: Take the example of an artificial satellite moving around the earth in a circular orbit. The work done by the gravitational force is zero as the gravitational force acting on the satellite (along the radius of the circle) and its displacement (along the tangent to the circle) are perpendicular to each other.

5. Write about the National Physical Laboratory.

Answer: The National Physical Laboratory, conceptualized in 1943 was situated in New Delhi and functions under the Council of Scientific and Industrial Research. The institute conducts basic research in the various branches of physics as well as helps various industries and institutes that are engaged in developmental work. The main objective of this institute is to establish national standards of various physical quantities.

6. Calculate the work done to take an object of mass 20 kg to a height of 10 m.

(g = 9 8 m/s2)

Answer: Suppose mass of the object (m) = 20 kg, while height (s) = 10 m

given that g = 9.8 m/s2

The value for F = mg

= 20 × (-9 8)

(Because the displacement is opposite the direction of the force, the negative sign is mentioned.)

Hence, F = -196 N

Now, W = F s

= -196 ×10

Thus, W = -1960 J

(The negative sign appears because the direction of force is opposite to the direction of displacement so that the work done is negative.)

7. Pravin has applied a force of 100 N on an object, at an angle of 600 to the horizontal. The object gets displaced in the horizontal direction and 400 J work is done. What is the displacement of the object?

(cos 600 = 1/2)

Answer: Given here is θ=600 and F=100N, while W= 400J , SO s=?

Take the equation W = F s Cos θ

Replace with values 400= 100 × s × ½

Hence, 400 / 100 = ½ × s

4 × 2 = s

Therefore, s=8 m.

The displacement of the object = 8m.

8. Define energy. What is its unit?

Answer: Energy is the capacity of a body to perform work. Unit of energy is the same as that of work. In the SI system, the unit is joule, while in the CGS system, it is erg.

9. What is kinetic energy?

Answer: Kinetic energy is the energy that an object has because of its motion. The kinetic energy gained by the object is the work done by a force to displace a stationary object through a distance s. Hence, Kinetic energy = work done on the object. That is

K E = F × s.

10. What is the potential energy?

Answer: Potential energy of the object is the energy stored in it because of its specific state or position.

11. Ajay and Atul have been asked to determine the potential energy of a ball of mass m kept on a table as shown in the figure. What answers will they get? Will they be different? What do you conclude from this? Answer: As seen from the picture, the heights of the ball with respect to Ajay

and Atul are different. Also, given that the potential energy is relative, the potential energy with respect to them will also be different.

12. Into which form are the chemical energy stored in the exploding firecrackers converted to?

Answer: Energy is transformed from one form to another. In the example stated, the exploding firecrackers convert the chemical energy stored in them into light, sound and heat energy.

13. What is the law of conservation of energy?

Answer: ‘Energy can neither be created nor destroyed. It can be converted from one form into another. Thus, the total amount of energy in the universe remains constant’.

14. Arrive at the Expression for Kinetic Energy.

Answer: Given that a stationary object of mass m moves because of an applied force with an initial velocity (here u = 0) and applied force be F. This generates an acceleration a in the object, and, following time t, the final velocity of the object becomes equal to v. The displacement during this time is s. Hence, The work done on the object, W = F . s. That is, W = F × s.

Now, as per Newton’s second law of motion, F=ma, similarly based on Newton’s second equation of motion, s= ut + ½ at2. Now, the initial velocity here is 0.

Hence, s=0t + ½ at2= 0 + ½ at2= ½ at2

Therefore, given that W = F × s,

Replace it and you get, W= ma × ½ at2.

And W= ½ m(at)2.

Now if we were to use Newton’s First equation of motion, you get

v= u + at , replace the initial velocity with value and

v=0 + at , Hence, v= at , now v2 = (at)2.

Now, we have already deduced that W= ½ m(at)2, replacing (at)2

W= ½ mv2

Since, kinetic energy of an object is the amount of work done on it, KE = W. Thus, arriving at the expression KE= ½ mv2.

15. A stone having a mass of 250 gm is falling from a height. How much kinetic

energy does it have at the moment when its velocity is 2 m/s?

Answer: Here, mass(m) = 250 gm = 0.25kg and v= 2 m/s. So, use the equation to get KE. KE =½ mv2. Hence, KE = ½ × 0.25 × (2)2 = 0.25 × 2 = 0.5 J.

16. Write the expression for potential energy.

Answer: In order to carry an object of mass ‘m’ to a ‘h’ height above the earth’s surface, a force ‘mg’ has to be applied against the direction of the gravitational force. Here, the amount of work done can be calculated as follows.

Work = force x displacement, replacing the values

W = mg × h= mgh

Hence, the amount of potential energy stored in the object because of its displacement

I.e., P.E. = mgh ( given that W = P.E.). Thus, we can arrive at the conclusion that Displacement to height h causes energy equal to mgh to be stored in the object.

17. 500 kg water is stored in the overhead tank of a 10 m high building. Calculate

the amount of potential energy stored in the water.

Answer: Here, the h is given as 10m , while mass(m) =500kg

Now, given that g=9.8m/s2, calculate PE using the equation.

Therefore, PE= mgh

So, PE= 500 × 9.8 × 10= 49000 J.

18. Draw a diagram for transformation of energy. 19. Explain free fall.

Answer: When we release an object from a height, it is pulled towards the earth because of the gravitational force. An object is said to be in free fall or falling freely, when they fall solely under the influence of gravitational force.

20. Define power.

Answer: The rate at which work is done is known as power. Now, given that W is the work done in time ‘t’, then Power = work / time. Hence, P = W / t.

21. What is the unit of power?

Answer: The unit of work in the SI system is J. Hence, the power of the unit is known as J / s called watt. 1 Watt = 1 Joule / 1 sec. Meanwhile, in the industrial sector the unit used to measure power is termed as ‘horse power.’ 1 horse power = 746 watt.

22. Swaralee takes 40 s to carry a bag weighing 20 kg to a height of 5 m. How much power has she used?

Answer: Suppose mass(m) = 20 kg, h=5m and time(t) = 20s. Hence, to calculate the force that has to be applied by Swaralee, use the equation F=mg given that g = 9.8 m/s2

Hence, F= mg = 20 × 9.8 = 19 N. Now, to get the work done by Swaralee to carry the bag to a height of 5m, use W = Fs, so replacing the values you get

W = F × s = 196 × 5 = 980 J. Now, get Power using the equation P= work divided by time. Therefore, P= W/t = 980 / 40 = 24.5 Watt.

23. A 25 W electric bulb is used for 10 hours every day. How much electricity does it consume each day?

Answer: Here, P is given as 25 W with W = 0.025 kW. Now, given that

Energy consumed= power × time

Energy = 0.025 × 10 =0.25 kW hr.

24. Explain the difference between kinetic energy and potential energy.

Answer: The energy present in the body due to property of motion is called kinetic energy, whereas potential energy is the energy present due to the property of the state. While kinetic energy can be easily transferred from one body to the other, potential energy is not transferable. Now, find the differences between kinetic and potential energy more in-depth from the link.

25. Prove that the kinetic energy of a freely falling object on reaching the ground is nothing but the transformation of its initial potential energy.

Answer: See, the figure given below: 1. In this image, the point A is at a height h from the ground, while the point B is at a distance x, vertically below A and the point C is on the ground directly below A and B. Now, when the object is at stationary A, the initial velocity u= 0. Here, K = ½ × mass × velocity2. Hence, KE= ½ mu2. Replacing u with the value, you get KE= ½ × m × (0)2= 0. Meanwhile, potential energy is mgh. Now, given that total energy is equal to kinetic energy plus potential energy, the equation would be, Total Energy = KE + PE. Therefore, total energy = 0 + mgh = mgh.
2. Now, take the instance of the object at B after a fall of distance x with velocity vB Here, u = 0, s = x and a=g. Given that v2= u2+2as, replace the equation with values. Thus, vB2= 0 + 2gx= 2gx. Now replacing values for the equation,

KE= ½ mv2= ½ mvB2= ½ m2gx. Hence, KE=mgx. Now, estimating the height of the object when at B= h-x. Therefore, KE= mg(h-x)= mgh-mgx, and so

total energy(TE) = KE + PE = mgx + (mgh-mgx). Hence TE=mgh.

1. Given that velocity of the object will be vC when it reaches the ground, near point C. Here, u = 0, s = h, a = g. Suppose, v2= u2 + 2as, replace the equation with values. vC 2= o2 + 2gh. Since, KE= ½ m vC 2= ½ m(2gh) = mgh. Now, the height of the object (h) from the ground at point C= 0. Hence, PE= mgh = mg × 0= 0. Now, total energy (TE) = KE + PE = mgh + 0= mgh.

Equations (1), (2) and (3) prove that the total energy of the object is the same at the three points A, B and C. Thus, every object has potential energy when at a height above the ground and it keeps getting converted to kinetic energy as the object falls towards the ground. On reaching the ground (point C), all the potential energy gets converted to kinetic energy.

26. Determine the amount of work done when an object is displaced at an angle of 300 with respect to the direction of the applied force.

Answer: Work done is the product of force and displacement, i.e, W = Fs. Now, if the force (F) is applied at an angle (a) to the displacement, then the product of the component of force in the direction of displacement is to be determined. This is given as W = F x s x cos a. Now, given that a= 30˚, W = F x s x cos 30˚. Hence, Work done (W) = Fs √3/2.

27. Why is the work done on an object moving with uniform circular motion zero?

Answer: Work done on an object is denoted as the product of force applied and the displacement covered in the direction of force applied. Hence, the work done in a uniform circular motion is always zero. In circular motion θ is 90° hence, work done by force is zero.

28. An electric pump has 2 kW power. How much water will the pump lift every minute to a height of 10 m?

Answer: Here, P = 2 kW = 2 x 103 watt, whereas t = 1 m = 60 seconds and h=10m. (g=9.8m/s2). Now, the amount of the water lifted by the pump is the mass of water(m).

Given the formula P = W /t = mgh / t and applying the given values to calculate,

2 x 103= m x 9. 8 x 10 / 60

m x 9. 8 x 10= 2 x 103 x 60

Hence, m = 2 x 103 x 60 / 9. 8 x 10= 2 x 102 x 60 / 9.8= 12000/ 9.8 =1224.5 kg

Thus, the amount of water lifted by the pump is 1224.5 kg.

29. If the energy of a ball falling from a height of 10 metres is reduced by 40%, how high will it rebound?

Answer: Suppose the ball is dropped from a height, h of 10 m, with g = 9.8m/s2.

Now, the energy of the ball on falling from is potential energy, Hence

E = mgh (PE). Hence, the potential energy at h = m×9.8 ×10 = 98m J

Now, on reaching the ground level, the ball loses 40% of its initial energy and rebounds back to h1 with its new energy E1. E1== 60% × E . Mass though remains constant. Hence,

E1= mgh1

=m × 9.8 × h1 = 60 / 100 × 9.8 , where the m’s cancel out on division.

Thus, 9.8 × h1= 0.6 × 98

h1= 0.6 × 10 . Hence, h1,= 6m.

30. Ravi applied a force of 10 N and moved a book 30 cm in the direction of the force. How much was the work done by Ravi?

Answer: Given that Force (F) = 10 N and displacement (s) is 30cm= 30/ 100= 0.03m ,calculate the work done. W=?

Apply the given values to the equation W = Force × displacement. Hence,

W= Fs = 10 × 0.03 = 3 Nm or Joules.

Meanwhile, students studying in the 9th standard should know the syllabus thoroughly to excel in their exam. Keep visiting BYJU’S to get the detailed Maharashtra Board syllabus, Maharashtra state Board books, Question papers, Sample papers, etc.

## Frequently Asked Questions on Maharashtra State Board Solutions for Class 9 Science Chapter 2 Work and Energy

### Where can I get Maharashtra State Board Class 9 Science Solutions Chapter 2?

We have provided the Maharashtra State Board Solutions for Class 9 Science Chapter 2 Work and Energy as a scrollable PDF, and we have also mentioned the clickable link for the students to access. Meanwhile, the questions and the solutions  are also available online on our webpage.

### Why is Maharashtra State Board Class 9 Science Solutions Chapter 2 important?

These solutions are significant because they are the perfect guide for the students who are preparing for the Class 9 exams. These solutions cover the topics that forms the basis for the questions that are likely to be asked in the board exams. Preparing with the help of these solutions help students to ace the exams.

### How to study for Maharashtra State Board Class 9 Science Solutions Chapter 2?

Students can solve the questions and refer back to the solutions to identify mistakes and avoid them for exams. Timing the process, helps to manage time better. Solutions are the best tool to self asses your exam preparations.