Selina Concise Solutions for ICSE Class 10 Physics Chapter 1 Force

Question 1. Define work. Is work a scalar or a vector?

Answer: Work is said to be done only when the force applied on a body makes the body move. It is a scalar quantity.
 

Question 2. How is the work done by a force measured when (i) force is in direction of displacement, (ii) force is at an angle to the direction of displacement?

Answer: (i) When force is in direction of displacement, then work done, \(W = F \times S\)
(ii) When force is at an angle to the direction of displacement, then work done, \(W = FS \cos \theta\)
 

Question 3. A force F acts on a body and displaces it by a distance S in a direction at an angle θ with the direction of force. (a) Write the expression for the work done by the force. (b) what should be the angle between the force and displacement to get the (i) zero work (ii) maximum work?

Answer: (a) When force is at an angle to the direction of displacement, then work done, \(W = FS \cos \theta\)
(b) (i) For zero work done, the angle between force and displacement should be \(90^\circ\) as \(\cos 90^\circ = 0\)
\(W = FS \cos 90^\circ = FS \times 0 = 0\)
(ii) For maximum work done, the angle between force and displacement should be \(0^\circ\) as \(\cos 0^\circ = 1\)
Hence, \(W = FS \cos 0^\circ = FS\)
 

Question 4. A body is acted upon by a force. State two condition when the work done is zero.

Answer: Two conditions when the work done is zero are:
(i) When there is no displacement (\(S=0\)) and,
(ii) When the displacement is normal to the direction of the force (\(\theta = 90^\circ\)).
 

Question 5. State the condition when the work done by a force is (i) positive, (ii) negative. Explain with the help of examples.

Answer: (i) If the displacement of the body is in the direction of force, then work done is positive. Hence, \(W = F \times S\)
For example: A coolie does work on the load when he raises it up against the force of gravity. The force exerted by coolie (=mg) and displacement, both are in upward direction.
(ii) If the displacement of the body is in the direction opposite to the force, then work done is negative. Hence, \(W = -F \times S\)
For example: When a body moves on a surface, the force of friction between the body and the surface is in direction opposite to the motion of the body and so the work done by the force of friction is negative.
 

Question 6. A body is moved in a direction opposite to the direction of force acting on it. State whether the work is done by the force or work is done against the force.

Answer: Work is done against the force.
 

Question 7. When a body moves in a circular path, how much work is done by the body? Give reason.

Answer: When a body moves in a circular path, no work is done since the force on the body is directed towards the centre of circular path (the body is acted upon by the centripetal force), while the displacement at all instants is along the tangent to the circular path, i.e., normal to the direction of force.
 

Question 8. A satellite revolves around the earth in a circular orbit. What is the work done by the force of gravity? Give reason.

Answer: Work done by the force of gravity (which provides the centripetal force) is zero as the force of gravity acting on the satellite is normal to the displacement of the satellite.
 

Question 9. In which of the following cases, is work being done? (i) A man pushing a wall. (ii) a coolie standing with a load of 12 kgf on his head. (iii) A boy climbing up a staircase.

Answer: Work is done only in case of a boy climbing up a stair case.
 

Question 10. A coolie carrying a load on his head and moving on a frictionless horizontal platform does no work. Explain the reason.

Answer: When a coolie carrying some load on his head moves, no work is done by him against the force of gravity because the displacement of load being horizontal, is normal to the direction of force of gravity.
 

Question 11. The work done by a fielder when he takes a catch in a cricket match, is negative Explain.

Answer: Force applied by the fielder on the ball is in opposite direction of displacement of ball. So, work done by the fielder on the ball is negative.
 

Question 12. Give an example when work done by the force of gravity acting on a body is zero even though the body gets displaces from its initial position.

Answer: When a coolie carries a load while moving on a ground, the displacement is in the horizontal direction while the force of gravity acts vertically downward. So the work done by the force of gravity is zero.
 

Question 13. What are the S.I. and C.G.S units of work? How are they related? Establish the relationship.

Answer: S.I unit of work is Joule. C.G.S unit of work is erg. Relation between joule and erg : \(1 \text{ joule} = 1\text{N} \times 1\text{m}\) But \(1\text{N} = 10^5 \text{ dyne}\) And \(1\text{m} = 100 \text{ cm} = 10^2 \text{ cm}\) Hence, \(1 \text{ joule} = 10^5 \text{ dyne} \times 10^2 \text{ cm}\) \( = 10^7 \text{ dyne} \times \text{ cm} = 10^7 \text{ erg}\) Thus, \(1 \text{ Joule} = 10^7 \text{ erg}\)
 

Question 14. State and define the S. I. unit of work.

Answer: S.I unit of work is Joule. 1 joule of work is said to be done when a force of 1 newton displaces a body through a distance of 1 meter in its own direction.
 

Question 15. Express joule in terms of erg.

Answer: Relation between joule and erg : \(1 \text{ joule} = 1\text{N} \times 1\text{m}\) But \(1\text{N} = 10^5 \text{ dyne}\) And \(1\text{m} = 100 \text{ cm} = 10^2 \text{ cm}\) Hence, \(1 \text{ joule} = 10^5 \text{ dyne} \times 10^2 \text{ cm}\) \( = 10^7 \text{ dyne} \times \text{ cm} = 10^7 \text{ erg}\)
 

Question 17. A boy of mass \(m\) climbs up a staircase of vertical height \(h\). (a) What is the work done by the boy against the force of gravity? (b) What would have been the work done if he uses a lift in climbing the same vertical height?

Answer: Let a boy of mass \(m\) climb up through a vertical height \(h\) either through staircase or using a lift. The force of gravity on the boy is \(F = mg\) acting vertically downwards and the displacement in the direction opposite to force (i.e., vertical) is \(S = -h\). Therefore the work done by the force of gravity on the boy is \(W = FS = -mgh\) or, the work \(W = mgh\) is done by the boy against the force of gravity.
 

Question 18. Define the term energy and state its S.I. unit.

Answer: The energy of a body is its capacity to do work. Its S.I unit is \(\text{Joule (J)}\).
 

Question 19. What physical quantity does the electron volt (\(\text{eV}\)) measure? How is it related to the S.I. unit of that quality?

Answer: \(\text{eV}\) measures the energy of atomic particles. \(1\text{eV} = 1.6 \times 10^{-19} \text{ J}\)
 

Question 16. A body of mass \(m\) falls down through a height \(h\). Obtain an expression for the work done by the force of gravity.

Answer: Let a body of mass \(m\) fall down through a vertical height \(h\) either directly or through an inclined plane e.g. a hill, slope or staircase. The force of gravity on the body is \(F = mg\) acting vertically downwards and the displacement in the direction of force (i.e., vertical) is \(S = h\). Therefore the work done by the force of gravity is \(W = FS = mgh\)
 

Question 20. Complete the following sentence: \(1 \text{ J} = \text{Calorie}\)

Answer: \(1 \text{ J} = 0.24 \text{ calorie}\)
 

Question 21. Name the physical quantity which is measured in calorie. How is it related to the S.I. unit of the quality?

Answer: Calorie measures heat energy. \(1\text{calorie} = 4.18 \text{ J}\)
 

Question 22. Define a kilowatt hour. How is it related to joule?

Answer: \(1\text{kWh}\) is the energy spent (or work done) by a source of power \(1\text{kW}\) in \(1\text{h}\). \(1\text{kWh} = 3.6 \times 10^6 \text{ J}\)
 

Question 23. Define the term power. State its S.I. unit.

Answer: The rate of doing work is called power. The S.I. unit of power is \(\text{watt (W)}\).
 

Question 24. State two factors on which power spent by a source depends. Explain your answer with examples.

Question 25. Differentiate between work and power.

Answer:

Question 26. Differentiate between energy and power.

Answer:

Question 27. State and define the S.I. unit of power.

Answer: S.I unit of power is watt (W). If 1 joule of work is done in 1 second, the power spent is said to be 1 watt.
 

Question 28. What is horse power (H.P)? How is it related to the S.I. unit of power?

Answer: Horse power is another unit of power, largely used in mechanical engineering. It is related to the S.I unit watt as : 1 H.P = 746 W
 

Question 29. Differentiate between watt and watt hour.

Answer: Watt (W) is the unit of power, while watt hour (Wh) is the unit of work, since power \( \times \) time = work.
 

Question 30. Name the quality which is measured in (a) kWh (b) kW (c) Wh (d) eV

Answer: a. Energy is measured in kWh b. Power is measure in kW c. Energy is measured in Wh d. Energy is meaused in eV

In simple words: Energy has bigger units like kWh (kilowatt hour) and Wh (watt hour). Similarly bigger unit of power is kW (kilo watt). The energy of atomic particles is very small, and hence, it is measured in eV (electron volt).
 

Question 1. One horse power is equal to:

Answer: 746 W
 

Question 2. kWh is the unit of:

Answer: The unit kWh is the unit of energy.
 

Question 1. A body, when acted upon by a force of 10 kgf, gets displaced by 0.5 m. Calculate the work done by the force, when the displacement is (i) in the direction of force, (ii) at an angle of 60 \(^\circ\) with the force, and (iii) normal to the force. (g = 10 N kg\(^{-1}\))

Answer: Force acting on the body = 10 kgf = 10 \(\times\) 10 N = 100 N
Displacement, S = 0.5 m
Work done = force x displacement in the direction of force
(i) W = F \(\times\) S
W = 100 \(\times\) 0.5 = 50 J
(ii) Work = force x displacement in the direction of force
W = F \(\times\) S cos \(\theta\)
W = 100 \(\times\) 0.5 cos60 \(^\circ\)
W = 100 \(\times\) 0.5 \(\times\) 0.5 (cos 60 \(^\circ\) = 0.5)
W = 25 J
(iii) Normal to the force:
Work = force x displacement in the direction of force
W = F \(\times\) S cos \(\theta\)
W = 100 \(\times\) 0.5 cos90 \(^\circ\)
W = 100 \(\times\) 0.5 \(\times\) 0 = 0 J (cos90 \(^\circ\) = 0)
 

Question 2. A boy of mass 40 kg runs upstairs and reaches the 8 m high floor in 5 s. Calculate: (i) the force of gravity acting on the boy. (ii) the work done by him against gravity. (iii) the power spent by boy. (Take g = 10 m s\(^{-2}\))

Answer: Mass of boy = 40 kg
Vertical height moved, h = 8m
Time taken, t = 5s.
(i) Force of gravity on the boy
F = mg = 40 \(\times\) 10 = 400N
(ii) While climbing, the boy has to do work against the force of gravity.
Work done by the boy in climbing = Force \(\times\) distance moved in the direction of force
Or, W = F \(\times\) S = 400 \(\times\) 8 = 3200 J
(iii) Power spent = \(\frac{\text{work done}}{\text{time taken}}\) = \(\frac{3200}{5}\) = 640W
 

Question 3. It takes 20 s for a person A to climb up the stairs, while another person B does the same in 15 s. Compare the (i) Work done and (ii) power developed by the persons A and B.

Answer: (i) The work done by persons A and B is independent of time. Hence both A and B will do the same amount of work. Hence,
 

Question 4. A boy weighing 350 N runs up a flight of 30 steps, each 20 cm high in 1 minute, Calculate: (i) the work done and (ii) power spent.

Answer: Total distance covered in 30 steps, S = 30 x 20 cm = 600 cm = 6 m
Work done by the boy in climbing = Force x distance moved in direction of force
Work, W = F x S = 350 x 6 = 2100 J
Power developed = \(\frac{\text{work done}}{\text{time taken}}\) = \(\frac{2100\text{J}}{60\text{s}}\) = 35 W
 

Question 5. A man spends 6.4 KJ energy in displacing a body by 64 m in the direction in which he applies force, in 2.5 s Calculate: (i) the force applied and (ii) the power Spent (in H.P) by the man.

Answer: Work done by man = 6.4 kJ
Distance moved, S = 64 m
(i) Work done by the man = Force x distance moved in direction of force
Work, W = F × S
\(6.4 \times 10^3\) = F × 64
\(F = \frac{6.4 \times 10^3}{64}\) = 100 N
(ii) Power spent = \(\frac{6.4 \times 10^3}{2.5}\) = 2560 W
1 H.P = 746 W
\(1\text{W} = \frac{1}{746} \text{ H.P}\)
\(2560\text{ W} = \frac{2560}{746} \text{ H.P}\) = 3.43 H.P
 

Question 6. A weight lifter a load of 200 kgf to a height of 2.5 m in 5 s. Calculate: (i) the work done, and (ii) the power developed by him. Take \(g = 10 \text{ N kg}^{-1}\)

Answer: Force = mg = 200 × 10 = 2000 N
Distance, S = 2.5 m
Time, t = 5 s
(i) Work done, W = F S
W = 2000 × 2.5 m = 5000 J
Power developed = \(\frac{\text{work done}}{\text{time taken}}\) = \(\frac{5000\text{J}}{5\text{s}}\) = 1000 W
 

Question 7. A machine raises a load of 750 N through a height of 16 m in 5 s. calculate: (i) energy spent by machine, (ii) power at which the machine works.

Answer: (i) Energy spent by machine or work done = F S
Work, W = 750 × 16 = 12000 J
(ii) Power spent = \(\frac{\text{work done}}{\text{time taken}}\) = \(\frac{12000\text{J}}{5\text{s}}\) = 2400 W
 

Question 8. An electric heater of power 3 KW is used for 10 h. How much energy does it consume? Express your answer in (i) kWh, (ii) joule.
 

Question 9. A boy of mass 40 kg runs up a flight of 15 steps each 15 cm high in 10 s. Find: (i) the work done and (ii) the power developed by him Take g = 10 \(N kg^{-1}\)

Answer: Force of gravity on boy F= mg = 40 × 10 = 400 N
Total distance covered in 15 steps , S = 15 × 15cm = 225cm = 2.25m
Work done by the boy in climbing = Force x distance moved in direction of force
Work, W = F × S = 400 × 2.25 = 900J
Power developed = \(\frac{\text{work done}}{\text{time taken}}\) = \(\frac{900J}{10s}\) = 90W
 

Question 10. A water pump raises 50 litres of water through a height of 25 m in 5 s. Calculate the power which the pump supplies. (Take g = 10 \(N kg^{-1}\) and density of water = 1000 \(kg m^{-3}\))

Answer: Volume of water = 50 L = 50 × \(10^{-3}\) \(m^3\)
Density of water = 1000 \(kg m^{-3}\)
Mass of water = Volume of water × density of water
= 50 × \(10^{-3}\) × 1000 = 50kg
Work done in raising 50kg water to a height of 25m against the force of gravity is:
W = mg × h = mgh
Power P = \(\frac{\text{work done}}{\text{time taken}}\) = \(\frac{mgh}{t}\) = \(\frac{50 \times 10 \times 25J}{5s}\) = 2500W
 

Question 11. A man raises a box of mass 50 kg to a height of 2 m in 2 minutes, while another man raises the same box to the same height in 5 minutes. Compare: (i) the work done and (ii) the power developed by them.

Answer: (i) Work done in raising a 50kg mass to a height 2m against the force of gravity is:
W = mg × h = mgh
Hence both men will do the same amount of work. Hence,
\(\frac{\text{work done by A}}{\text{work done by B}}\) = \(\frac{mgh}{mgh}\) = \(\frac{50 \times 10 \times 2}{50 \times 10 \times 2}\) = \(\frac{1}{1}\) = 1 : 1
(ii) First man A takes 2 minutes to raise 50 kg mass
Second man B takes 5 minutes to raise 50 kg mass.
Power developed by man A = \(\frac{\text{work done}}{\text{Time taken}}\) = \(\frac{mgh}{t}\) = \(\frac{50 \times 10 \times 2 J}{120 S}\) = \(\frac{25}{3}\) w
Power developed by man B = \(\frac{\text{work done}}{\text{Time taken}}\) = \(\frac{mgh}{t}\) = \(\frac{50 \times 10 \times 2 J}{5 \times 60 s}\) = \(\frac{10}{3}\) w
Power developed \(\propto \frac{1}{\text{time taken}}\)
\(\frac{\text{Power developed by A}}{\text{Power developed by B}}\) = \(\frac{\frac{25}{3}}{\frac{10}{3}}\) = \(\frac{25}{10}\) = \(\frac{5}{2}\) = 5 : 2
 

Question 12. A pump is used to lift 500 kg of water from a depth of 80 m in 10 s. calculate: (a) the work done by the pump (b) the power at which the pump works, (c) the power rating of the pump if its efficiency is 40% (Take g = 10 \(m s^{-2}\))

Answer: Work done in raising a 500kg mass to a height of 80m against the force of gravity is:
(a) W = mg × h = mgh
W = 500 × 10 × 80 = 4 × \(10^5\) J
(b) Power at which pump works = \(\frac{\text{work done}}{\text{Time taken}}\) = \(\frac{mgh}{t}\) = \(\frac{500 \times 10 \times 80J}{10S}\) = \(\frac{4 \times 10^5}{10}\) = 40 KW
(c) Efficiency = \(\frac{\text{useful power}}{\text{power input}}\)
Efficiency = 40 % = 0.4
0.4 = \(\frac{40 KW}{\text{Power input}}\)
Power input = \(\frac{40 KW}{0.4}\) = 100kw
 

Question 13. An ox can apply a maximum force of 1000 N. It is taking part in a cart race and is able to pull the cart at a constant speed of 30 m s\(^{-1}\) while making its best effort. Calculate the power developed by the ox.

Answer: Given, force = 1000N, velocity = 30m/s Power, P = force \( \times \) velocity P = 1000 \( \times \) 30 = 30,000W = 30 kW
 

Question 14. If the power of a motor is 40 kw, at what speed can it raise a load of 20,000 N?

Answer: Power = 40kW Force = 20,000N Power = force \( \times \) velocity Velocity = \( \frac{\text{Power}}{\text{Force}} \) = \( \frac{40 \text{ KW}}{20,000} \) = \( \frac{40,000}{20,000} \) = 2 m/s
 

Question 1. What are the two forms of mechanical energy?

Answer: Two forms of mechanical energy are: (i) Kinetic energy (ii) Potential energy
 

Question 2. Name the forms of energy which a wound-up watch spring possesses.

Answer: Elastic potential energy is possessed by wound up watch spring
 

Question 3. Name the type of energy (kinetic energy K or potential energy U) possessed in the following cases: (a) A moving cricket ball (b) A compressed spring (c) A moving bus (d) The bob of a simple pendulum at its extreme position. (e) The bob of a simple pendulum at its mean position. (f) A piece of stone placed on the roof.

Answer: (a) Kinetic energy (K) (b) Potential energy (U) (c) Kinetic energy (K) (d) Potential energy (U) (e) Kinetic energy (K) (f) Potential energy (U)
 

Question 4. When an arrow is shot from a bow, it has kinetic energy in it. Explain briefly from where does it get its kinetic energy?

Answer: When the string of a bow is pulled, some work is done which is stored in the deformed state of the bow in the form of its elastic potential energy. On releasing the string to shoot an arrow, the potential energy of the bow changes into the kinetic energy of the arrow which makes it move
 

Question 5. Define the term potential energy of a body. State its different forms and give one example of each.

Answer: Potential energy: The energy possessed by a body by virtue of its specific position (or changed configuration) is called the potential energy. Different forms of P.E. are as listed below: (i) Gravitational potential energy: The potential energy possessed by a body due to its position relative to the centre of Earth is called its gravitational potential energy. Example: A stone at a height has gravitational potential energy due to its raised height.
 

Question 6. A ball is placed on a compressed spring. What form of energy does the spring possess? On releasing the spring, the ball flies away. Give a reason.

Answer: The compressed spring has elastic potential energy due to its compressed state. When it is released, the potential energy of the spring changes into kinetic energy which does work on the ball if placed on it and changes into kinetic energy of the ball due to which it flies away.
 

Question 7. What is meant by the gravitational potential energy? Derive expression for it.

Answer: Gravitational potential energy is the potential energy possessed by a body due to its position relative to the centre of earth. For a body placed at a height above the ground, the gravitational potential energy is measured by the amount of work done in lifting it up to that height against the force of gravity. Let a body of mass m be lifted from the ground to a vertical height h. The least upward force F required to lift the body (without acceleration) must be equal to the force of gravity (=mg) on the body acting vertically downwards. The work done W on the body in lifting it to a height h is \(W = \text{force of gravity} (mg) \times \text{displacement} (h) = mgh\). This work is stored in the body when it is at a height h in the form of its gravitational potential energy. Gravitational potential energy \(U = mgh\)
 

Question 8. Write an expression for the potential energy of a body of mass m places at a height h above the earth’s surface.

Answer: The work done W on the body in lifting it to a height h is \(W = \text{force of gravity} (mg) \times \text{displacement} (h) = mgh\). This work is stored in the body when it is at a height h in the form of its gravitational potential energy. Gravitational potential energy \(U = mgh\)
 

Question 9. Name the form of energy which a body may possess even when it is not in motion. Give an example to support your answer.

Answer: Potential energy is possessed by the body even when it is not in motion. For example: a stone at a height has the gravitational potential energy due to its raised position.
 

Question 10. What do you understand by the kinetic energy of a body?

Answer: A body in motion is said to possess the kinetic energy. The energy possessed by a body by virtue of its state of motion is called the kinetic energy.
 

Question 11. A body of mass m is moving with a velocity v. Write the expression for its kinetic energy.

Answer: Kinetic energy \( = \frac{1}{2} \times \text{mass} \times (\text{velocity})^2 = \frac{1}{2} m v^2\)
 

Question 13. A body of mass m is moving with a uniform velocity u. A force is applied on the body due to which its velocity changes from u to v. How much work is being done by the force.

Answer: Body of mass \(m\) is moving with a uniform velocity \(u\). A force is applied on the body due to which its velocity changes from \(u\) to \(v\) and produces an acceleration \(a\) in moving a distance \(S\). Then, Work done by the force = force \(\times\) displacement. \(W = F \times S\) --------- (i) From relation : \(v^2 = u^2 + 2aS\). Displacement, \(S = \frac{v^2 - u^2}{2a}\). And force, \(F = ma\). From equation (i), \(W = ma \times \left( \frac{v^2 - u^2}{2a} \right)\). \(W = \frac{1}{2} m (v^2 - u^2)\). This can also be written as \(W = K_f - K_i\). Where \(K_i\) is the initial kinetic energy \(K_i = \frac{1}{2} m u^2\). And \(K_f\) is the final kinetic energy \(K_f = \frac{1}{2} m v^2\). Thus work done on the body = increase in kinetic energy. \(W = \frac{1}{2} m (v^2 - u^2)\).
 

Question 14. A light mass and a heavy mass have equal momentum. Which will have more kinetic energy? (Hint : Kinetic energy K = P 2 /2m where P is the momentum)

Answer: Kinetic energy, \(k = \frac{p^2}{2m}\) where \(p\) is the momentum. Both the masses have same momentum \(p\). The kinetic energy, \(K\) is inversely proportional to mass of the body. Hence light mass body has more kinetic energy because smaller the mass, larger is the kinetic energy.
 

Question 15. Name the three forms of kinetic energy and give on example of each.

Answer: The three forms of kinetic energy are: (i) Translational kinetic energy - example: a freely falling body; (ii) Rotational kinetic energy - example: A spinning top; (iii) Vibrational kinetic energy - example: atoms in a solid vibrating about their mean position.
 

Question 16. Differentiate between the potential energy (U ) and the kinetic energy (K)

Answer: Potential energy (\(U\)) is the energy possessed by a body by virtue of its specific position or changed configuration. Its forms include gravitational potential energy and elastic potential energy. For example, a wound up watch spring has potential energy. In contrast, Kinetic energy (\(K\)) is the energy possessed by a body by virtue of its state of motion. Its forms include translational, rotational, and vibrational kinetic energy. For example, a moving car has kinetic energy.
 

Question 17. Complete the following sentences: (a) The kinetic energy of a body is the energy by virtue of its............. (b) The potential energy of a body is the energy by virtue of its ...................

Answer: (a) Motion. (b) Position.
 

Question 18. Is it possible that no transfer of energy may take place even when a force is applied to a body?

Answer: Yes, when force is normal to displacement, no transfer of energy takes place.
 

Question 19. Name the form of mechanical energy, which is put to use.

Answer: Kinetic energy.
 

Question 20. In what way does the temperature of water at the bottom of a waterfall differ from the temperature at the top? Explain the reason.

Answer: When water falls from a height, the potential energy stored in water at a height changes into the kinetic energy of water during the fall. On striking the ground, a part of the kinetic energy of water changes into the heat energy due to which the temperature of water rises.
 

Question 21. Name six different forms of energy?

Answer: The six different forms of energy are: 1. Solar energy 2. Heat energy 3. Light energy 4. Chemical or fuel energy 5. Hydro energy 6. Nuclear energy
 

Question 22. Energy can exist in several forms and may change from one form to another. For each of the following, state the energy changes that occur in: (a) the unwinding of a watch spring (b) a loaded truck when started and set in motion. (c) a car going uphill (d) photosynthesis in green leaves (e) Charging of a battery, (f) respiration, (g) burning of a match stick (h) explosion of crackers.

Answer: (a) Potential energy of wound up spring converts into kinetic energy. (b) Chemical energy of petrol or diesel converts into mechanical energy (kinetic energy) (c) Kinetic energy to potential energy (d) Light energy changes into chemical energy (e) Electrical energy changes into chemical energy (f) Chemical energy changes into heat energy (g) Chemical energy changes into heat and light energy (h) Chemical energy changes into heat, light and sound energy
 

Question 23. State the energy changes in the following cases while in use: (a) loudspeaker (b) a steam engine (c) microphone (d) washing machine (e) an electric bulb (f) burning coal (g) a solar cell (h) bio-gas burner (i) an electric cell in a circuit (j) a petrol engine of a running car (k) an electric toaster (l) a photovoltaic cell (m) an electromagnet.

Answer: (a) Electrical energy into sound energy (b) Heat energy into mechanical energy (c) Sound energy into electrical energy (d) Electrical energy to mechanical energy (e) Electrical energy into light energy
 

Question 1. A body at a height possesses:

Answer: (b) Potential energy

In simple words: When an object is lifted up, it stores energy because of its position. This stored energy is called potential energy. Think of a ball held high up, ready to fall.
 

Question 2. In an electric cell which in use, the change in energy is from:

Answer: (d) Chemical to electrical

In simple words: An electric cell, like a battery, has special chemicals inside. When you use it, these chemicals react and change their stored chemical energy into electrical energy to power your toys or devices.
 

Question 1. Two bodies of equal masses are placed at heights h and 2h. Find the ratio of their gravitational potential energies.

Answer: Height \(H_1 = h\)
Height \(H_2 = 2h\)
Mass of body 1 = \(m\)
Mass of body 2 = \(m\)
Gravitational potential energy of body 1 = \(mgH_1 = mgh\)
Gravitational potential energy of Body 2 = \(mgH_2 = mg(2h)\)
Ratio of gravitational potential energies = \(\frac{mgh}{mg(2h)} = \frac{1}{2} = 1 : 2\)
 

Question 2. Find the gravitational potential energy of 1 kg mass kept at a height of 5 m above the ground if g = 10 m s\(^{-2}\).

Answer: Mass, \(m = 1 \text{ kg}\)
Height, \(h = 5 \text{ m}\)
Gravitational potential energy = \(mgh\)
= \(1 \times 10 \times 5 = 50 \text{ J}\)
 

Question 3. A box of weight 150 kgf has gravitational potential energy stored in it equal to 14700 J. Find the height of the box above the ground. (Take g = 9.8 N kg\(^{-1}\))

Answer: Gravitational potential energy = \(14700 \text{ J}\)
Force of gravity = \(mg = 150 \times 9.8 \text{ N/kg} = 1470 \text{ N}\)
Gravitational potential energy = \(mgh\)
\(14700 = 1470 \times h\)
\(h = 10 \text{ m}\)
 

Question 4. A body of mass 5 kg falls from a height of 10 m to 4 m. Calculate: (i) the loss in potential energy of the body, (ii) the total energy possessed by the body at any instant? (Take g = 10 m s\(^{-2}\))

Answer: (i) Mass of the body = \(5 \text{ kg}\)
P.E. at height \(10 \text{ m}\) = \(mgh\) = \(5 \times 10 \times 10 = 500 \text{ J}\)
P.E. at height \(4 \text{ m}\) = \(mgh\) = \(5 \times 10 \times 4 = 200 \text{ J}\)
Loss in P.E. = \((500 - 200) \text{ J} = 300 \text{ J}\)
(ii) The total energy possessed by the body at any instant remains constant for free fall. It is equal to the sum of P.E. and K.E.
At height \(10 \text{ m}\), i.e., at topmost point, K.E. = \(0\)
Total energy = P.E. + K.E.
Total energy = \(500 + 0 = 500 \text{ J}\)
 

Question 5. Calculate the height through which a body of mass 0.5 kg is lifted if the energy spent in doing so is 1.0 J. Take g = 10 m s\(^{-2}\)

Answer: Mass = \(0.5 \text{ kg}\)
Energy = \(1 \text{ J}\)
Gravitational potential energy = \(mgh\)
\(1 = 0.5 \times 10 \times h\)
\(1 = 5h\)
Height, \(h = 0.2 \text{ m}\)
 

Question 6. A boy weighing 25 kgf climbs up from the first floor at height 3 m above the ground to the third floor at height 9m above the ground. What will be the increase in his gravitational potential energy? (Take g = 10 N kg\(^{-1}\))

Answer: Force of gravity on boy = \(mg = 25 \times 10 = 250 \text{ N}\)
Increase in gravitational potential energy = \(Mg(h_2 - h_1)\)
= \(250 \times (9-3)\)
= \(250 \times 6 = 1500 \text{ J}\)
 

Question 7. A vessel containing 50 kg of water is placed at a height 15 m above the ground. Assuming the gravitational potential energy at ground to be zero, what will be the gravitational potential energy of water in the vessel? (g = 10 m s\(^{-2}\))

Answer: Mass of water, \(m = 50 \text{ kg}\)
Height, \(h = 15 \text{ m}\)
Gravitational potential energy = \(mgh\)
= \(50 \times 10 \times 15\)
= \(7500 \text{ J}\)
 

Question 8. A man of mass 50 kg climbs up a ladder of height 10 m. Calculate: (i) the work done by the man, (ii) the increase in his potential energy. (g = 9.8 m s\(^{-2}\))

Answer: Mass of man = \(50 \text{ kg}\)
Height of ladder, \(h_2 = 10 \text{ m}\)
(i) Work done by man = \(mgh_2\)
= \(50 \times 9.8 \times 10 = 4900 \text{ J}\)
(ii) Increase in his potential energy:
Height, \(h_2 = 10 \text{ m}\)
Reference point is ground, \(h_1 = 0 \text{ m}\)
Gravitational potential energy = \(Mg(h_2 - h_1)\)
= \(50 \times 9.8 \times (10 - 0)\)
= \(50 \times 9.8 \times 10 = 4900 \text{ J}\)
 

Question 9. A block A, whose weight is 200 N, is pulled up a slope of length 5 m by means of a constant force F (= 150 N) as illustrated in Fig 2.13

Answer: [Diagram: Fig 2.13 illustration of block A being pulled up a slope]
The solution for this question is not provided in the extracted text.
 

Question 9. (a) what is the work done by the force F in moving the block A, 5 m along the slope? (b) By how much has the potential energy of the block A increased? (c) Account for the difference in work done by the force and the increase in potential energy of the block.

Answer: Given, \(F = 150\text{N}\)
(a) Work done by the force in moving the block 5m along the slope = Force x displacement in the direction of force = \(150 \times 5 = 750\text{ J}\)
(b) The potential energy gained by the block \(U = mgh\) where \(h = 3\text{m}\) = \(200 \times 3 = 600\text{ J}\)
(c) The difference i.e., \(150\text{ J}\) energy is used in doing work against friction between the block and the slope, which will appear as heat energy.
 

Question 10. Find the kinetic energy of a body of mass 1 kg moving with a uniform velocity of 10 m s^-1.

Answer: Mass, \(m = 1\text{kg}\)
Velocity, \(v = 10\text{ m s}^{-1}\)
Kinetic energy = \(\frac{1}{2} \times \text{mass} \times (\text{velocity})^2\)
= \(\frac{1}{2} \times 1 \times (10)^2\)
= \(\frac{1}{2} \times 1 \times 100\)
= \(50\text{J}\)
 

Question 11. If the speed of a car is halved, how does its kinetic energy change?

Answer: If the speed is halved (keeping the mass same), the kinetic energy decreases, it becomes one-fourth (since kinetic energy is proportional to the square of velocity, i.e., \(K \propto v^2\)).
 

Question 12. Two bodies of equal masses are moving with uniform velocities v and 2v. Find the ratio of their kinetic energies.

Answer: Given, velocity of first body \(v_1 = v\)
And velocity of second body, \(v_2 = 2v\)
Since masses are same, kinetic energy is directly proportional to the square of the velocity (\(K \propto v^2\)).
Hence, ratio of their kinetic energies is:
\(\frac{K_1}{K_2} = \frac{v_1^2}{v_2^2}\)
= \(\frac{v^2}{(2v)^2}\)
= \(\frac{v^2}{4v^2}\)
= \(\frac{1}{4}\)
= \(1 : 4\)
 

Question 13. A car is running at a speed of 15 km h^-1 while another similar car is moving at a speed of 30 km h^-1. Find the ratio of their kinetic energies.

Answer: Given, velocity of first car, \(v_1 = 15 \text{ km/h}\)
And velocity of second car, \(v_2 = 30 \text{ km/h}\)
Since masses are same, kinetic energy is directly proportional to the square of the velocity (\(K \propto v^2\)).
Hence, ratio of their kinetic energies is:
\(\frac{K_1}{K_2} = \frac{v_1^2}{v_2^2}\)
= \(\frac{15^2}{(30)^2}\)
= \(\frac{15 \times 15}{30 \times 30}\)
= \(\frac{1}{4}\)
= \(1 : 4\)
 

Question 14. A bullet of mass 0.5 kg slows down from a speed of 5 m s^-1 to that of 3 m s^-1. Calculate the change in kinetic energy of the ball.

Answer: Mass of ball = \(0.5\text{kg}\)
Initial velocity = \(5\text{ m s}^{-1}\)
Initial kinetic energy = \(\frac{1}{2} \times \text{mass} \times (\text{velocity})^2\)
= \(\frac{1}{2} \times 0.5 \times (5)^2\)
= \(\frac{1}{2} \times 0.5 \times 25\)
= \(6.25\text{ J}\)
Final velocity of the ball = \(3\text{ m s}^{-1}\)
Final kinetic energy of the ball = \(\frac{1}{2} \times \text{mass} \times (\text{velocity})^2\)
= \(\frac{1}{2} \times 0.5 \times (3)^2\)
= \(\frac{1}{2} \times 0.5 \times 9\)
= \(2.25\text{ J}\)
Change in kinetic energy = Initial kinetic energy - Final kinetic energy
= \(6.25\text{ J} - 2.25\text{ J}\)
= \(4.00\text{ J}\)
 

Question 15. A cannon ball of mass 500 g is fired with a speed of 15 \(m s^{-1}\). Find: (i) its kinetic energy and (ii) its momentum.

Answer: Mass of canon ball = 500g = 0.5 kg
Speed, v = 15\(m/s\)
(a) Kinetic energy of ball = \( \frac{1}{2} \times \text{mass} \times (\text{velocity})^2 \)
= \( \frac{1}{2} \times 0.5 \times (15)^2 \)
= \( \frac{1}{2} \times 0.5 \times 225 \)
= 56.25 J
(b) Momentum of the ball = mass \(\times\) velocity
= 0.5 \(\times\) 15 = 7.5 \(kgm/s\)
 

Question 16. A bullet of mass 50 g is moving with a velocity of 500 \(m s^{-1}\). It penetrated 10 cm into a still target and comes to rest. Calculate : (a) the kinetic energy possessed by the bullet, (b) the average retarding force offered by the target.

Answer: Mass of bullet = 50g = 0.05kg
Velocity = 500\(m/s\)
Distance penetrated by the bullet = 10cm = 0.1m
(a) Kinetic energy of the bullet = \( \frac{1}{2} \times \text{mass} \times (\text{velocity})^2 \)
= \( \frac{1}{2} \times 0.05 \times (500)^2 \)
= \( \frac{1}{2} \times 0.05 \times 500 \times 500 \)
= 6250 J
(b) Work done by the bullet against the material of the target = resistive force \(\times\) distance
6250 = resistive force \(\times\) 0.1m
Resistive force = 62500 N
 

Question 17. A body of mass 10 kg is moving with a velocity 20 \(m s^{-1}\). If the mass of the body is doubled and its velocity is halved, find the ratio of the initial kinetic energy to the final kinetic energy.

Answer: Let initial Mass, \(m_1\) = 10kg and velocity, \(v_1\) = 20 \(m/s\)
Final mass, \(m_2\) = 2 \(\times\) 10 = 20 kg and velocity, \(v_2\) = 20/2 = 10\(m/s\)
Initial kinetic energy, \(K_1\) = \( \frac{1}{2} \times \text{mass} \times (\text{velocity})^2 \)
= \( \frac{1}{2} \times 10 \times (20)^2 \)
= \( \frac{1}{2} \times 10 \times 20 \times 20 \)
= 2000 J
Final kinetic energy, \(K_2\) = \( \frac{1}{2} \times \text{mass} \times (\text{velocity})^2 \)
= \( \frac{1}{2} \times 20 \times (10)^2 \)
= \( \frac{1}{2} \times 20 \times 10 \times 10 \)
= 1000J
\( \frac{K_1}{K_2} = \frac{2000}{1000}