Chapter 5 Energy In Action, Work, Power, And Energy Very Short Answer Type
Question 1. State the mathematical expression of work done.
Answer: Work = Force x displacement of the point of application in the direction of the force.
W=Fxd
W = Work done;
F = Force acting on body.
d = displacement of the body from the point of application of the force.
Question 2. Define work done by a force.
Answer:
Work done by a force: If the displacement of the point of application of a force takes place in the direction of the force then the force is said to have done the work.
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Question 3. Define work done against a force.
Answer:
Work done against a force: When the displacement of the point of application of a force takes place in the direction opposite to the applied force, then work is said to be done against the force.
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Question 4. What is the practical unit of power?
Answer:
Practical unit: The practical unit of power in both CGS and SI units is Watt.
Question 5. Define watt.
Answer:
Watt: The power of doing 1 erg of work in 1 second is known as 1 watt.
Question 6. Define horsepower.
Answer:
Horse Power (HP): The power to work at the rate of 550 ft-lb per second is called 1 horsepower.
Question 7. Define potential energy.
Answer:
Potential energy: The energy of an object because of its position or shape is called its potential energy.
Question 8. Define Gravitational potential energy.
Answer:
Gravitational Potential energy: The potential energy due to height above the earth’s surface is called gravitational potential energy.
Question 9. Define elastic potential energy.
Answer:
Elastic Potential Energy: When a spring is stretched or compressed from its natural length, it gets extra energy. It can return to its natural length by performing some work. The extra energy stored in a stretched or compressed spring is called its elastic potential energy.
Question 10. Define Kinetic energy.
Answer:
Kinetic energy: It is the energy of a body by its motion.
Question 11. What is mechanical energy?
Answer:
Mechanical energy: The sum of the kinetic energy and the potential energy of an object is called its mechanical energy.
Question 12. State the law of conservation of energy.
Answer:
Law of conservation of energy: Energy can not be created or destroyed. Only one form of energy is transformed to another type of energy. The total amount of energy is constant in the universe.
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Question 13. A heavier body and a lighter body are moving with the same velocity. Which of them can do more work?
Answer: A heavier body can do more work.
Question 14. What is the SI unit of energy?
Answer: Joule.
Question 15. What is the SI unit of power?
Answer: Wait.
Question 16. Is energy a scalar or a vector quantity?
Answer: Energy is a scalar quantity.
Question 17. Give an example of a scalar quantity where three fundamental units are used.
Answer: Power.
Question 18. Is work done due to the rotation of the earth around the sun?
Answer: A force does no work in a direction perpendicular to its direction of application. Here no work is done.
Question 19. Does applied force always perform work?
Answer:
No workforce:
(1) If there is no displacement of the point of application of the force.
(2) Again, if any force acts perpendicularly to the direction of the displacement then no work is done.
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Question 20. A man swims in a river against the current water. If he remains fixed with respect to the shore, does he do work with respect to the shore?
Answer: Here work done by the man is zero as his displacement with respect to the shore is zero.
Question 21. Give the expression of gravitational potential energy.
Answer: Gravitational potential energy = mass of the body (m) x acceleration due to gravity (g) x height of the body above the surface of the earth (h).
Question 22. What is the absolute unit of force in the SI system?
Answer: 1 joule.
Question 23. What is power?
Answer: Work done in unit time is called power.
Question 24. What is the relation between power and force?
Answer: Power = force x velocity.
Question 25. What is the energy of a body?
Answer: Energy is defined as the capacity of a body to perform work.
Question 26. Write down the absolute and gravitational SI units of work.
Answer:
The absolute SI unit of work: Joule.
The gravitational SI unit of work: kg-m.
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Question 27. What is the relationship between work and power?
Answer: Work done (W) = Power (P) x Time (t).
Question 28. What is the gravitational SI unit of power?
Answer: The gravitational SI unit of Power: is kg-m/sec.
Question 29. How many kinds of mechanical energy are there?
Answer: Mechanical energy is of two types (1) Potential energy and (2) Kinetic energy.
Question 30. How is work measured?
Answer: Work done (W) = Force (F) x Displacement (S).
Question 31. What is the practical unit of work?
Answer: The practical unit of work is Joule.
Question 32. Give the expression of kinetic energy.
Answer: Kinetic energy = ½ x mass x\(\left(\text { velocity) }{ }^2\right.\)
Question 33. A bullet of mass 10 g travels at 400 m/s velocity. Find its kinetic energy.
Answer:
Kinetic energy = ½ x mass x \(\left(\text { velocity) }{ }^2\right.\)
= ½×10 g×\((40000 \mathrm{~cm})^2\)=800 J.
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Question 34. The power of an engine is 5KW. Find the work done by it in 1 hour.
Answer:
5 KW = 5 x 1000 Watt = 5000 J/s (∴ 1 watt = 1 d/s)
∴ Work done by the engine in 3600s = (5000 x 3600) J
= 18 x \({10}^6\) J.
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Question 35. How much work is done by a force of 5 N in moving a body through a distance of 2 m in the direction of the force?
Answer:
Here, force (F) = 5 N and displacement (d) = 2 m.
∴ Work done, W = F.d = (5 x 2) J = 10 J.
Question 36. A machine does 2160 joule work in 4 minutes. What is the power of the machine?
Answer:
Here, work done (W) = 2160 J, time taken (t) = 4 minutes = 240 s.
∴ Power of the machine (P)\(=\frac{W}{t}\)=\(\frac{2160 \mathrm{~J}}{240 \mathrm{~s}}\) = 9 watt.
Question 37. A force of 20 N moves a body with a constant velocity of 4 m/s. Calculate the power of the agent.
Answer:
Here, force (F) = 20 N, velocity (v) = 4 m/s.
∴ Power = Force x velocity= (20 x 4) watt = 80 watt.
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Question 38. A 180 kg car engine develops 400 watts for each kg. What force does it exert in moving the car at a speed of 20 m/s?
Answer:
Here, the power of the car engine = (180 x 400) watt = 72000 watts, velocity = 20 m/s.
∴ Power = Force x velocity
∴72000 = Force x 20
∴ Force\(=\frac{72000}{20}\)=3600N
Chapter 5 Energy In Action, Work, Power, And Energy 2 Marks Questions And Answers:
Question 1. When is a work said to be done
(1) by a force and
(2) against a force?
Answer:
(1) When we apply a force on a body at rest and the body is displaced towards the point of application of the force, work is said to be done by the force.
(2) If a heavy body is lifted from the ground and put on the table, then work is done against the force of gravity.
Question 2. What is the relation between Joule and erg?
Answer:
Relation between Joule and erg
We know that, 1Joule = 1 Newton x 1 metre = \({10}^5\) dynes x \({10}^2\) cm
= \({10}^7\) dynes x cm = \({10}^7\) ergs.
Question 3. Define Mechanical energy. State its different forms.
Answer:
Mechanical energy: The energy possessed by a body due to its: state of rest or motion is called mechanical energy.
It is found in two forms :
(1) The potential energy and
(2) Kinetic energy.
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Question 4. Define potential energy. Explain with examples.
Answer:
Potential energy: The energy that a body possesses by virtue of its position or configuration is called the potential energy of the body.
Example : (1) A wound-up watch spring has potential energy because of the wound-up State of its coils. As the spring unwinds, it does work in moving the hands of the watch.
(2) The water placed at a height has the potential energy stored in it and if it is allowed to fall, the falling water can do work like turning a wheel.
Question 5. What is gravitational potential energy? How is it measured?
Answer:
Gravitational potential energy
The gravitational potential energy of a body at a height above the ground is measured by the amount of work done in lifting it up to that height against the force of gravity.
If a body of mass (m) is lifted from the ground to a vertical height (h), then the gravitational potential energy = Force (mg) x displacement (h). Or, Potential energy = mgh.
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Question 6. What is the difference between Power and Energy?
Answer:
Difference between Power and Energy
The power of a body is the work that it can perform in a unit of time. Energy is the total work that the body can do. The S.I. unit of power is 1 Watt while the S.I. unit of energy is 1 Joule. Power and energy both are scalar quantities.
Question 7. A boy weighing 300 N runs up a flight of 30 steps of a staircase each 20 cm high. Calculate the work done.
Answer:
The distance traveled by the boy = 30 x 20 cm = 600 cm=6m. Work done = Force x distance = 300 Nx 6 m= 1800 J.
Question 8. A boy of mass 50 kg runs up flights of 30 steps each 20 cm high in 10 seconds. Find
(1) The work done by the boy
(2) The power developed. (g =9.8 m/\(\sec ^2\) ).
Answer:
The force of gravity on the boy F = mg = 50 x 9.8 = 490 N.
(1) Work done by the boy in climbing = Force x distance
= 490 x 6 = 2940 J.
(2) Power developed \(=\frac{\text { Work done }}{\text { time taken }}\)=\(\frac{2940}{10}\)=294 watt.
Question 9. A man rowing a boat upstream is at rest with respect to the shore. Is he doing any work?
Answer:
The water current is acting a force on the boat opposite to the motion. The man rowing the boat applies an equal amount of force to proceed. Here there is no displacement of the point of application of force. Hence, no work is done.
Question 10. Calculate the kinetic energy of a body of mass 6 kg moving with a speed of 10 m/\(\sec ^2\).
Answer:
The mass of the body m = 6 kg, velocity v = 10 m/\(\sec ^2\)
Therefore, Kinetic energy \(=\frac{1}{2} m v^2\) = ½x 6 x \((10)^2\) = 300J.
Question 11. A man is moving with a box along a horizontal plane from one place to another. Is he doing any work?
Answer:
No work is done by the man because the point of application of the force does not change by carrying the box from one place to another place. Also the vertical component of the weight in the horizontal direction is zero.
Question 12. How is the kinetic energy of a moving truck affected if
(1) Its mass is doubled
(2) Its velocity is doubled?
Answer:
The kinetic energy \(=\frac{1}{2} m v^2\)
(1) If the mass m is doubled (keeping the velocity same), the kinetic energy is doubled.
(2) If the velocity is doubled (keeping the mass the same); the kinetic energy increases four times.
Question 13. State an unbalanced force and indicate the direction and the work it does.
Answer: Two unequal oppositely directed forces acting on a body create an unbalanced force which is equal to the difference between the forces. The unbalanced force causes displacement and work on the body in the direction of the greater force.
Question 14. Are watt and watt-hour the same physical quantities? Explain.
Answer:
Watt and watt-hour are not the same physical quantities. Watt, which means 1 Joule/s is the unit of power, i.e., work done per unit of time, but watt-hour (work/time x time) equals to work or energy. 1 watt-hour is the work performed by a machine or an agent that works at the rate of 1 Joule per second.
Question 15. If an agent does a large amount of work, does it always mean that the agent possesses large power?
Answer:
Only the magnitude of work performed does not clearly indicate the power of anybody. If the time in which the work is performed is mentioned, it gives the idea of power clearly. A large quantity of work done, but in a long time, results to a small power but a smaller quantity of work done in a very short period of time results in a greater power.
Question 16. Define energy. Why is the unit of energy the same as that of work?
Answer:
Energy
Energy is something that can produce mechanical work directly or after some intermediate transformation.
When some work is done on a system or on a body to change its position, shape, or configuration, some energy is stored in the body. If the body returns to its original shape, the same amount of work is released from it.
So, the energy possessed by a body or a system is measured by the quantity of mechanical work actually obtained or obtainable from it. Hence, the unit of energy is same as that of work.
Question 17. Establish the relation between power, force, and velocity.
Answer:
Relation between power, force, and velocity :
⇒ \(\text { Power }=\frac{\text { Work }}{\text { time }}\)
⇒ \(=\frac{\text { force } \times \text { displacement }}{\text { time }}\)
⇒ \(=\text { force } \times \frac{\text { displacement }}{\text { time }}\)
= force x velocity
∴Power = force x velocity
Question 18. A particle starting from rest acquires kinetic energy ‘E’ in time t and its momentum is p. Find its acceleration in terms of E, p, and t.
Answer: Let the acceleration be f, velocity acquired in time ‘t’ be v and the mass of the body be m.
∴ f\(=\frac{v-0}{t}\) or f\(=\frac{v}{t}\). Again, E\(=\frac{1}{2} m v^2\)
or 2E\(=m v^2\)=mv.v=p.v.=p.ft
( ∴The body moves from rest, its initial velocity, u = 0, hence, from the relation, v = u +ft, v =ft).
Question 19. Find the kinetic energy (K.E.) of a particle of mass ‘m’ gram and momentum ‘p’ g-cm/s in terms of ‘m’ and ‘p’.
Answer:
Let the particle move with velocity v cm/s.
KE\(=\frac{1}{2} m v^2\)erg
⇒ \(=\frac{1}{2} \frac{m^2}{m} v^2\)erg
⇒ \(=\frac{1}{2 m}(m v)^2\)erg
⇒ \(=\frac{p^2}{2 m}\)erg. This is the required relation.
Question 20. The momentum of a heavy body is equal to that of a light body. Which of the two possesses greater kinetic energy?
Answer:
Let the mass of the heavy body be M and its velocity be V. The corresponding quantities of the light body are m and v. So, by the given condition, MV = mv. Since M is greater than m, so v must be proportionally greater than V, to keep equality between MV and mv.
Now, if mv is multiplied with the larger quantity v and MV with the smaller quantity V, the first product, mv. v will be larger than the second product MV. V.
Hence, mv.v >MV.V, or \(\frac{1}{2} m v^2>\frac{1}{2} M v^2\)
This means the kinetic energy of the light body > kinetic energy of the heavy body.
Question 21. Can a body possess energy without momentum? Is the reverse possible?
Answer: A body remaining motionless at a certain position other than its normal position on the surface of the earth may have potential energy but it is without momentum. The body also has no kinetic energy.
In the reverse case, if the body has momentum, it must have velocity and thus kinetic energy, hence momentum without kinetic energy is not possible. Of course, the body may have momentum without potential energy if it confines its motion on the surface of the earth.
Question 22. If a constant force acts on a body, prove that the power of the body is directly proportional to its velocity.
Answer: We know, power \(=\frac{\text { Work }}{\text { time }}\)=\(\frac{\text { force } \times \text { displacement }}{\text { time }}\)
⇒ \(=\text { force } \times \frac{\text { displaceme } n t}{\text { time }}\)= force × velocity.
∴ Power = force x velocity. If force is constant, power α velocity.
Question 23. Show that the law of conservation of energy is not violated in machines.
Answer:
It must not be misunderstood that a machine violates the law of conservation of energy since it overcomes a force or resistance. The work done by the force applied to the machine is always equal to the sum of the work done by the machine and the work performed to overcome the resistive frictional forces existing at different moving parts of the machine.
Question 24. In a flat race, the competitor coming first has more power than the other coming second, though both of them have the same weight. Why?
Answer:
Since the body weights of both the competitors are equal and also since they have covered equal distances in the race, the two have done equal work. Now, the competitor coming first has done the same work in a bit shorter time than that taken by the second competitor. Hence, the power of the competitor coming first is more than that of the second competitor.
Question 25. Why is it easier to climb a hill by a sloping way than along a steep way?
Answer:
A sloping segment of a road cut on the surface of a hill may be considered as an inclined plane. Now, the effort needed to raise some weight to a certain height with the help of an inclined plane is less than that when the weight is tried to be raised to the same height vertically up. For this reason, it is easier to climb a hill by a sloping way than to do the same along a steep way.
Thus a weight is raised or someone may walk up to a certain height on a hill along a sloping path with greater ease than following a steep vertical path to do the same. In fact, the work done in some steps, when the curve path is followed, is the same as that done along a steep path for the same purpose.
Question 26. In nature, any system tends to be in a position where it possesses minimum energy. Explain with an example.
Answer:
Explanation: In nature, a system tends to be in such a position or state so that it possesses minimum potential or kinetic energy. For this reason, a body possessing some potential energy tends to shift to another position or state to minimize its energy.
Example: A lifted body seeks the ground level where the potential energy of anybody is zero with respect to that level.
Question 27. What is a kilowatt? What is the relation between Horsepower and kilowatt?
Answer:
Kilowatt
Kilowatt is a bigger unit of power in the absolute system of power.
1 kilowatt = 1000 watt.
1 H.P. = 746 Joule/sec. = 746 watt.
Now, 746 watt \(=\frac{746}{1000}\)
= 0.746 kilowatt.
Question 28. In which of the following cases work is being done?
(1) A boy carrying a bag on his head and moving on a frictionless surface.
(2) A boy climbing up a staircase.
(3) A boy pushing a wall.
(4) A boy is standing with a load of 10 kg on his head.
Answer:
(2) A boy climbing up a staircase.
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Question 29. What is the transformability of energy?
Answer:
Transformability of energy
Energy may transform from one form to different forms. Every form of energy ultimately transforms to heat energy, directly or after some intermediate transformation. If we rub our palms together, they are warmed. In this case, the mechanical energy expended in rubbing transforms to heat energy.
Question 30. What is the practical unit of work in SI system? Define it.
Answer:
The practical unit of work in SI system is the joule.
Definition of Joule: When 1 Newton force is applied on a body and if the displacement of the body be 1 meter in the direction of force, the work done is 1 Joule.
Question 31. What is the relation between Work and Power?
Answer:
Relation between Work and Power
If w work is done by a body or a system in time t, then the power P of the body or
The system is given by,P\(=\frac{W}{t}\)
Question 32. Find the mass in grams which is equivalent to 5.4 megajoules of energy.
Answer: E = 5.4 megajoule = 5.4 x \({10}^6\) x \({10}^7\)ergs. Now, from Einstein’s equation E \(=\mathrm{mc}^2\) or,
m\(=\frac{E}{c^2}\)\(=\frac{5.4 \times 10^{13}}{\left(3 \times 10^{10}\right)^2}\)=\(\frac{5.4 \times 10^{13}}{9 \times 10^2}\)
=6×\({10}^{-8}\)
So, the required mass is 6 x\({10}^{-8}\) grams.
Question 33. A man carrying a bag stands in an ascending lift. Explain
(1) Does the man do work on the bag
(2) Is there any change of energy occurring in the bag?
Answer:
Explanation
(1) Since the weight does not displace with respect to the man, so the man does not do any work on it.
(2) The height of the weight from the ground gradually increases, so its potential energy increases. Also, the weight moves upward with the same velocity as that of the lift, so it has kinetic energy.
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Question 34. Is work done when the Earth moves around the sun?
Answer:
Explanation: The centripetal force (the force with which the sun attracts the earth) of the sun on the earth does no work since the direction of displacement of the earth at every point of its orbit is perpendicular to it. So, it is a no-work force.
Question 35. In a tug of war, one team wins over the other. Which team does work and how?
Answer:
Explanation: In a tug of war the winning team works against the force applied by the losing team who are displaced through some distance.
Question 36. A moving object suddenly comes to rest. Which energy converts to which energy in this case?
Answer:
In moving conditions, the body possesses kinetic energy. When it stops, the kinetic energy is expended in doing mechanical work against the opposing forces which stop the body.
Question 37. A man carrying a briefcase in hand walks on. a flat horizontal surface. Does he do work?
Answer:
Reason: The man does not do work on the briefcase. Because, as the man holds the briefcase in hand, he exerts a force on it vertically upwards but the briefcase is made to move horizontally. So, the force and its point of application are perpendicular to each other, i.e., no work is done.
Question 38. Is any work done when someone pushes against a heavy immovable object? Or, A large stone did not move in spite of pushing hard. Is any work done? Is any energy spent? Explain.
Answer:
As there is no displacement of the object (stone), no useful work is done on the object. But the work done by our body is not zero. Energy is lost in straining our muscles and moving blood to these strained muscles.
Question 39. A boy holds a stone in his outstretched hand. Is he doing any work?
Answer:
As the boy as well as the stone is at a position of rest, no useful work is done by the boy. But to hold the stone against the force of gravity, the boy is to spend some energy. The muscles in the hand are being stretched to hold the load. So, a change in shape takes place. Then blood is pumped to the stretched muscles. As a result, energy is spent.
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Question 40. A car is moving with uniform velocity. Is the engine or the car doing any work under this condition?
Answer:
The engine of the car is to do work to overcome the frictional force of the ground. No useful work is done.
Question 41. Two boys of the same mass climb to the top of a hill in one hour and two hours respectively. Compare :
(1) the work done by the boys against the gravitational force due to the Earth
(2) the power of the two boys.
Answer:
(1) The work done by each of the two boys for climbing the hill will be the same because work done against the earth’s gravity is mgh. (Here m and h for the boys are the same).
(2) The power of the boy who climbs in one hour will be double of that of the boy who climbs in two hours.
Question 42. Deduce the relation between joule and erg.
Answer:
Relation between joule and erg: 1 joule = 1 Newton x 1 metre
= \({10}^5\)dyne x 100 cm
= (\({10}^5\) x \({10}^2\)) dyne-cm
=\({10}^7\)
1 joule = \({10}^7\) erg.
Question 43. What is a no-work force? Give an example.
Answer:
No-work force: When a force acts perpendicular to the direction of displacement of a particle, then the force does not do any work and is known as a no-work force.
Example: A porter while carrying luggage does no work since he moves horizontally and perpendicular to the force of gravity. The porter, of course, did work at the beginning while he lifted the luggage against gravity.
Question 44. What do you understand by ‘work done against a force’? Give one example.
Answer:
Work done against a force: When the displacement of the point of application of the force takes place in the direction opposite to the applied force, then work is said to be done against the force.
Example: When a body is lifted vertically upwards, displacement of the body is in the direction opposite to the gravitational force, and work is done against the gravitational force.
Question 45. What do you understand by ‘work done by a force’? Give one example.
Answer:
Work done by a force: If the displacement of the point of application of the force takes place in the direction of the force then the force is said to have done the work.
Example: When a body is dropped vertically the gravitational force attracts it downward. The displacement of the body takes place in the direction of the gravitational force. Her work is done by the gravitational force.
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Question 46. Show with an example that work is done by overcoming some resistive force.
Answer:
When a body lying on a surface is pushed, it moves to overcome the opposing frictional force. Here the body when moving, does some work by overcoming the frictional force offered by the surface.
Question 47. Will work be done by the applied force in all cases?
Answer:
No, work will not be done in all cases. Some conditions are given below: No workforce:
(1) If there is no displacement of the point of application of the force, then the force is called a no workforce.
(2) Again, if any force acts perpendicularly in the direction of the displacement, then the force does not do any work. Because in this case θ= 90° and cos 90° = 0.
Question 48. Define work.
Answer:
Work: Work is said to be done when the point of application of a force moves and is measured by the product of the force and the distance moved in the direction of the force.
Question 49. How is work measured?
Answer:
Measurement of work:
(1) If the force F acts on a body and causes a displacement S, then the amount of work done (w)=FxS.
(2) If the direction of displacement makes an angle with the direction of the applied force, then the component of the force along the direction of displacement will be F cosθ. So, work done (w) = F cosθ. S = FS cosθ
Question 50. Express the kinetic energy of a body in terms of momentum and velocity.
Answer:
Kinetic energy: Kinetic energy is the energy that a body possesses by reason of its motion.
kinetic energy(KE)\(=\frac{1}{2} \times m \times v^2\) (m=mass of the body)
∴ KE\(=\frac{1}{2} m v \cdot v\) , (v= velocity of the moving body)
⇒ \(=\frac{1}{2} p \cdot v\) (p=mv= momentum of the body)
So, kinetic energy =½ x momentum x velocity.
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Question 51. A particle starting from rest requires kinetic energy ‘E’ in time ‘t’ and its momentum is ‘p’. Find its acceleration in terms of E, p, and t.
Answer:
Deduction: Let acceleration be f, velocity acquired in time t be v, and mass of the body be m.
∴f\(=\frac{v-0}{t}\) or ,\(f=\frac{V}{t}\) or, r=ft
Again,E\(=\frac{1}{2} m v^2\) or,2E=\(m v^2\)=mv.v=p.v=pft
∴ \(f\)=\(\frac{2 E}{p t}\)This is the required relation.
Question 52. What is potential energy? Write down a mathematical expression for the potential energy of a certain mass raised to a certain height above ground.
Answer:
Potential energy: Potential energy is the energy that a body possesses by virtue of its position of configuration.
Calculation: The weight of the body of mass ‘m’ is mg, where ‘g’ is the acceleration due to gravity at the place. To raise the body, an upward. force mg has to be applied to it. If it is raised to ‘h’ distance above,
work done = mg x h (Force x displacement)
∴ work done = mgh.
This work remains stored in the body as potential energy.
Question 53. Which energy converts to which other when the spring in a toy is compressed?
Answer:
While the spring of a toy is brought to a compressed condition, mechanical energy is expended. This energy remains stored in the spring as potential energy is created due to changes in the configuration of different parts of the spring.
Question 54. A machine does 1920 J of work in 240 s. What is the power of the machine?
Answer:
Given, W = 1920 J; time, t = 240 s.
∴ Power\(=\frac{w}{t}\)=\(\frac{1920}{240}\)=8W.
Question 55. An electric bulb consumes 3600 J of energy in one minute. Calculate its power.
Answer:
Given, Energy = 3600 J; time, t = 1 min = 60 s.
Power of the bulb \(=\frac{E}{t}\)=\(\frac{3600}{60}\)=60W.
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Question 56. A force of 10 N moves a body with a constant velocity of 2 m\({s}^{-1}\). Calculate the power of the body.
Answer:
Given, F = 10 N; V = 2 m\({s}^{-1}\)
Now powerP\(=\frac{w}{t}\)\(=\frac{F s}{t}\)=\(f \times \frac{s}{t}\)=Fv=10×2=20w.
Question 57. A car and a truck have the same speed of 30 m\({s}^{-1}\). If their masses are in the ratio of 1 : 3, find the ratio of their kinetic energies.
Answer:
Given \(v_1\)\(=v_2\) = 30 m\({s}^{-1}\); \(m_1\)\(=m_2\) = 1: 3.
∴\(\frac{\mathrm{KE} \text { of car }}{\mathrm{KE} \text { of truck }}\)
⇒ \(=\frac{\frac{1}{2} m_1 v_1^2}{\frac{1}{2} m_2 v_2^2}\)\(=\frac{m_1}{m_2}\)=\(\frac{1}{3}\).
Question 58. A body of mass 2 kg is moving with a speed of 20 m\({s}^{-1}\). Find its kinetic energy
Answer:
Given, m=2kg; v=20m\({s}^{-1}\).
K.E=\(\frac{1}{2} m v^2\)\(=\frac{1}{2} \times 2 \times(20)^2\)=400J.
Question 59. A force is applied on a car of mass 1500 kg so that its speed increases from 54 km\({h}^{-1}\) to 72 km\({h}^{-1}\). Find the work done by the force on the car.
Answer:
Given, m = 1500 kg; u = 54 km\({h}^{-1}\) \(=54 \times \frac{5}{18}\) = 15 m\({s}^{-1}\)
v=72Km\({h}^{-1}\)\(=72 \times \frac{5}{18}\)=20m\({s}^{-1}\)
∴ Work done by the force,W\(=\frac{1}{2} m v^2-\frac{1}{2} m u^2\)=\(\frac{1}{2} m\left(v^2-u^2\right)\)
⇒ \(=\frac{1}{2} \times 1500\left(20^2-15^2\right)\)=1.3×\(10^{5}\)J.
Question 60. A moving body of mass 15 kg has 60 J of kinetic energy. Calculate its speed
Answer:
m=15kg; \(E_k\)=60 J
EK\(=\frac{1}{2} m v^2\)or,\(v^2=\frac{2 E_K}{m}\)\(=\frac{2 \times 60}{15}\)=8
∴V=2.83 m\({s}^{-1}\)
Question 61. A rifle bullet of mass 15 g is traveling at a speed of 300 meters/sec. Find the kinetic energy of the bullet.
Answer:
Kinetic energy = ½ x mass x \((\text { velocity })^2\)
⇒ \(=\frac{1}{2} \times m \times v^2\), m=15,g=0.015 kg
⇒ \(=\frac{1}{2} \times 0.015 \times(300)^2\), v=300 m/sec
= 675 joules KE =?
Question 62. The power of a motor car is 10 kW. What is the amount of work done by the car in 1hr?
Answer:
Power of the car = 10 kW = 10 x \(10^{3}\) J/sec.
∴ work done by the car in 1 hr
= 10 x\(10^{3}\) x 3600 J
=3.6 x \(10^{7 }\)J , 1 hr = 60 x 60 =3600 sec
= 3.6 x \(10^{7}\)J, 1 kW = \(10^{3}\) W
Question 63. If 150-joule work is done for the displacement of 15 m of a body, calculate the amount of force applied here.
Answer:
Work is done here = 150 joule
displacement =50 m.
We know ,applied force\(=\frac{\text { Work done }}{\text { displacement }}\)
⇒ \(=\frac{150}{50}\)=30 Newton.
Question 64. If 10 Newton force is applied on a body then 5 m displacement of the body occurs towards the direction of the force. What will be the work done?
Answer:
We know, Work done = Force x displacement
= 10 x 5 Newton−meter
= 50 joules.
Wbbse Madhyamik Class 9 Physical Science And Environment
Question 65. If 100 g mass of a body is dropped from 80 cm high place then calculate the amount of potential. the energy lost during this process.
Answer:
Amount of potential energy lost =m x g xh
= 0.1 x 9.8 x 0.8 joule, m = 100g = 0.1 kg
= 0.784 joule, g = 9.8 m/\({sec}^{2}\)
h=80cm=0.8m
Question 66. The power of a machine is 40 W. What is the amount of work done by the machine in 30 sec?
Answer:
The power of the machine = 40 W.
= 40 joule/sec.
∴Work done by the machine
= (40 x 30) joule
= 1200 joule.
Wbbse Madhyamik Class 9 Physical Science And Environment
Question 67. A man carries a bag of 20 g weight. Calculate the work he does when he moves 25 cm in
(1) Horizontal direction and
(2) Vertical upward direction.
Answer:
(1) In this case the man does not do any work since the displacement of the bag is perpendicular to the force of gravity.
(2) Work = force x displacement = 20 g-wt x 25 cm
= 500 gram-centimeter = (500 x 980) erg
= 49 x \(10^{4}\)erg. (g is taken as 980 cm/\({s}^{2}\))
Question 68. A man weighs 40 kg-wt and climbs a staircase 20 m high. If he takes 16 seconds to go to the top of the staircase find
(1) The work the man performs and
(2) His power (Given g = 9.8 m/\(\left.s^2\right)\)
Answer:
(1) Work the man performs = force x displacement
= 40 kg-wt x 20 m = 800 kg-m
= 800 x9.8N-m or, J = 7840 J.
(2)Power of the man\(=\frac{\text { work he performed }}{\text { time taken }}\)
=\(\frac{7840 \mathrm{~J}}{16 \mathrm{~s}}\)
=490 Watt.
Question 69. A group can lift 1000 liters of water per hour to a height 10 m. Find the work done by the pump and also its power. (1 litre of water has a mass 1 kg)
Answer:
Work done by the pump = weight of water x height to which it is raised.
= 1000 kg-wt x 10m = 104 kg-m = 104 x 9.8 J.
Power\(=\frac{\text { work done }}{\text { time of work }}\)
=\(\frac{10^4 \times 9.8 \mathrm{~J}}{3600 \mathrm{~s}}\)
=27.22 J/s (approx)
=27.22 Watt.
Wbbse Madhyamik Class 9 Physical Science And Environment
Question 70. A porter lifts a box of 20 kg from the ground and puts it on his head 1.2 m above the ground. Calculate the work done by him on the box. (Take g = 9.8 m/\({s}^{2}\)).
Answer:
Mass of the box (m) = 20 kg, displacement (d) = 1.2 m.
∴ Work done, W = F.d = mg.d = 20 x 9.8 x 1.2 J = 235.2 J.
Question 71. A man weighing 60 kg runs up a staircase having 30 steps each of 0.2 m in 30 s. Calculate his power. (Given, g = 10/s)
Answer:
Here, mass (m) = 60 kg, height raised (h) = (30 0.2)m = 6 m, time taken (t) = 30s.
∴ Power\(=\frac{\text { work done }}{\text { time taken }}\)
⇒ \(=\frac{\text { Force }(\mathrm{mg}) \times \text { displacement }(\mathrm{h})}{\text { time taken }(\mathrm{t})}\)
⇒ \(=\frac{60 \times 10 \times 6 \mathrm{~J}}{30 \mathrm{~s}}\)
=120 Watt.
Question 72. Find the potential energy of a body of mass 5 kg kept at a height of 10 m if acceleration due to gravity at that place is 9.8 m/s.
Answer:
Here, mass (m) = 5 kg, acceleration due to gravity (g) = 9.8 m/\(s^2\) and height (h)=10m.
∴Potential energy of the body \(\left(E_p\right)\) = mgh =5 x 9.8 x 10 J = 490 J.
∴ The required potential energy = 490 J
Wbbse Madhyamik Class 9 Physical Science And Environment
Question 73. A bag of rice weighs 100 kg. To what height should it be raised so that its essential energy becomes 4900 J ? (Given g = 9.8 m/\(s^2\))
Answer:
Here, mass (m) = 100 kg,
acceleration due to gravity (g) = 9.8 m\(s^2\),
Potential energy \(\left(E_p\right)\)=4900J
because \(\left(E_p\right)\)=mgh
h\(=\frac{E_p}{m g}\)
⇒ \(=\frac{4900}{100 \times 9.8}\)
⇒ \(=\frac{4900}{980}\) m=5m
∴The bag should be raised to a height of 5 m.
Wbbse Class 9 Physical Science Solutions
Question 74. A body of mass 1 kg is moving with a speed of 40 m/s. Calculate its kinetic energy.
Answer:
Given
A body of mass 1 kg is moving with a speed of 40 m/s.
Here, mass (m) = 1 kg, Speed (v) = 40 m/s
∴ Kinetic energy \(\left(E_K\right)\) \(=\frac{1}{2} m v^2\)
⇒ \(=\frac{1}{2} \times 1 \times(40)^2\)
=800 J
The kinetic energy of the body = 800 J.
Question 75. A moving body of mass 50 kg has 400 J kinetic energy. Calculate its velocity
Answer:
Here, mass (m) = 50 kg, Kinetic energy\(\left(E_K\right)\) = 400 J.
because \(\left(E_K\right)\)\(=\frac{1}{2} m v^2\) or,
v\(=\sqrt{\frac{2 E_k}{m}}\)
∴ velocity of the body\(=\sqrt{\frac{2 \times 400}{50}}\)
=4 m/s.
Question 76. A car and a bus have equal speed. If their masses are in the ratio 2 : 3, find the ratio of their kinetic energies.
Answer:
Given, \(v_1\)=\(v_2\) and \(\frac{m_1}{m_2}\)=\(\frac{2}{3}\)
∴\(\frac{\text { Kinetic energy of the car }\left(E_1\right)}{\text { Kinetic energy of the bus }\left(E_2\right)}\)
⇒ \(=\frac{\frac{1}{2} m_1 v_1^2}{\frac{1}{2} m_2 v_2^2}\)=\(\frac{m_1}{m_2}\)
∴(\(v_1=v_2\)) \(=\frac{2}{3}\)
∴ The ratio of their kinetic energies = 2 : 3.
Question 77. A state bus of mass 8,000 kg is moving with a velocity of 54 km/h. Calculate the work done to stop this bus.
Answer:
Here, mass (m) = 8000 kg
velocity (v) = 54 Km/h \(=\left(54 \times \frac{5}{18}\right)\)m/s= 15m/s.
∴ Work done to stop the bus = Change in kinetic energy of the bus
⇒ \(=\left(\frac{1}{2} m v^2-0\right)\)
⇒ \(=\frac{1}{2} m v^2\)
⇒ \(\frac{1}{2} \times 8000 \times(15)^2\) J
= 900000 J
= 9×\(10^5\) J.
∴ Required work done to stop the bus = 9 x \(10^5\) J.
Wbbse Class 9 Physical Science Solutions Chapter 5 Energy In Action, Work, Power And Energy 3 Marks Questions And Answers:
Question 1. Define work. How is it measured?
Answer:
Work: Work is said to be done only when the force applied to a body makes the body to move. Or, If the position of a body is changed by the application of a force on it, then work is said to be done by the force.
The amount of work done depends upon :
(1) The magnitude and direction of the force
(2) The displacement of the body produced by the force.
The work done by a force is measured by
Work = Force x displacement of the point of application in the direction of the force.
If a force F displaces a body from position A to position B in the direction of the force, then the displacement by the force AB = d and the work done W =F xd.
Question 2. A body is acted upon by a force. State the conditions when the work done is zero.
Answer:
If a force acts on a body and the body does not move, no work is done.
Let a force F is acting on a body along AB and its displaces the body such that the point of application of the force moves from A to C. The displacement AC = d makes an angle ¢
to the direction of the force.
Then, AB=AC or, cosθ= d cosθ.
Hence, work done = F x AB =F x dcos9.
Now, we consider two cases :
(1) If θ= 90° , then cos 90° = 0. Hence, work done = 0.
That is, if ‘the displacement is normal to the direction of force, work done is zero.
(2) If θ = 0° , then cos 0° = 1. Hence, work done = F x d.
Thus, work done is positive and maximum if the displacement is in the direction of force.
(3) If the displacement is in a direction opposite to the force, i.e.,
θ = 180°, then cos 180° =−1. Work done =− F x d.
Hence, the work done is negative.
Question 3. State and define the units of work.
Answer:
Joule: One Joule of work is said to be done when a force of 1 Newton displaces a body through 1 meter in its own direction.
Hence, 1 Joule = 1 Newton x 1 meter.
The bigger units of work are Kilo Joule (KJ) and Mega Joule 2 (MJ):
Where 1 KJ = \(10^3\) J and 1 MJ = \(10^3\) J.
The C.G.S. unit of work is the erg.
Erg: One erg of work is said to be done when a force of 1 dyne a body through 1 cm in its own direction.
Hence, 1 erg = 1 dyne x 1 cm.
Question 4. Define power. How is it measured
Answer:
Power: The rate of change of doing work is called Power. Power is measured as the amount of work done in one second.
Thus, Power P \(=\frac{\text { Work done }}{\text { time taken }}\) or,P\(=\frac{W}{t}\)
Since, work done W = F x d
∴ Power\(=\frac{W}{t}\)\(=\frac{W \times d}{t}\)=F×v (where, v=d/t)
Thus, Power = Force x average velocity
Power is a scalar quantity.
Question 5. State and define the units of power. Find the relation between them.
Answer: The S.I. unit of power is Watt.
Watt: If one joule of work is done in one second, then power is said to be 1 watt.
1 Watt \(=\frac{1 \text { Joule }}{1 \text { second }}\) 1= J/sec
The bigger units of power are Kilowatts (KJ) and Megawatts (MJ).
1 kilowatt (KW) = \(10^3\) W and 1 megawatt (MW) = \(10^6\) W.
The C.G.S. unit of power is erg per second.
Relation between S.I unit and C.G.S. units of power: 1 Watt = 1 J/sec = \(10^7\) ergs/sec.
Question 6. Define energy. State its units.
Answer:
Energy: The energy of a body is the capacity to do work.
Energy is a scalar quantity.
The units of energy are the same as that of work.
The S.1. unit of energy is Joule (J) and its C.G.S. unit is erg.
The bigger units of energy are watt-hours (Wh) and kilowatt-hours (kWh).
1 watt-hour = 1 watt x 1 hour
= 1 J/sec x 3600 sec = 3600 J.
1 Kilo watt hour = 1000 x 3600 J = 3.6 X \(10^6\) J.
Question 7. How is kinetic energy measured?
Answer:
The kinetic energy of a body:
= work done by the retarding force to stop the body
= Force (F) x displacement (s)
or, K.E.=F× s…………………..(1)
From the equation of motion,\(v^2\)=\(u^2+2 f s\)
Since, final velocity v= 0, hence 0=\(u^2+2 f s\) (f is retardation)
or, s\(=\frac{u^2}{2 f}\)………….(2) and
Force F= m.f ……………(3)
Now, substituting the values of s and F in relation (1), we get
KE. = Fx s
= m.f ×\(\frac{U^2}{2 f}\)=\(\frac{1}{2} m u^2\)
Hence K. E. \(=\frac{1}{2} \text { mass } \times(\text { velocity })^2\).
Question 8. Explain that potential energy changes into kinetic energy when it is put into use.
Answer:
Potential energy changes to kinetic energy which does work as is explained in the following examples:
(1) A stretched bow has potential energy because of its stretched position. On releasing the bow, the potential energy changes into kinetic energy which works on the arrow and makes it move.
(2) A hydroelectric power station converts the potential energy of falling water into kinetic energy and thus generates electricity. A hydroelectric power station uses water that is stored in a reservoir behind a dam at a certain height. The water is allowed to fall through a tunnel or pipe to the plant’s water turbine. The kinetic energy stored in the falling water spins the turbine shaft, which drives the electric generator.
Question 9. Show that the total sum of potential energy and kinetic energy of a freely falling body under gravity is constant.
Answer:
Let a body of mass m fall freely under gravity from a height h above the ground.
At the top, kinetic energy =0 (since the body is at rest), and the potential energy = mgh.
Hence, the total energy at the top height = 0 + mgh = mgh.
After falling a height of h, let the velocity of the body at the bottom be, say, v.
Hence, by the equation of motion,
⇒ \(v^2\)=\(u^2+2 g h\)
Since u = 0 at the top, \(v^2\)=0+2 g h= 2gh……… (1)
At the bottom, potential energy = 0,
and the kinetic energy \(=\frac{1}{2} m v^2\)
⇒ \(=\frac{1}{2} m \times 2 g h\)= mgh. (from ……..1)
Total energy at the bottom = 0 + mgh = mgh.
During fall, at every point, the sum of potential energy and kinetic energy remains = mgh. Thus, the total mechanical energy (sum of potential energy and kinetic energy) always remains constant at each point of motion and it is equal to mgh.
Question 10. State the differences between potential energy and kinetic energy.
Answer:
(1) The Potential energy of a body is the energy that the body possesses by virtue of its position while kinetic energy is the energy it gains by virtue of its motion.
(2) Potential energy and kinetic energy are both scalar quantities and measured in the same unit.
(3) Potential energy = mgh, where m is the mass of the body and h is the height of the body.
Kinetic energy \(=\frac{1}{2} m v^2\) , where v is the velocity of the body.
Wbbse Physical Science And Environment Class 9 Solutions
Question 11.
(1) What is meant by work? Define work.
(2) How is it measured?
(3) Is work a vector or scalar quantity?
(4) Find out the dimension of work.
Answer:
(1) Meaning of work : The term ‘work’ is restricted to those types of activities in which a force acting on a body causes change of position, change of velocity, change of shape, etc. in the body. Thus, when a man carries luggage or a horse pulls a carriage or a laborer draws water from a well, each of them does some work.
Definition of work: Work is said to be done by a force when it succeeds in moving the body it acts upon, i.e., when there is displacement of a body by the action of a force.
(2) Measurement of work :
The amount of work done by a force when it acts on a body is measured by the product of the magnitude of the force and the displacement of its point of application.
Let a body be displaced from position A to position B in the horizontal direction by a force ‘P’. This displacement will be AB = S in the direction of the force.
Work (W) done by the force = applied force (F) x displacement (S), i.e., W = F x S.
(3) Work is a scalar quantity: Work is expressed by W = F x S. As FS is the product of two vectors quantities, (force and displacement) work is a scalar quantity. Thus, if a force of 5N acting towards the east displaces a body through a distance of s in the east, it does exactly the same amount of work as a 5 N force acting towards the north displaces a body through a distance of s in the north.
(4) Dimension of work : Work = Force x Distance’= (Mass x Acceleration) x (Distance)
⇒ \(=\left(\text { mass } \times \frac{\text { Distance }}{\text { Time }^2}\right) \times(\text { Distance })\)
⇒ \(=\left(M \cdot \frac{L}{T^2}\right)(L)\)
⇒ \(=\left[M^1 L^2 T^{-2}\right]\)
So, the dimensions of work are 1, 2, and -2 in mass, length, and time respectively.
Question 12. (1) State and define the absolute units of power in C.G.S. and S.I. systems. Or The units of which physical quantity are erg-second and watt? Define them. Or, Define the SI unit of power.
(2) State the relationship between the S.I. and C.G.S. units of power.
Answer:
(1) Absolute unit of power in C.G.S. system: The absolute unit of power in the C.G.S. system is erg per second (erg \(s^{-1}\)). It is defined as the rate of doing one erg of work per second.
Absolute unit of power in M. K S. (or S.1.) system: The absolute unit of power in the M.K.S. (or S.I.) system is watt and is abbreviated as w. If one joule of work is done in 1 second, the power is said to be 1 watt.
(2) Relationship between the S.I. unit and C.G.S. unit :
1 watt=1 joule \(s^{-1}\), since 1 joule =\(10^{7}\)
=\(10^{7}\) erg \(s^{-1}\).
Question 13. A piece of stone is allowed to fall from the top of a house without applying any force. What type of energy is there in the stone
(1) When it is not dropped?
(2) When it comes down but remains at some height above the ground?
(3) Just before it touches the ground?
(4) When it falls to the ground?
Give reasons for your answer.
Answer:
When a body is raised it performs some work against gravitational force. This work is stored in the body as gravitational potential energy.
(1) Now when the body is not dropped it contains only potential energy, but no kinetic energy. This is because with respect to the ground the body is at a higher position.
(2) In this position the body possesses both kinetic and potential energy, sum total of which is equal to the amount of energy when it was on the roof.
(3) Just before touching the ground, the body possesses only kinetic energy.
(4) When the body strikes the ground, its kinetic energy is lost in the form of sound energy, heat energy, etc.
Question 14. Two bodies of unequal masses have the same kinetic energy. How do their momenta compare in terms of their masses?
Answer:
Explanation: Let masses of the bodies \(m_1 \text { and } m_2\) be and their corresponding velocities be \(v_1 \text { and } v_2\)
⇒ \(\frac{1}{2} m_1 v_1^2\)=\(\frac{1}{2} m^2 v_2^2\)
or, \(\frac{m_1^2}{m_1} v_1^2\)=\(\frac{m_2^2}{m^2} v_2^2\)
or, \(\left(\frac{m_1 v_1}{m_2 v_2}\right)\)=\(\frac{m_1}{m_2}\)
or, \(\frac{p_1}{p_2}\)=\(\frac{\sqrt{m_1}}{\sqrt{m_2}}\)
Thus, the momenta of the two bodies are directly proportional to the square root of their corresponding masses.
Question 15. A heavy body and a light body have equal kinetic energy; which of them possesses greater momentum?
Answer:
Explanation: Let the mass of the heavier body be \(\mathrm{m}_1\)and that of the lighter body be \(\mathrm{m}_2\) also let the velocity of the heavier body be \(\mathrm{v}_1\) and that of the lighter body be \(\mathrm{v}_2\).
∴ \(\frac{1}{2} m_1 v_1^2\)=\(\frac{1}{2} m_2 v_2^2\)
or, \(m_1 v_1^2\)=\(m_2, v_2^2\) ………………… (1)
since \(\mathrm{m}_1\) >\(\mathrm{m}_2\) to keep equality
⇒ \(v_1^2,<v_2{ }^2 \text {, i.e., } v_2>v_1, \text { so } \frac{v_2}{v_1}>1\)…………..(2)
Again, from the relation (1) we get, \(m_1 v_1 \cdot v_1\)=\(m_2 v_2 \cdot v_2\) or, \(\frac{m_1 v_1}{m_2 v_2}\)=\(\frac{v_2}{v_1}\)
From the relation (2) we get, \(\frac{m_1 v_1}{m_2 v_2}>1 \text {, i.e., } m_1 v_1>m_{22} v\).
So, the momentum of the heavier body\(\left(m-1 v_1\right)\) is greater than the momentum of the lighter body\(\left(m-2 v_2\right)\).
Question 16. Examine whether work is done in the following cases :
(1) A football is kicked.
(2) A boy picks up a brick from the ground.
(3) A piece of wood is floating on the stream.
(4) A ripe fruit falls from the tree on the ground.
(5) A jar of water is lifted from a well.
(6) A boat is sailed.
Answer:
(1) The ball has been displaced on kicking (force), so work will be done.
(2) As the piece of brick is picked up against the force of gravitation, work will be done.
(3) The piece of wood is displaced by the current of the stream (force). So work is done
(4) Work is done here because there is a displacement of the fruit by the force of gravitation.
(5) Here work is done because there is a displacement against the force of gravitation.
(6) Here wind acts as a source of force and thereby the boat moves in the direction of wind. So work is done.
Wbbse Physical Science And Environment Class 9 Solutions
Question 17. An apple of mass m falls freely under gravity from a height h. What is its kinetic
and potential energy.
(1) At the beginning of the fall and
(2) At a height of h/2?
Answer:
(1) At the beginning of the fall
KE. of the apple = 0
P.E. of the apple = mgh.
(2) At a height h/2
K.E. of the apple = ½ mgh
P.E. of the apple = ½ mgh.
Question 18. Distinguish between potential energy and kinetic energy.
Answer:
Difference between potential energy and kinetic energy
| Potential energy | Kinetic energy |
| (1)Energy acquired by a body when it is taken to some different position from its normal position is called its potential energy. | (1) Energy acquired by a body by virtue of its motion is called its kinetic energy. |
| (2) Gravitational potential energy increases as the body is raised from the surface of the earth. | (1) Kinetic energy of a body increases with its velocity. |
| (3) Potential energy remains stored in a body due to its special position or configuration. | (3) Kinetic energy is generated in a body due to its motion. |
| (4) A body can have potential energy while it is at rest, but it cannot have kinetic energy at rest because it contains the component of velocity. | (4) A body has kinetic energy while in motion but it may or may not have potential energy. |
| (5) Gravitational potential energy of a body = mass of the body x its height from the surface of the earth x acceleration due to gravity | (5) Kinetic energy of a body = x mass of the body x (velocity of the body)2. |
| (6) Gravitational potential energy of a body at any place depends on the acceleration due to the gravity of the place. | (6) Kinetic energy of a body does not depend on the acceleration due to the gravity of a place. |
| (7) It is a scalar quantity. | (7) It is a scalar quantity. |
| 8) SI unit of it is Joule. | (8) SI unit of it is Joule. |
Question 19. Distinguish between energy and power
Answer:
Difference between energy and power
| Energy | Power |
| (1) The capacity of a body or person to do work is its energy. | (1) Power is the rate of doing work. |
| (2) Energies of two persons may be the same, but their powers may be different. | (2) The quantity energy of two persons may be different, although they have the same power. |
| (3) Power is a quantity dependent on time. | (3) Energy is not related to time. |
| (4) It is a scalar quantity. | (4) It is a scalar quantity. |
Question 20. A body remains at rest at a certain height. State the type or types of energy the body possesses in the following steps :
(1) When the body is at rest at the top of the height.
(2) The body falls through some distance.
(3) Just before it touches the ground.
(4) It touches the ground and becomes stationary.
Answer:
A body of mass ‘mm’ is at rest above the ground at a height ‘h’. We notice the change of its mechanical energies, i.e. kinetic and potential energies at the following stages.
(1) Just before it falls. Here, the body has the potential energy ‘mgh’, and since it is at rest, its kinetic energy is zero. So, at mgh this stage total energy of the body is mgh.
(2) The body descends through ‘a’, where, a < h. The height of the body above ground is (h − a), so its potential energy is mg (h − a). The potential energy thus decreases by mga. While the body descends, it acquires some velocity, v (say). So its
kinetic energy is \(\frac{1}{2} m v^2\) which can be proved equal to mga.
Thus, a decrease of potential energy equals a gain in kinetic energy. So, the total energy ground
= mg (h− a)+mga
= mgh− mga + mga = mgh.
(3) The body is just above the ground before touching it. Here ‘f is almost zero, so the potential energy is negligible. The body possesses only kinetic energy which may be proved equal to mgh.
Thus, at all stages, the total energy of the body, before it touches au is constant and equal to mgh.
(4) When the body touches the ground, ‘h’ is zero, so potential energy is zero. Since it comes to rest, its kinetic energy is also zero. The energy of the body dissipates in the environment as heat energy and sound energy.
Wbbse Physical Science And Environment Class 9 Solutions
Question 21. Define and exemplify ‘work done by a force’ and ‘work done against a force’.
Answer:
Work done by a force: If the point of application moves in the direction of the applied force, work is said to be done by the force.
Example: A stone is released from a certain height. Here gravitational force acts on the stone downwards and the stone also falls in the same direction.
Work done against a force: If a body has its displacement in the opposite direction of the applied force, work is said to be done against the force.
Example: Let us suppose a body is raised from the ground to a certain height. Here gravitational force acts on the body clean Ceti but the body is displaced in the opposite direction.
Work is a scalar quantity. Work may be in the direction of force and also against the direction of force. The former is called positive work and the latter is called negative work.
For example, when a body falls from a certain height, work is done on the body in the direction of the force of gravity. So this is a positive work. When a body is lifted from the ground, work is done against the force, this may be called negative work.
Question 22. Define potential energy. How is it measured?
Answer:
Potential energy of an object placed at a height: Let an object of mass m is lifted from the surface of the earth against the force of gravity to a height h. The work done in the process remains in it as the potential energy of the body.
Force acting on the body = weight of the body = mg acting vertically downward.
The height through which the body is raised = h
Thus work done in raising the body W = force-displacement = mg x h = mgh
This work remains stored in the body as the gravitational potential energy \(E_p\)
Thus, \(E_p\)= mgh.
The gravitational potential energy depends on the difference in heights of the initial position and final position of the body but is independent of the path followed by the body while going from the initial position to the final position. For a body raised above the surface of the earth, the earth is taken as zero.
Question 23. If a pump lifts 500 liters of water per hour to a height of 30 m. Then find the work done by the pump and its power in watts. (mass of 1 liter of water = 1 kg)
Answer:
Given
If a pump lifts 500 liters of water per hour to a height of 30 m.
Work done by the pump
= weight of water x displacement of water
= 5000 x 1 9.8 x 30J.
=14.7 x \(10^5\) J
Power of the pump \(=\frac{\text { Work done }}{\text { Time }}\)=\(\frac{14.7 \times 10^5 \mathrm{~J}}{60 \times 60 \mathrm{sec}}\)= 408 W.
Question 24. One body has a mass of 50 g and a velocity of 10 cm/sec. The other body has a mass of 20 g and a velocity 20 cm/sec. Compare their velocities and kinetic energies.
Answer:
Given
One body has a mass of 50 g and a velocity of 10 cm/sec. The other body has a mass of 20 g and a velocity 20 cm/sec.
We know, The momentum of a body = mass of the body x velocity.
Here, the velocity of the first body: velocity of the second body
= (50 x 10): (20 x 20)=5:4
We also know, Kinetic energy \(=\frac{1}{2} \times \text { mass } \times(\text { velocity })^2\)
Here, the kinetic energy of the first body: kinetic energy of the second body
⇒ \(=\frac{1}{2} \dot{\times} 50 \times(10)^2: \frac{1}{2} \times 20 \times(20)^2\)
The ratio of the velocities of the two bodies = 5: 8
Question 25. A particle of mass ‘m’ at rest is acted upon by a force ‘p’ for a time ‘f. Show that the particle acquires the kinetic energy \(\frac{p^2 t^2}{2 m}\) after the time ‘f.
Answer:
Given
A particle of mass ‘m’ at rest is acted upon by a force ‘p’ for a time ‘f.
It is known from Newton’s second law of motion that, force (p) = mass (m) x acceleration (f).
∴ f\(=\frac{p}{m}\) Also, \(f=\frac{\text { final velocity }(v)-\text { initial velocity }(u)}{t i m e}\)
∴\(f=\frac{v-0}{t}\)
(because The body was initially at rest, u = 0) or, v = ft
⇒ \(=\frac{p}{m} t\) ( because f\(=\frac{p}{m}\))
So, Kinetic energy \(=\frac{1}{2} m \times v^2 \times \frac{1}{2} m\left(\frac{p t}{m}\right)^2\)=\(\frac{p^2 t^2}{2 m}\)
Question 26. A body moves 80 m under the action of an applied force of 10 N on a floor having a frictional force of 2 N. Compute the
(1) work done by the applied force,
(2) work done by the force of friction and
(3) work done by the net force.
Answer:
(1) Work done by the applied force = F.d = 10 x 80 = 800 J.
(2) Work done by the force of friction = Fs.d = (2 x 80) = 160 J.
(3) Work done by the net force = (F −Fs).d = (10−2) x 80 = 640 J.
WBBSE Solutions for Class 9 Physical Science And Environment
- Chapter 1 Measurement
- Chapter 2 Force And Motion
- Chapter 3 Matter: Structure And Properties
- Chapter 4 Matter: Atomic Structure; Physical & Chemical Properties of Matter
- Chapter 5. Energy In Action: Work, Power & Energy
- Chapter 6 Heat
- Chapter 7 Sound