Work, Energy and Power: ICSE Class 10 Physics answers, notes
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Summary
In physics, work is done when a force causes an object to move over a distance. If you push a car and it moves, you have done work. However, if you push against a wall and it does not move, no work is done, even though you might feel tired. The amount of work depends on both the size of the force and the distance the object moves. Work can be positive if the force helps the motion, or negative if the force opposes the motion, like when friction slows something down.
Energy is the ability to do work.When work is done, energy is transferred from one object to another. Work and energy are measured in the same units, called joules. Energy exists in many forms. The two main types are potential energy and kinetic energy. Potential energy is stored energy. A stretched rubber band or a rock at the top of a cliff has potential energy. Kinetic energy is the energy of motion.A moving car or a rolling ball has kinetic energy.These two forms are part of what is called mechanical energy.
A very important rule is the law of conservation of energy, which states that energy cannot be created or destroyed; it only changes from one form to another. For example, as a ball falls, its potential energy (due to its height) changes into kinetic energy (due to its motion). When a pendulum swings back and forth, it constantly changes its energy between potential and kinetic, but the total amount of mechanical energy stays the same if we ignore air resistance. Other forms of energy include chemical energy stored in food and batteries, electrical energy that powers our homes, and heat, light, and sound energy.
Power is the rate at which work is done, or how fast energy is used. If two people lift the same load, the person who does it faster uses more power. Power is measured in a unit called the watt, where one watt is equal to one joule of work done per second. Another unit for power is horsepower, which is often used for engines. A powerful engine can do a lot of work in a short amount of time.
Textbook solutions
Exercise A
MCQs
1. In vector form, work done W is written as :
(a) W→ = F→S→
(b) W = F→S→
(c) W = F→ × S→ cos θ
(d) W→ = F × S cos θ
Answer: (b) w = F→S→
2. For work done to be maximum, the angle between the force and displacement should be :
(a) 0°
(b) 30°
(c) 45°
(d) 90°
Answer: (a) 0°
3. When a body moves in a circular path in a horizontal plane, the work done is :
(a) Positive
(b) Negative
(c) Zero
(d) Maximum
Answer: (c) Zero
4. When a ball of mass m is thrown upwards to a height h, the work done by the force of gravity is :
(a) mgh
(b) -mgh
(c) mg
(d) mgh²
Answer: (b) -mgh
5. A coolie A takes 1 minute to lift a load to the roof of a bus, whereas coolie B takes 2 minutes to lift the same load to the roof of the same bus. The work done by B is:
(a) equal to work done by A
(b) greater than work done by A
(c) less than work done by A
(d) none of the above
Answer: (a) equal to work done by A
6. One horse power is equal to :
(a) 1000 W
(b) 500 W
(c) 764 W
(d) 746 W
Answer: (d) 746 W
7. 1 Joule is equivalent to ………. erg.
(a) 10⁵
(b) 10⁷
(c) 10⁹
(d) 1
Answer: (b) 10⁷
8. The energy of electron is 2.5 eV. When expressed in Joule, it is :
(a) 1.6 × 10⁻¹⁹ J
(b) 4.0 × 10⁻¹⁹ J
(c) 3.2 × 10⁻¹⁹ J
(d) 0
Answer: (b) 4.0 × 10⁻¹⁹ J
9. A block of mass 20 kg is pulled up a slope as shown in the figure given below with a constant speed by applying a force of 200 N parallel to the slope from the initial position A to the final position B. Work done in moving the block from A to B is :
(a) 600 J
(b) 500 J
(c) 60 J
(d) 200 J
Answer: (a) 600 J
10. Assertion (A): When the displacement is normal to the direction of force, the work done is zero.
Reason (R) : Work done depends on the angle between force and displacement.
(a) both A and R are true and R is the correct explanation of A
(b) both A and R are True and R is not the correct explanation of A
(c) assertion is false but reason is true
(d) assertion is true but reason is false
Answer: (a) both A and R are true and R is the correct explanation of A
Very Short Questions
1. 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 so that the work done is (i) zero, (ii) maximum?
Answer:
(a) The expression for the work done W by the force is W = F × S cos θ.
(b) (i) For the work done to be zero, the angle θ between the force and displacement should be 90°, because cos 90° = 0, making W = 0.
(ii) For the work done to be maximum, the angle θ between the force and displacement should be 0°, because cos 0° = 1, making W = F × S, which is the maximum positive work.
2. 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: When a body is moved in a direction opposite to the direction of force acting on it, work is done against the force. For example, if a body of mass m goes up through a vertical height h, the work W = -mgh is done by the force of gravity on the body, or the work W = mgh is done by the body against the force of gravity.
3. State whether work is done or not, by writing yes or no, in the following cases:
(a) A man pushes a wall.
(b) A coolie stands with a box on his head for 15 min.
(c) A boy climbs up 20 stairs.
Answer: (a) No. If a man tries to push a wall and is unable to move it, then scientifically, no work is being done by him.
(b) No. A coolie does no work while standing with a heavy load on his head since the displacement of the load is zero.
(c) Yes. A boy climbing up stairs exerts a force which produces motion, so he does work.
4. Give an example when work done by the force of gravity acting on a body is zero even though the body gets displaced from its initial position.
Answer: An example is when a coolie walks on a horizontal ground while carrying a load on his head. In this case, no work is done by the force of gravity because the displacement of the load is horizontal, which is normal (at 90°) to the direction of the force of gravity (which is vertically downwards). Since cos 90° = 0, the work done by the force of gravity is zero.
5. A boy of mass m climbs up the stairs 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:
(a) The work done by the boy against the force of gravity is mgh.
(b) If he uses a lift in climbing the same vertical height h, the work done by the body against the force of gravity would still be mgh. The work done by the force of gravity is the same whether the body comes down or goes up from a certain height using stairs or slope or a lift.
6. What physical quantity does the electron volt (eV) measure? How is it related to the S.I. unit of that quantity?
Answer: The electron volt (eV) measures energy, particularly the energy transfer in the case of atomic particles, which is very small.
It is related to the S.I. unit of energy, joule (J), as 1 eV = 1.6 × 10⁻¹⁹ joule.
7. Complete the following sentences:
(a) 1 J = ______ calorie.
(b) 1 kWh = ______ J.
Answer: (a) 1 J = 0.24 calorie.
(b) 1 kWh = 3.6 × 10⁶ J.
8. Name the physical quantity which is measured in calorie. How is it related to the S.I. unit of that quantity?
Answer: The physical quantity measured in calorie is heat energy.
It is related to the S.I. unit of energy, joule (J), as 1 calorie = 4.18 J.
9. (a) Name the physical quantity measured in terms of horse power.
(b) How is horse power related to the S.I. unit of power?
Answer: (a) The physical quantity measured in terms of horse power is power.
(b) Horse power (H.P.) is related to the S.I. unit of power, watt (W), as 1 H.P. = 746 W.
10. Name the quantity which is measured in
(a) kWh (b) kW (c) Wh (d) eV.
Answer: (a) kWh (kilowatt hour) measures energy or work.
(b) kW (kilowatt) measures power.
(c) Wh (watt hour) measures energy or work.
(d) eV (electron volt) measures energy.
11. Is it possible that no transfer of energy takes place even when a force is applied to a body?
Answer: Yes, it is possible that no transfer of energy takes place even when a force is applied to a body. There is no transfer of energy if a body is acted upon by a force normal to the direction of its displacement, because in such a case, the work done is zero. For example, for a body moving in a circular path, the centripetal force is normal to its displacement and work done is zero, meaning there is no transfer of energy. Also, if there is no displacement of the body even when a force acts on it, no work is said to be done, and thus no energy is transferred.
Short Answers
1. Define work. When is work said to be done by a force?
Answer: Work is said to be done only when the force applied on a body makes the body move (i.e., there is a displacement of the body). Work is said to be done only when a body moves under the influence of a force.
2. How is the work done by a force measured when (i) force is in the direction of displacement, (ii) force is at an angle to the direction of displacement?
Answer: The work done by a force is measured as follows:
(i) When the force is in the direction of displacement, suppose a constant force F displaces a body from position A to position B along its own direction. Then the displacement of the body is AB (= S), and the work done is W = F X S.
(ii) When the force is at an angle to the direction of displacement, the work done W = FS cos θ. Alternatively, work done can be expressed as W = Force × component of displacement in the direction of force, or Work = Component of force in the direction of displacement × displacement.
3. A body is acted upon by a force. State two conditions when the work done is zero.
Answer: The amount of work done by a force is zero in the following two situations:
(1) when there is no displacement (i.e. S = 0).
(2) when the displacement is normal to the direction of force (i.e., θ = 90°).
4. When a body moves in a circular path, how much work is done by the body? Give reason.
(Hint: The body is acted upon by the centripetal force)
Answer: When a body moves in a circular path in a horizontal plane, no work is done. This is because the centripetal force on the body at any instant is directed towards the centre of the circular path and the displacement at that instant is along the tangent to the circular path, i.e., normal to the direction of force on the body (i.e. θ = 90°). Thus, the work done W = 0.
5. A satellite revolves around the earth in a circular orbit. What is the work done by the force of gravity on the satellite? Give reason.
Answer: The work done by the force of gravity on the satellite is zero. The reason is that the force of gravity on the satellite is normal to its displacement.
6. A coolie X carrying a load on his head climbs up a slope and another coolie Y carrying the identical load on his head moves the same distance on a frictionless horizontal platform. Who does more work? Explain the reason.
Answer: Coolie X does more work. Coolie X does work against the force of gravity while coolie Y does no work because his displacement is normal to the force of gravity.
7. The work done by a fielder when he takes a catch in a cricket match is negative. Explain.
Answer: The work done by a fielder when he takes a catch in a cricket match is negative because the fielder applies force opposite to the direction of displacement of the ball.
8. What are the S.I. and C.G.S units of work? How are they related? Establish the relationship.
Answer: The S.I. unit of work is joule (J). The C.G.S. unit of work is erg.
They are related as 1 joule = 10⁷ erg.
To establish the relationship:
1 joule = 1 N × 1 m
But 1 N = 10⁵ dyne and 1 m = 10² cm
Therefore, 1 joule = 10⁵ dyne × 10² cm = 10⁷ dyne × cm = 10⁷ erg.
9. State and define the S.I. unit of work.
Answer: The 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 metre in its own direction.
10. Express joule in terms of erg.
Answer: 1 joule = 1N × 1m
We know, 1 N = 105 dyne and 1 m = 102 cm
Therefore, 1 joule = 105 dyne x 102 cm
⇒ 1 joule = 107 dyne cm
⇒ 1 joule = 107 erg.
11. Define the term energy and state its S.I. unit.
Answer: The energy of a body is its capacity to do work. The S.I. unit of energy is joule (J).
12. Define kilowatt hour. How is it related to joule?
Answer: One kilowatt hour (1 kWh) is the energy spent (or work done) by a source of power 1 kW in 1 h.
It is related to joule as: 1 kilowatt hour (kWh) = 1 kilowatt × 1 hour = 1000 J s⁻¹ × 3600 s = 3.6 × 10⁶ J.
13. 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 watt (W).
14. Differentiate between work and power.
Answer: The differences between work and power are:
| Work | Power |
| 1.Work done by a force is equal to the product of force and displacement in the direction of force. | 1.Power of a source is the rate of work done by the source. |
| 2.Work done does not depend on time | 2.. Power spent depends on the time in which work is done |
| 3. S.I. unit of work is joule (J). | 3. S.I. unit of power is watt (W). |
15. Differentiate between energy and power.
Answer: The differences between energy and power are:
| Energy | Power |
| 1. Energy of a body is its capacity to do work. | 1. Power of a source is the rate at which energy is supplied by it. |
| 2. Energy spent does not depend on time. | 2. Power depends on the time in which energy is spent. |
| 3. S.I. unit of energy is joule (J). | 3. S.I. unit of power is watt (W). |
16. State and define the S.I. unit of power.
Answer: The S.I. unit of power is watt. If 1 joule of work is done in 1 second, the power spent is said to be 1 watt. That is, 1 watt = 1 joule / 1 second.
17. Differentiate between watt and watt hour.
Answer: Watt (W) is the S.I. unit of power. Watt hour (Wh) is a unit of work or energy. One watt hour is the energy spent (or work done) by a source of power 1 W in 1 h. The key difference is that watt measures the rate of doing work (power), while watt hour measures the total amount of work done or energy consumed (energy), since power × time = work or energy.
Long Questions
1. State the condition when the work done by a force is (a) positive, (b) negative. Explain with the help of examples.
Answer: (a) The work done by a force is positive if the displacement is in the direction of force, i.e., θ = 0°, then cos 0° = 1, so W = F × S. Examples:
(1) In free fall of a body of mass m under gravity through a height h, the force of gravity F (= mg) is in the direction of displacement S (= h) and the work done by the force of gravity is W = FS = mgh.
(2) A coolie does work on the load when he raises it up against the force of gravity. Both the force exerted by the coolie (= mg) and the displacement (= h), are in upward direction. The work done by the coolie in raising the load is W = mgh.
(b) The work done by a force is negative if the displacement is in a direction opposite to the force, i.e., θ = 180°, then cos 180° = -1, so W = -F × S. This is usually the case when the force opposes the motion or it tries to stop a moving body. Examples:
(1) When a body moves on a surface (rolling ball), the force of friction between the body and the surface is in a direction opposite to the motion of the body (i.e. θ = 180°). Thus, the work done on the body by the force of friction is negative.
(2) When a ball of mass m is thrown upwards from A to B to a height h, the displacement h (upwards) is opposite to the direction of force of gravity mg (downwards), so the work done on the body by the force of gravity mg in displacement h is W = -mgh i.e., negative.
2. 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 be moved down through a vertical height h. The force of gravity on the body is F (= mg) which is acting vertically downwards and the vertical displacement of the body in the direction of force is S = h. Therefore, the work done by the force of gravity on the body is W = FS = mgh.
3. State two factors on which power spent by a source depends. Explain your answer with examples.
Answer: Power spent by a source depends on the following two factors:
(1) the amount of work done by the source, and
(2) the time taken by the source to do the said work.
If a machine (or a person) does a given amount of work in less time, more power is spent by it (or the person). Example: If coolie A takes 1 minute to lift a load to the roof of a bus, while another coolie B takes 2 minutes to lift the same load to the roof of the same bus, the work done by both the coolies is same, but the power spent by coolie A is twice the power spent by coolie B because coolie A does work at double the rate (i.e., in half the time).
Numericals
1. A body, when acted upon by a force of 20 kgf, gets displaced by 1 m. Calculate the work done by the force, when the displacement is (i) in the direction of force, (ii) at an angle of 60° with the force, and (iii) normal to the force. (g = 10 N kg⁻¹)
Answer:
Given:
Force = 20 kgf
Displacement S = 1 m
Acceleration due to gravity g = 10 N kg⁻¹
First, convert the force from kgf to N:
Force F = 20 kgf
=> F = 20 × 10 N = 200 N
The formula for work done is W = F S cos θ.
(i) Displacement is in the direction of force (θ = 0°):
W = F S cos 0°
=> W = 200 N × 1 m × 1 = 200 J.
(ii) Displacement is at an angle of 60° with the force (θ = 60°):
W = F S cos 60°
=> W = 200 N × 1 m × 0.5 = 100 J.
(iii) Displacement is normal to the force (θ = 90°):
W = F S cos 90°
=> W = 200 N × 1 m × 0 = 0 J.
2. A boy of mass 40 kg climbs up the stairs and reaches the roof at a height 8 m in 5 s. Calculate :
(i) the force of gravity acting on the boy,
(ii) the work done by him against the force of gravity,
(iii) the power spent by the boy.
(Take g = 10 m s⁻²)
Answer:
Given:
Mass of the boy m = 40 kg
Height h = 8 m
Time t = 5 s
Acceleration due to gravity g = 10 m s⁻²
(i) The force of gravity acting on the boy:
Force F = mg
=> F = 40 kg × 10 m s⁻² = 400 N.
(ii) The work done by him against the force of gravity:
Work done W = Fh
=> W = 400 N × 8 m = 3200 J.
(iii) The power spent by the boy:
Power P = W / t
=> P = 3200 J / 5 s = 640 W.
3. A man spends 7.4 kJ energy in displacing a body by 74 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:
Given:
Energy spent (Work done) W = 7.4 kJ = 7400 J
Displacement S = 74 m
Time t = 2.5 s
(i) The force applied:
W = F × S
=> F = W / S
=> F = 7400 J / 74 m = 100 N.
(ii) The power spent (in H.P.) by the man:
Power P = W / t
=> P = 7400 J / 2.5 s = 2960 W.
Since 1 H.P. = 746 W,
=> Power in H.P. = 2960 / 746 ≈ 3.97 H.P.
4. A weight lifter lifted 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 N kg⁻¹.
Answer:
Given:
Load = 200 kgf
Height h = 2.5 m
Time t = 5 s
g = 10 N kg⁻¹
Force F = 200 kgf
=> F = 200 × 10 N = 2000 N.
(i) The work done:
Work done W = Fh
=> W = 2000 N × 2.5 m = 5000 J.
(ii) The power developed by him:
Power P = W / t
=> P = 5000 J / 5 s = 1000 W.
5. A machine raises a load of 750 N through a height of 16 m in 5 s. Calculate :
(i) the energy spent by the machine,
(ii) the power of the machine if it is 100% efficient.
Answer:
Given:
Load F = 750 N
Height h = 16 m
Time t = 5 s
(i) The energy spent by the machine:
Energy spent (W) = Force × height
=> W = 750 N × 16 m = 12000 J.
(ii) The power of the machine if it is 100% efficient:
Power P = W / t
=> P = 12000 J / 5 s = 2400 W.
6. 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.
Answer:
Given:
Power P = 3 kW
Time t = 10 h
(i) Energy consumed in kWh:
Energy (E) = Power × time
=> E = 3 kW × 10 h = 30 kWh.
(ii) Energy consumed in joule:
Since 1 kWh = 3.6 × 10⁶ J,
=> E = 30 × 3.6 × 10⁶ J
=> E = 1.08 × 10⁸ J.
7. A water pump raises 50 litre of water through a height of 25 m in 5 s. Calculate the power of the pump used (assuming 100% efficiency).
(Take g = 10 N kg⁻¹ and density of water = 1000 kg m⁻³).
Answer:
Given:
Volume of water V = 50 litre = 0.05 m³
Height h = 25 m
Time t = 5 s
g = 10 N kg⁻¹
Density of water ρ = 1000 kg m⁻³
Mass m = Density × Volume
=> m = 1000 kg m⁻³ × 0.05 m³ = 50 kg.
Work done W = mgh
=> W = 50 kg × 10 N kg⁻¹ × 25 m = 12500 J.
Power P = W / t
=> P = 12500 J / 5 s = 2500 W.
8. A pump is used to lift 600 kg of water from a depth of 75 m in 10 s. Calculate :
(a) the work done by the pump,
(b) the power at which the pump works, and
(c) the power rating of the pump if its efficiency is 40%. (Take g = 10 m s⁻²)
Answer:
Given:
Mass of water m = 600 kg
Depth h = 75 m
Time t = 10 s
Efficiency η = 40% = 0.40
g = 10 m s⁻²
(a) The work done by the pump:
Work done W = mgh
=> W = 600 kg × 10 m s⁻² × 75 m = 450,000 J.
(b) The power at which the pump works (useful power output):
Power (output) P_out = W / t
=> P_out = 450,000 J / 10 s = 45,000 W or 45 kW.
(c) The power rating of the pump (power input):
Efficiency η = Power output / Power input
=> Power input P_in = Power output / Efficiency
=> P_in = 45,000 W / 0.40 = 112,500 W or 112.5 kW.
9. 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 F = 1000 N
Speed v = 30 m s⁻¹
Power P = Force × speed
=> P = 1000 N × 30 m s⁻¹ = 30,000 W or 30 kW.
10. The power of a motor is 40 kW. At what speed can the motor raise a load of 20,000 N?
Answer:
Given:
Power P = 40 kW = 40,000 W
Load F = 20,000 N
Power P = Force × speed
=> Speed v = P / F
=> v = 40,000 W / 20,000 N = 2 m s⁻¹.
11. Rajan exerts a force of 150 N in pulling a cart at a constant speed of 10 m s⁻¹. Calculate the power exerted.
Answer:
Given:
Force F = 150 N
Speed v = 10 m s⁻¹
Power P = Force × speed
=> P = 150 N × 10 m s⁻¹ = 1500 W.
12. A boy weighing 350 N climbs up 30 steps, each 20 cm high in 1 minute. Calculate :
(i) the work done, and (ii) the power spent.
Answer:
Given:
Weight of the boy F = 350 N
Number of steps = 30
Height of one step = 20 cm = 0.2 m
Time t = 1 minute = 60 s
Total height h = 30 steps × 0.2 m/step = 6 m.
(i) The work done:
Work done W = F × h
=> W = 350 N × 6 m = 2100 J.
(ii) The power spent:
Power P = W / t
=> P = 2100 J / 60 s = 35 W.
13. It takes 20 s for a person A of mass 50 kg to climb up the stairs, while another person B of same mass does the same in 15 s. Compare the (i) work done, and (ii) power developed by the persons A and B.
Answer:
Given:
For person A: m_A = 50 kg, t_A = 20 s
For person B: m_B = 50 kg, t_B = 15 s
Height h is the same for both.
(i) Comparison of work done:
Work done W = mgh. Since m, g, h are the same for both,
=> W_A = W_B.
Ratio of work done = W_A / W_B = 1/1, or 1:1.
(ii) Comparison of power developed:
Power P = W / t.
Ratio of power = P_A / P_B = (W_A / t_A) / (W_B / t_B).
Since W_A = W_B,
=> P_A / P_B = t_B / t_A = 15 / 20 = 3/4.
The ratio is 3:4.
14. A boy weighing 40 kgf climbs up a stair of 30 steps each 20 cm high in 4 minutes and a girl weighing 30 kgf does the same in 3 minutes. Compare:
(i) the work done by them, and
(ii) the power developed by them.
Answer:
Given:
For the boy: F_boy = 40 kgf, t_boy = 4 min = 240 s
For the girl: F_girl = 30 kgf, t_girl = 3 min = 180 s
Total height h = 30 steps × 20 cm/step = 6 m.
(i) Comparison of the work done by them:
W_boy = F_boy × h = 40 kgf × 6 m = 240 kgf·m.
W_girl = F_girl × h = 30 kgf × 6 m = 180 kgf·m.
Ratio of work done = W_boy / W_girl = 240 / 180 = 4/3.
=> The ratio is 4:3.
(ii) Comparison of the power developed by them:
P_boy = W_boy / t_boy = 240 kgf·m / 240 s = 1 kgf·m/s.
P_girl = W_girl / t_girl = 180 kgf·m / 180 s = 1 kgf·m/s.
Ratio of power = P_boy / P_girl = 1 / 1 = 1.
=> The ratio is 1:1.
15. A man raises a box of mass 50 kg to a height of 2 m in 20 s, while another man raises the same box to the same height in 50 s.
(a) Compare: (i) the work done, and (ii) the power developed by them.
(b) Calculate: (i) the work done, and (ii) the power developed by each man. Take g = 10 N kg⁻¹.
Answer:
Given:
Mass m = 50 kg, Height h = 2 m
Time for man 1, t₁ = 20 s
Time for man 2, t₂ = 50 s
g = 10 N kg⁻¹
(a) Comparison:
(i) The work done:
Work done W = mgh. Since m, g, h are the same, the work done is the same.
=> Ratio of work done = W₁ / W₂ = 1:1.
(ii) The power developed:
Power P = W / t. Since W is the same, power is inversely proportional to time.
=> Ratio of power = P₁ / P₂ = t₂ / t₁ = 50 / 20 = 5/2. The ratio is 5:2.
(b) Calculation:
(i) The work done by each man:
W = mgh
=> W = 50 kg × 10 N kg⁻¹ × 2 m = 1000 J.
(ii) The power developed by each man:
Power of man 1, P₁ = W / t₁
=> P₁ = 1000 J / 20 s = 50 W.
Power of man 2, P₂ = W / t₂
=> P₂ = 1000 J / 50 s = 20 W.
16. A boy takes 3 minutes to lift a 20 litre water bucket from a 20 m deep well, while his father does it in 2 minutes.
(a) Compare: (i) the work, and (ii) power developed by them.
(b) How much work each does? Take density of water = 10³ kg m⁻³ and g = 9.8 N kg⁻¹.
Answer:
Given:
Volume V = 20 litre = 0.02 m³
Depth h = 20 m
Time for boy, t_boy = 3 min = 180 s
Time for father, t_father = 2 min = 120 s
Density of water ρ = 10³ kg m⁻³
g = 9.8 N kg⁻¹
Mass m = ρ × V
=> m = 1000 kg m⁻³ × 0.02 m³ = 20 kg.
(a) Comparison:
(i) The work:
Since both lift the same mass to the same height, the work done is the same.
=> Ratio of work done = W_boy / W_father = 1:1.
(ii) The power developed:
Power P = W / t. Since W is the same, power is inversely proportional to time.
=> Ratio of power = P_boy / P_father = t_father / t_boy = 120 / 180 = 2/3.
(b) How much work each does:
Work done W = mgh
=> W = 20 kg × 9.8 N kg⁻¹ × 20 m = 3920 J.
Exercise B
MCQs
1. The energy possessed by a body due to the force of attraction of earth on it is called its :
(a) Rotational kinetic energy
(b) Kinetic energy
(c) Gravitational potential energy
(d) Elastic potential energy
Answer: (c) Gravitational potential energy
2. The energy possessed by a body in the deformed state due to change in its size or shape is called:
(a) Rotational kinetic energy
(b) Kinetic energy
(c) Gravitational potential energy
(d) Elastic potential energy
Answer: (d) Elastic potential energy
3. According to work energy theorem, the increase in the kinetic energy of a moving body is equal to _________ by a force acting in the direction of the moving body.
(a) displacement
(b) velocity
(c) acceleration
(d) work done
Answer: (d) work done
4. A wire clamped at both the ends is struck in the middle. It will possess :
(a) rotational kinetic energy
(b) translational kinetic energy
(c) vibrational kinetic energy
(d) none of these
Answer: (c) vibrational kinetic energy
5. The wheel of a moving vehicle possesses :
(a) rotational kinetic energy
(b) translational kinetic energy
(c) vibrational kinetic energy
(d) both rotational and translational kinetic energy
Answer: (d) both rotational and translational kinetic energy
6. The velocity of a moving cart is reduced to 1/3rd of the initial velocity. Its kinetic energy is reduced by:
(a) 1/2
(b) 1/3
(c) 1/6
(d) 1/9
Answer: (d) 1/9
7. A body of mass 10 kg is moving with a velocity of 20 m/s. The mass of the body is doubled and velocity in halved. Its initial kinetic energy is:
(a) 200 J
(b) 4000 J
(c) 2000 J
(d) 1000 J
Answer: (c) 2000 J
8. The kinetic energy K and momentum p are related as:
(a) p = √2K
(b) p = √2gK
(c) K = p²/2m
(d) K = √2mp
Answer: (c) K = p²/2m
9. On doubling the velocity of a moving body, the quantity which gets doubled is:
(a) acceleration
(b) momentum
(c) kinetic energy
(d) weight
Answer: (b) momentum
10. In a photoelectric cell, the change in energy is from :
(a) electrical to mechanical
(b) light energy to electrical energy
(c) electrical energy to light energy
(d) chemical energy to light energy
Answer: (b) light energy to electrical energy
11. In a petrol vehicle while in motion, the change in energy is from:
(a) kinetic energy to potential energy
(b) chemical energy to kinetic energy
(c) potential energy to kinetic energy
(d) chemical energy to potential energy
Answer: (b) chemical energy to kinetic energy
12. In nuclear fission:
(a) chemical energy changes to electrical energy
(b) mechanical energy changes to chemical energy
(c) nuclear energy changes to electrical energy
(d) electrical energy changes to nuclear energy
Answer: (c) nuclear energy changes to electrical energy
13. In an electrical cell while in use, the change in energy is from:
(a) electrical to mechanical
(b) electrical to chemical
(c) chemical to mechanical
(d) chemical to electrical
Answer: (d) chemical to electrical
Very Short Questions
1. What are the two forms of mechanical energy?
Answer: The two forms of mechanical energy are (1) potential energy, and (2) kinetic energy.
2. Name the form of energy which a wound up watch spring possesses.
Answer: A wound up watch spring possesses elastic potential energy.
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) A stretched wire
(e) An arrow shot out of a bow
(f) A piece of stone placed on the roof.
Answer: The type of energy possessed is:
(a) A moving cricket ball: Kinetic energy (K)
(b) A compressed spring: Elastic potential energy (U)
(c) A moving bus: Kinetic energy (K)
(d) A stretched wire: Elastic potential energy (U)
(e) An arrow shot out of a bow: Kinetic energy (K)
(f) A piece of stone placed on the roof: Gravitational potential energy (U)
4. (a) A body of mass m is moving with a velocity v. Write the expression for its kinetic energy.
(b) Show that the quantity 2K/v² has the unit of mass, where K is the kinetic energy of the body.
Answer: (a) The expression for its kinetic energy K is K = ½ mv².
(b) Given K is the kinetic energy of the body, K = ½ mv².
Multiplying both sides by 2, we get 2K = mv².
Dividing both sides by v², we get 2K/v² = m.
Since m represents mass, the quantity 2K/v² has the unit of mass.
5. Two bodies A and B of masses m and M (M >> m) have same kinetic energy. Which body will have more momentum?
Answer: If a light body A of mass m and a heavy body B of mass M have the same kinetic energy K, then the heavy body B will have more momentum than the light body A. Therefore, body B will have more momentum.
6. 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 ………. .
(c) The conversion of part of energy into an undesirable form is called ………. .
Answer: (a) The kinetic energy of a body is the energy by virtue of its motion.
(b) The potential energy of a body is the energy by virtue of its position or size and shape.
(c) The conversion of part of energy into an undesirable form is called degradation of energy (or dissipation of energy).
7. 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: The energy changes that occur are:
(a) In the unwinding of a watch spring, elastic potential energy changes into kinetic energy which does work in moving the hands of the watch.
(b) In a loaded truck when started and set in motion, the chemical energy of petrol (or diesel) changes into mechanical energy (or kinetic energy).
(c) In a car going uphill, chemical energy of fuel changes into kinetic energy and gravitational potential energy.
(d) In photosynthesis in green leaves, the light energy from the Sun is absorbed by the green plants and they change it in the form of chemical energy (food).
(e) In charging of a battery, electrical energy changes into the chemical energy of the cell.
(f) In respiration, the chemical energy is converted into heat energy.
(g) In burning of a match stick, the chemical energy changes into heat and light energies.
(h) In explosion of crackers, the chemical energy changes into heat, light and sound energies.
8. State the energy changes in the following cases while in use:
(a) loudspeaker
(b) a steam engine
(c) microphone
(d) washing machine
(e) a glowing 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 iron
(l) a ceiling fan
(m) an electromagnet.
Answer: The energy changes are:
(a) In a loudspeaker, electrical energy changes into sound energy.
(b) In a steam engine, the heat energy of steam changes into the kinetic energy of piston (mechanical energy).
(c) In a microphone, sound energy converts into electrical energy in form of varying electric signals.
(d) In a washing machine, electrical energy changes into mechanical energy.
(e) In a glowing electric bulb, electrical energy changes into heat and light energies.
(f) In burning coal, chemical energy changes into heat energy.
(g) In a solar cell, light (or solar) energy changes into electrical energy.
(h) In a bio-gas burner, chemical energy of bio-gas changes into heat and light energy.
(i) In an electric cell in a circuit, the chemical energy stored in it changes into electrical energy.
(j) In a petrol engine of a running car, the chemical energy of petrol changes into mechanical energy (or kinetic energy).
(k) In an electric iron, electrical energy changes into heat energy.
(l) In a ceiling fan, electrical energy changes into mechanical energy.
(m) In an electromagnet, electrical energy changes into magnetic energy.
9. Name the process used for producing electricity from nuclear energy.
Answer: The process used for producing electricity from nuclear energy is nuclear fission, where the energy released in the process of nuclear fission is nuclear energy which is converted into electrical energy in a nuclear reactor.
10. Is it practically possible to convert a form of energy completely into another useful form ? Explain your answer.
Answer: No, it is not practically possible to convert a form of energy completely into another useful form. While transforming energy from one form to another desired form, the entire energy does not change into the desired form. A part of this energy changes either to some other undesirable form (usually heat due to friction) or a part is lost to the surroundings due to radiation and is thus of no use to us. This conversion of energy to the undesirable (or non-useful) form is called dissipation of energy. Since this part of energy is not available to us for any productive purpose, so we call this part of energy as degraded form of energy.
Short Questions
1. Define the term potential energy of a body. Name its two forms and give one example of each.
Answer: The energy possessed by a body at rest due to its position or size and shape is called potential energy. Its two forms are:
(i) Gravitational potential energy: This is due to the changed position of the object. An example is a body placed at a height above the ground.
(ii) Elastic potential energy: This is due to the changed size and shape of the object. An example is a wound up watch spring.
2. 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: A body may possess potential energy even when it is not in motion. For example, a body placed at a height above the ground possesses gravitational potential energy.
3. Write an expression for the potential energy of a body of mass m placed at a height h above the earth’s surface. State the assumptions made, if any.
Answer: The expression for the gravitational potential energy U of a body of mass m placed at a height h above the earth’s surface is U = mgh.
The assumption made is that when the body is on the earth’s surface (or ground), its gravitational potential energy (Uh) is zero.
4. What do you understand by the kinetic energy of a body?
Answer: The kinetic energy of a body is the energy possessed by a body due to its state of motion.
5. State the work-energy theorem.
Answer: According to the work energy theorem, the increase in kinetic energy of a moving body is equal to the work done by a force acting in the direction of the moving body. Alternatively, when a force is applied in the direction of motion of a body, it accelerates the motion and thus increases the kinetic energy of the body. This increase in kinetic energy is equal to the work done by the force on the body.
6. A body of mass m is moving with a uniform velocity u. A force is applied on the body due to which its velocity increases from u to v. How much work is being done by the force ?
Answer: The work done by the force is equal to the increase in its kinetic energy, which is W = ½m (v² – u²).
7. A light mass and a heavy mass have equal momentum. Which will have more kinetic energy ? [Hint : kinetic energy K = p²/2m where p is the momentum]
Answer: If a light body A of mass m and a heavy body B of mass M, both have same momentum p, then the kinetic energy of the light body A will be more than that of the heavy body B. Thus, the lighter mass will have more kinetic energy.
8. Name the three forms of kinetic energy and give one example of each.
Answer: The three forms of kinetic energy are:
(i) Translational kinetic energy: The kinetic energy of the body due to motion in a straight line path. Example: A car moving on a straight path.
(ii) Rotational kinetic energy: The kinetic energy of the body due to rotational motion. Example: A spinning top.
(iii) Vibrational kinetic energy: The kinetic energy of the body due to its vibrational motion. Example: A wire clamped at both the ends, when struck in the middle, vibrates.
9. State two differences between potential energy and kinetic energy.
Answer: Two differences between potential energy and kinetic energy are:
| Potential energy | Kinetic energy |
| 1. It is the energy possessed by a body due to its changed position or changed size and shape. | 1. It is the energy possessed by a body due to its state of motion. |
| 2. It is equal to the work done in bringing the body to its changed state. | 2. It is equal to the work that a moving body can do before coming to rest. |
10. Name six different forms of energy.
Answer: Six different forms of energy are: solar energy, heat energy, light energy, chemical energy, hydro energy, and electrical energy.
11. What is degraded energy ?
Answer: Degraded energy is a part of energy that changes into an undesirable or non-useful form during energy transformation. While transforming energy from one form to another desired form, the entire energy does not change into the desired form. A part of this energy changes either to some other undesirable form (usually heat due to friction) or a part is lost to the surroundings due to radiation and is thus of no use to us. This conversion of energy to the undesirable (or non-useful) form is called dissipation of energy. Since this part of energy is not available to us for any productive purpose, so we call this part of energy as degraded form of energy.
Long Questions
1. What is meant by gravitational potential energy ? Derive an expression for it for a body placed at a height above the ground.
Answer: Gravitational potential energy is the potential energy possessed by a body due to the force of attraction of Earth on it.
To derive an expression for it, let a body of mass m be lifted from the ground (or the earth’s surface) to a vertical height h. The least upward force required to lift the body (without acceleration) must be equal to the force of gravity (i.e., F = mg) on the body acting vertically downwards. The work done W on the body in lifting it to a height h is given by:
W = Force of gravity (mg) × displacement (h)
W = mgh
This work done in lifting the body up gets stored in the body in the form of its gravitational potential energy when it is at height h. Thus, the gravitational potential energy U at height h is given by:
U = mgh.
2. 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. Thus, the arrow gets its kinetic energy from the elastic potential energy stored in the stretched bow.
3. 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: A compressed spring possesses elastic potential energy due to its compressed state.
On releasing the spring, the ball flies away because the potential energy of the spring changes into kinetic energy which does work on the ball placed on it and changes into the kinetic energy of the ball due to which it flies away.
4. A pebble is thrown up. It goes to a height and then comes back on the ground. State the different changes in form of energy during its motion.
Answer: When a pebble is thrown up, its initial kinetic energy (imparted at the time of throwing) starts converting into gravitational potential energy as it gains height. Simultaneously, its kinetic energy decreases. At the highest point of its trajectory, its kinetic energy becomes momentarily zero (if thrown vertically upwards) and its gravitational potential energy is maximum.
As the pebble comes back on the ground, its gravitational potential energy starts converting back into kinetic energy. Its potential energy decreases as its height decreases, and its kinetic energy increases. Just before hitting the ground, its kinetic energy is maximum (ideally equal to the initial kinetic energy if air resistance is neglected), and its potential energy is minimum (or zero if the ground is taken as the reference level).
5. 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: The temperature of water at the bottom of a waterfall is higher than the temperature at the top.
The reason is that when water falls from a height, the potential energy stored in water at that height changes into the kinetic energy of water during the fall. On striking the ground (or bottom), a part of the kinetic energy of water changes into heat energy due to which the temperature of water rises.
Numericals
1. Two bodies of equal masses are placed at heights h and 2h. Find the ratio of their gravitational potential energies.
Answer:
Let the mass of each body be ‘m’.
Let the acceleration due to gravity be ‘g’.
Potential energy of the first body at height h, (PE₁) = mgh
Potential energy of the second body at height 2h, (PE₂) = mg(2h) = 2mgh
Ratio of their gravitational potential energies = PE₁ / PE₂
=> Ratio = (mgh) / (2mgh)
=> Ratio = 1 / 2
Thus, the ratio of their gravitational potential energies is 1 : 2.
2. Find the gravitational potential energy of 1 kg mass kept at a height of 5 m above the ground. Calculate its kinetic energy when it falls and hits the ground. Take g = 10 m s⁻².
Answer:
Given:
Mass (m) = 1 kg
Height (h) = 5 m
Acceleration due to gravity (g) = 10 m s⁻²
Gravitational potential energy (PE) = mgh
=> PE = 1 kg × 10 m s⁻² × 5 m
=> PE = 50 J
According to the principle of conservation of energy, the potential energy at the maximum height is converted into kinetic energy just before hitting the ground.
Kinetic energy on hitting the ground (KE) = Potential energy at height 5 m
=> KE = 50 J
3. A box of mass 300 kg has gravitational potential energy stored in it equal to 29400 J. Find the height of the box above the ground. (Take g = 9.8 N kg⁻¹).
Answer:
Given:
Mass (m) = 300 kg
Potential Energy (PE) = 29400 J
Acceleration due to gravity (g) = 9.8 N kg⁻¹ = 9.8 m s⁻²
We know, PE = mgh
=> 29400 J = 300 kg × 9.8 m s⁻² × h
=> 29400 = 2940 × h
=> h = 29400 / 2940
=> h = 10 m
4. A body of mass 10 kg falls from a height of 20 m to 8 m. Calculate : (i) the loss in potential energy of the body, and (ii) the total energy possessed by the body at any instant ? (Take g = 10 m s⁻²).
Answer:
Given:
Mass (m) = 10 kg
Initial height (h₁) = 20 m
Final height (h₂) = 8 m
Acceleration due to gravity (g) = 10 m s⁻²
(i) The loss in potential energy:
Loss in PE = mg(h₁ – h₂)
=> Loss in PE = 10 kg × 10 m s⁻² × (20 m – 8 m)
=> Loss in PE = 100 × 12
=> Loss in PE = 1200 J
(ii) The total energy possessed by the body:
The total mechanical energy remains constant. It is equal to the initial potential energy at the height of 20 m.
Total Energy = PE at height h₁ = mgh₁
=> Total Energy = 10 kg × 10 m s⁻² × 20 m
=> Total Energy = 2000 J
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:
Given:
Mass (m) = 0.5 kg
Energy spent (Work done, W) = 1.0 J
Acceleration due to gravity (g) = 10 m s⁻²
The energy spent is equal to the gain in potential energy.
W = mgh
=> 1.0 J = 0.5 kg × 10 m s⁻² × h
=> 1.0 = 5 × h
=> h = 1.0 / 5
=> h = 0.2 m
6. A boy weighing 50 kgf climbs up from the first floor at a height of 3 m above the ground to the third floor at a height of 9 m above the ground. What will be the increase in his gravitational potential energy ? (Take g = 10 N kg⁻¹).
Answer:
Given:
Mass of the boy (m) = 50 kg (since 1 kgf corresponds to a mass of 1 kg)
Initial height (h₁) = 3 m
Final height (h₂) = 9 m
Acceleration due to gravity (g) = 10 N kg⁻¹ = 10 m s⁻²
Increase in potential energy (ΔPE) = mg(h₂ – h₁)
=> ΔPE = 50 kg × 10 m s⁻² × (9 m – 3 m)
=> ΔPE = 500 × 6
=> ΔPE = 3000 J
7. A vessel containing 50 kg of water is placed at a height of 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:
Given:
Mass of water (m) = 50 kg
Height (h) = 15 m
Acceleration due to gravity (g) = 10 m s⁻²
Gravitational potential energy (PE) = mgh
=> PE = 50 kg × 10 m s⁻² × 15 m
=> PE = 500 × 15
=> PE = 7500 J
8. A man of mass 50 kg climbs up a ladder of height 15 m. Calculate: (i) the work done by the man, and (ii) the increase in his potential energy. (g = 9.8 m s⁻²).
Answer:
Given:
Mass of the man (m) = 50 kg
Height (h) = 15 m
Acceleration due to gravity (g) = 9.8 m s⁻²
(i) The work done by the man against gravity:
Work done (W) = mgh
=> W = 50 kg × 9.8 m s⁻² × 15 m
=> W = 490 × 15
=> W = 7350 J
(ii) The increase in his potential energy:
Increase in potential energy (ΔPE) = mgh
=> ΔPE = 50 kg × 9.8 m s⁻² × 15 m
=> ΔPE = 7350 J
9. A block A, weighing 100 N, is pulled up a slope of length 5 m by means of a constant force F (= 100 N) as illustrated in Fig. 2.15.
(a) What is the work done by the force F in moving the block A, 5 m along the slope ?
(b) What is the increase in potential energy of the block A ?
(c) Account for the difference in the work done by the force and the increase in potential energy of the block.
Answer:
Given:
Weight of block A (mg) = 100 N
Length of the slope (distance, S) = 5 m
Applied force (F) = 100 N
From the figure, the vertical height (h) = 3 m
(a) Work done by the force F:
Work done (W) = Force × distance
=> W = 100 N × 5 m
=> W = 500 J
(b) Increase in potential energy of the block A:
Increase in PE (ΔPE) = Weight × vertical height
=> ΔPE = 100 N × 3 m
=> ΔPE = 300 J
(c) Account for the difference:
The work done by the force F (500 J) is greater than the increase in potential energy (300 J).
Difference = W – ΔPE = 500 J – 300 J = 200 J.
This difference of 200 J is the work done against the force of friction between the block and the slope, which is converted into heat energy.
10. Find the kinetic energy of a body of mass 5 kg moving with a uniform velocity of 10 m s⁻¹.
Answer:
Given:
Mass (m) = 5 kg
Velocity (v) = 10 m s⁻¹
Kinetic energy (KE) = ½ mv²
=> KE = ½ × 5 kg × (10 m s⁻¹)²
=> KE = ½ × 5 × 100
=> KE = 250 J
11. If the speed of a car is halved, how does its kinetic energy change?
Answer:
Let the initial mass be ‘m’ and initial speed be ‘v’.
Initial kinetic energy (KE_initial) = ½ mv²
New speed (v_new) = v / 2
New kinetic energy (KE_new) = ½ m(v_new)²
=> KE_new = ½ m(v / 2)²
=> KE_new = ½ m(v² / 4)
=> KE_new = (1/4) × (½ mv²)
=> KE_new = (1/4) × KE_initial
The kinetic energy becomes one-fourth of its initial value.
12. Two bodies of equal masses are moving with uniform velocities v and 2v. Find the ratio of their kinetic energies.
Answer:
Let the mass of each body be ‘m’.
Velocity of the first body (v₁) = v
Velocity of the second body (v₂) = 2v
Kinetic energy of the first body (KE₁) = ½ mv₁² = ½ mv²
Kinetic energy of the second body (KE₂) = ½ mv₂² = ½ m(2v)² = ½ m(4v²) = 4 × (½ mv²)
Ratio of their kinetic energies = KE₁ / KE₂
=> Ratio = (½ mv²) / (4 × ½ mv²)
=> Ratio = 1 / 4
Thus, the ratio of their kinetic energies is 1 : 4.
13. Two bodies have masses in the ratio 5 : 1 and kinetic energies in the ratio 125 : 9. Calculate the ratio of their velocities.
Answer:
Given:
Ratio of masses (m₁ / m₂) = 5 / 1
Ratio of kinetic energies (KE₁ / KE₂) = 125 / 9
We know, KE = ½ mv², so v² = 2KE / m
(v₁/v₂)² = (2KE₁ / m₁) / (2KE₂ / m₂)
=> (v₁/v₂)² = (KE₁ / KE₂) × (m₂ / m₁)
=> (v₁/v₂)² = (125 / 9) × (1 / 5)
=> (v₁/v₂)² = 25 / 9
=> v₁/v₂ = √(25 / 9)
=> v₁/v₂ = 5 / 3
The ratio of their velocities is 5 : 3.
14. A car is running at a speed of 15 km h⁻¹ while another similar car is moving at a speed of 45 km h⁻¹. Find the ratio of their kinetic energies.
Answer:
Given:
Speed of the first car (v₁) = 15 km h⁻¹
Speed of the second car (v₂) = 45 km h⁻¹
The cars are similar, so their masses are equal (m₁ = m₂ = m).
Ratio of kinetic energies (KE₁ / KE₂) = (½ m₁v₁²) / (½ m₂v₂²)
=> KE₁ / KE₂ = v₁² / v₂² = (v₁ / v₂)²
=> KE₁ / KE₂ = (15 / 45)²
=> KE₁ / KE₂ = (1 / 3)²
=> KE₁ / KE₂ = 1 / 9
The ratio of their kinetic energies is 1 : 9.
15. A ball of mass 0.5 kg slows down from a speed of 10 m s⁻¹ to that of 6 m s⁻¹. Calculate the change in the kinetic energy of the ball.
Answer:
Given:
Mass (m) = 0.5 kg
Initial speed (v_i) = 10 m s⁻¹
Final speed (v_f) = 6 m s⁻¹
Initial kinetic energy (KE_i) = ½ mv_i²
=> KE_i = ½ × 0.5 kg × (10 m s⁻¹)² = 0.25 × 100 = 25 J
Final kinetic energy (KE_f) = ½ mv_f²
=> KE_f = ½ × 0.5 kg × (6 m s⁻¹)² = 0.25 × 36 = 9 J
Change in kinetic energy (ΔKE) = KE_f – KE_i
=> ΔKE = 9 J – 25 J = -16 J
This represents a decrease of 16 J in kinetic energy.
16. 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:
Given:
Mass (m) = 500 g = 0.5 kg
Speed (v) = 15 m s⁻¹
(i) Kinetic energy (KE):
KE = ½ mv²
=> KE = ½ × 0.5 kg × (15 m s⁻¹)²
=> KE = 0.25 × 225
=> KE = 56.25 J
(ii) Momentum (p):
p = mv
=> p = 0.5 kg × 15 m s⁻¹
=> p = 7.5 kg m s⁻¹
17. A body of mass 10 kg is moving with a velocity of 20 m s⁻¹. If the mass of the body is doubled and its velocity is halved, find: (i) the initial kinetic energy, and (ii) the final kinetic energy.
Answer:
Given:
Initial mass (m_i) = 10 kg
Initial velocity (v_i) = 20 m s⁻¹
Final mass (m_f) = 2 × m_i = 20 kg
Final velocity (v_f) = v_i / 2 = 10 m s⁻¹
(i) Initial kinetic energy:
KE_i = ½ m_i v_i²
=> KE_i = ½ × 10 kg × (20 m s⁻¹)²
=> KE_i = 5 × 400
=> KE_i = 2000 J
(ii) Final kinetic energy:
KE_f = ½ m_f v_f²
=> KE_f = ½ × 20 kg × (10 m s⁻¹)²
=> KE_f = 10 × 100
=> KE_f = 1000 J
18. A truck weighing 1000 kgf changes its speed from 36 km h⁻¹ to 72 km h⁻¹ in 2 minutes. Calculate : (i) the work done by the engine, and (ii) its power. (g = 10 m s⁻²).
Answer:
Given:
Mass of the truck (m) = 1000 kg
Initial speed (v_i) = 36 km h⁻¹ = 36 × (1000/3600) m s⁻¹ = 10 m s⁻¹
Final speed (v_f) = 72 km h⁻¹ = 72 × (1000/3600) m s⁻¹ = 20 m s⁻¹
Time (t) = 2 minutes = 120 s
(i) The work done by the engine (Work-Energy Theorem):
Work done (W) = Change in kinetic energy = KE_f – KE_i
=> W = ½ m(v_f² – v_i²)
=> W = ½ × 1000 kg × ((20 m s⁻¹)² – (10 m s⁻¹)²)
=> W = 500 × (400 – 100)
=> W = 500 × 300
=> W = 150000 J or 1.5 × 10⁵ J
(ii) Its power:
Power (P) = Work done / Time taken
=> P = 150000 J / 120 s
=> P = 1250 W or 1.25 kW
19. A body of mass 60 kg has momentum of 3000 kg m s⁻¹. Calculate : (i) the kinetic energy, and (ii) the speed of the body.
Answer:
Given:
Mass (m) = 60 kg
Momentum (p) = 3000 kg m s⁻¹
(ii) The speed of the body:
p = mv
=> v = p / m
=> v = 3000 kg m s⁻¹ / 60 kg
=> v = 50 m s⁻¹
(i) The kinetic energy:
KE = p² / (2m)
=> KE = (3000)² / (2 × 60)
=> KE = 9000000 / 120
=> KE = 75000 J or 7.5 × 10⁴ J
20. How much work is needed to be done on a ball of mass 50 g to give it a momentum of 5 kg m s⁻¹?
Answer:
Given:
Mass (m) = 50 g = 0.05 kg
Momentum (p) = 5 kg m s⁻¹
Work needed is equal to the kinetic energy gained by the ball.
Work done (W) = Kinetic Energy (KE) = p² / (2m)
=> W = (5 kg m s⁻¹)² / (2 × 0.05 kg)
=> W = 25 / 0.1
=> W = 250 J
21. How much energy is gained by a box of mass 20 kg when a man
(a) carrying the box waits for 5 minutes for a bus?
(b) runs carrying the box with a speed of 3 m s⁻¹ to catch the bus ?
(c) raises the box by 0.5 m in order to place it inside the bus ? (g = 10 m s⁻²)
Answer:
Given:
Mass of the box (m) = 20 kg
g = 10 m s⁻²
(a) When the man waits, the displacement of the box is zero. Work done on the box is zero.
Energy gained = 0 J
(b) When the man runs, the box gains kinetic energy.
Speed (v) = 3 m s⁻¹
Energy gained = KE = ½ mv²
=> Energy gained = ½ × 20 kg × (3 m s⁻¹)²
=> Energy gained = 10 × 9
=> Energy gained = 90 J
(c) When the man raises the box, it gains potential energy.
Height (h) = 0.5 m
Energy gained = PE = mgh
=> Energy gained = 20 kg × 10 m s⁻² × 0.5 m
=> Energy gained = 200 × 0.5
=> Energy gained = 100 J
22. A bullet of mass 50 g is moving with a velocity of 500 m s⁻¹. It penetrates 10 cm into a still target and comes to rest. Calculate: (a) the kinetic energy possessed by the bullet, and (b) the average retarding force offered by the target.
Answer:
Given:
Mass (m) = 50 g = 0.05 kg
Velocity (v) = 500 m s⁻¹
Distance of penetration (d) = 10 cm = 0.1 m
(a) The kinetic energy possessed by the bullet:
KE = ½ mv²
=> KE = ½ × 0.05 kg × (500 m s⁻¹)²
=> KE = 0.025 × 250000
=> KE = 6250 J
(b) The average retarding force:
The work done by the retarding force is equal to the initial kinetic energy of the bullet.
Work done (W) = Force (F) × distance (d)
W = KE
=> F × d = 6250 J
=> F × 0.1 m = 6250 J
=> F = 6250 / 0.1
=> F = 62500 N
23. A spring is kept compressed by a small trolley of mass 0.5 kg lying on a smooth horizontal surface as shown in the adjacent Fig. 2.16. When the trolley is released, it is found to move at a speed v = 2 m s⁻¹. What potential energy did the spring possess when compressed ?
Answer:
Given:
Mass of the trolley (m) = 0.5 kg
Speed of the trolley (v) = 2 m s⁻¹
By the principle of conservation of energy, the potential energy stored in the compressed spring is converted into the kinetic energy of the trolley upon release.
Potential Energy of spring (PE_spring) = Kinetic Energy of trolley (KE_trolley)
=> PE_spring = ½ mv²
=> PE_spring = ½ × 0.5 kg × (2 m s⁻¹)²
=> PE_spring = 0.25 × 4
=> PE_spring = 1.0 J
Exercise C
MCQs
1. According to the principle of conservation of mechanical energy:
(a) Potential energy of a system is always greater than its kinetic energy
(b) Kinetic enregy of a system is always greater than its potential energy
(c) Mechanical energy is always converted into heat energy
(d) Total mechanical energy of a system remains constant if no external force is worked on it.
Answer: (d) Total mechanical energy of a system remains constant if no external force is worked on it.
2. A skater is gliding on a smooth ice surface. What happens to his mechanical energy as he skates ?
(a) It increases
(b) It decreases
(c) It becomes zero
(d). It remains constant
Answer: (d). It remains constant
3. A spring is compressed, storing potential energy. When the spring is released, it propels an object forward. What change would you notice in the total mechanical energy of the system ?
(a) It increases
(b) It decreases
(c) It becomes zero
(d) It remains constant
Answer: (d) It remains constant
4. A ball of mass m is thrown vertically up with an initial velocity so as to reach height h. The correct statement is:
(a) Potential energy of the ball at the ground is mgh.
(b) Kinetic energy of the ball at the ground is zero.
(c) Kinetic energy of the ball at the highest point is mgh.
(d) Potential energy of the ball at the highest point is mgh.
Answer: (d) Potential energy of the ball at the highest point is mgh.
5. A body is released from a height. During its motion in striking the ground, the quantity which does not change is:
(a) momentum
(b) kinetic energy
(c) potential energy
(d) mechanical energy
Answer: (d) mechanical energy
6. A pendulum is oscillating on either side of its rest position. The correct statement is :
(a) It has only kinetic enrgy at its each position.
(b) It has the maximum kinetic energy at its extreme position.
(c) It has the maximum potential energy at its mean position.
(d) The sum of its kinetic energy and potential energy remains constant throughout the motion.
Answer: (d) The sum of its kinetic energy and potential energy remains constant throughout the motion.
7. A simple pendulum while oscillating rises to a maximum vertical height of 7 cm from its rest position when it reaches its extreme position on one side. The mass of the bob of simple pendulum is 400 g and g = 10 m/s2. The total energy of the simple pendulum at any instant while oscillating is :
(a) 0.28 J
(b) 2.8 J
(c) 28 J
(d) 28000 J
Answer: (a) 0.28 J
Explanation:
- Mass of the bob (m) = 400 g = 0.4 kg
- Maximum vertical height (h) = 7 cm = 0.07 m
- Acceleration due to gravity (g) = 10 m/s²
Calculation:
Total Energy (E) = Potential Energy at extreme position (PE_max) + Kinetic Energy at extreme position (KE_max)
=> E = mgh + 0
=> E = 0.4 kg × 10 m/s² × 0.07 m
=> E = 4 × 0.07 J
=> E = 0.28 J
8. Assertion (A): When a hammer is made to fall on a nail fixed upright on a wooden piece, the nail begins to penetrate the wood.
Reason (R) : As the hammer starts falling, its kinetic energy begins to change into potential energy.
(a) both A and R are true and R is the correct explaination of A
(b) both A and R are True and R is not the correct explanation of A
(c) assertion is false but reason is true
(d) assertion is true but reason is false
Answer: (d) assertion is true but reason is false
Very Short Questions
1. A body is thrown vertically upwards. Its velocity keeps on decreasing. What happens to its kinetic energy as its velocity becomes zero?
Answer: When a body is thrown vertically upwards and its velocity becomes zero at the highest point, its kinetic energy changes to potential energy.
2. A body falls freely under gravity from rest. Name the kind of energy it will possess
(a) at the point from where it falls,
(b) while falling,
(c) on reaching the ground.
Answer: When a body falls freely under gravity from rest, the kind of energy it will possess is:
(a) at the point from where it falls, it possesses potential energy.
(b) while falling, it possesses potential energy and kinetic energy.
(c) on reaching the ground, it possesses kinetic energy.
Short Questions
1. State the principle of conservation of energy.
Answer: According to the principle of conservation of energy, energy can neither be created nor can it be destroyed. It only changes from one form to another. In the universe, energy occurs in various forms. The sum of all forms of energy in the universe remains constant. When there is a transformation of energy from one form to another, the total energy always remains the same i.e., it remains conserved.
2. What do you understand by the conservation of mechanical energy? State the condition under which mechanical energy is conserved.
Answer: If there is only an interchange between potential energy and kinetic energy, the total mechanical energy (i.e., the sum of kinetic energy K and potential energy U) remains constant.
The condition under which mechanical energy is conserved is in the absence of all kinds of frictional forces.
3. Name two examples in which the mechanical energy of a system remains constant.
Answer: Two examples in which the mechanical energy of a system remains constant are:
(i) A freely falling body under gravity (neglecting air resistance).
(ii) An oscillating simple pendulum (neglecting air friction).
4. A pendulum with bob of mass m is oscillating on either side from its resting position A between the extremes B and C at a vertical height h above A. What is the kinetic energy K and potential energy U when the pendulum is at positions (i) A, (ii) B, and (iii) C?
Answer: When a pendulum with bob of mass m is oscillating on either side from its resting position A between the extremes B and C at a vertical height h above A:
(i) At position A (mean position): Kinetic energy K = mgh, Potential energy U = 0 (assuming potential energy at mean position is zero).
(ii) At position B (extreme position): Kinetic energy K = 0, Potential energy U = mgh.
(iii) At position C (extreme position): Kinetic energy K = 0, Potential energy U = mgh.
Long Questions
1. Show that the sum of kinetic energy and potential energy (i.e., total mechanical energy) is always conserved in the case of a freely falling body under gravity (with air resistance neglected) from a height h by finding it when (i) the body is at the top, (ii) the body has fallen a distance x, (iii) the body has reached the ground.
Answer: Let a body of mass m be falling freely under gravity from a height h above the ground (i.e., from position A). As the body falls down, its potential energy changes into kinetic energy. At each point of motion, the sum of potential energy and kinetic energy remains unchanged.
(i) When the body is at the top (at position A at height h above the ground):
Initial velocity of body = 0 (since body is at rest at A)
Kinetic energy K = 0
Potential energy U = mgh
Hence, total energy = K + U = 0 + mgh = mgh …..(i)
(ii) When the body has fallen a distance x (at position B):
Let v₁ be the velocity acquired by the body at B after falling through a distance x. Then u = 0, S = x, a = g.
From equation v² = u² + 2aS, we have v₁² = 0 + 2gx = 2gx.
Kinetic energy K = ½ mv₁² = ½ m × (2gx) = mgx.
Now at B, height of body above the ground = h – x.
Potential energy U = mg (h – x).
Hence, total energy = K + U = mgx + mg(h – x) = mgh …..(ii)
(iii) When the body has reached the ground (at position C):
Let the velocity acquired by the body on reaching the ground be v. Then u = 0, S = h, a = g.
From equation v² = u² + 2aS, we have v² = 0 + 2gh = 2gh.
Kinetic energy K = ½ mv² = ½ m × (2gh) = mgh.
Potential energy U = 0 (at the ground when h = 0).
Hence total energy = K + U = mgh + 0 = mgh …..(iii)
Thus, from eqns. (i), (ii) and (iii), we note that the total mechanical energy (i.e., the sum of kinetic energy and potential energy) always remains constant at each point of motion and is equal to the initial potential energy at height h.
2. A pendulum is oscillating on either side of its rest position. Explain the energy changes that take place in the oscillating pendulum. How does mechanical energy remain constant in it? Draw the necessary diagram.
Answer: Let A be the resting (or mean) position of the bob of a simple pendulum when it has zero potential energy (by assumption). When the bob of the pendulum is displaced to B from its resting position A, the bob gets raised by a vertical height h, so its potential energy increases by mgh if m is the mass of the bob. At this extreme position B, its kinetic energy is zero as it momentarily comes to rest.
Now on releasing the bob at B, it moves back from B to A. Its vertical height decreases from h to zero, so its potential energy decreases from mgh to zero and it gets converted into kinetic energy. Therefore, at the point A, the bob acquires a velocity v = √2gh, so it moves towards C. At the mean position A, it has its total mechanical energy in the form of kinetic energy and potential energy is zero.
As the bob moves from A to C, its kinetic energy decreases and potential energy increases. When the bob gets raised by a vertical height h above the point A at the other extreme position C, again it acquires potential energy mgh and its kinetic energy becomes zero. So the bob momentarily comes to rest at the point C. But due to the force of gravity, the bob moves back from C to A.
As the bob swings back from C to A, its potential energy decreases and kinetic energy increases. At A (mean position), it has its total mechanical energy in the form of kinetic energy and potential energy is zero, so the bob swings again from A to B to repeat the process.
Thus, during the swing, at the extreme positions B and C, the bob has only potential energy, while at the mean position A, it has only kinetic energy. At an intermediate position (between A and B or between A and C), the bob has both kinetic energy and potential energy, and the sum of both (i.e., the total mechanical energy) remains constant throughout the swing. This is strictly true only in vacuum where there is no force of friction due to air.
Numericals
1. A ball of mass 0.20 kg is thrown vertically upwards with an initial velocity of 20 m s⁻¹. Calculate the maximum potential energy it gains as it goes up.
Answer: According to the principle of conservation of energy, the initial kinetic energy imparted to the ball will be converted into its potential energy at the maximum height.
Given:
Mass of the ball (m) = 0.20 kg
Initial velocity (u) = 20 m s⁻¹
First, we calculate the initial kinetic energy (K.E.) of the ball.
K.E. = (1/2)mu²
=> K.E. = (1/2) × 0.20 kg × (20 m s⁻¹)²
=> K.E. = 0.10 × 400 J
=> K.E. = 40 J
Therefore, the maximum potential energy gained by the ball is equal to its initial kinetic energy.
Maximum Potential Energy (P.E.) = 40 J.
2. A stone of mass 500 g is thrown vertically upwards with a velocity of 15 m s⁻¹. Calculate: (a) the potential energy at the greatest height, (b) the kinetic energy on reaching the ground, and (c) the total energy at its half way point.
Answer:
Given:
Mass of the stone (m) = 500 g = 0.5 kg
Initial velocity (u) = 15 m s⁻¹
First, we calculate the initial kinetic energy, which is the total mechanical energy of the system (assuming no air resistance).
Initial K.E. = (1/2)mu²
=> Initial K.E. = (1/2) × 0.5 kg × (15 m s⁻¹)²
=> Initial K.E. = 0.25 × 225 J
=> Initial K.E. = 56.25 J
(a) At the greatest height, the entire initial kinetic energy is converted into potential energy.
Potential energy at the greatest height = Initial K.E.
=> P.E. = 56.25 J.
(b) When the stone reaches the ground, all its potential energy is converted back into kinetic energy.
Kinetic energy on reaching the ground = Potential energy at the greatest height
=> K.E. = 56.25 J.
(c) The total mechanical energy of the stone remains constant throughout its motion.
Total energy at its half way point = Initial K.E.
=> Total Energy = 56.25 J.
3. A metal ball of mass 2 kg is allowed to fall freely from rest from a height of 5 m above the ground.
(a) Taking g = 10 m s⁻², calculate :
(i) the potential energy possessed by the ball when it is initially at rest.
(ii) the kinetic energy of the ball just before it hits the ground ?
(b) What happens to the mechanical energy after the ball hits the ground and comes to rest ?
Answer:
Given:
Mass of the ball (m) = 2 kg
Height (h) = 5 m
Acceleration due to gravity (g) = 10 m s⁻²
(a) (i) The potential energy possessed by the ball at rest at height h is:
P.E. = mgh
=> P.E. = 2 kg × 10 m s⁻² × 5 m
=> P.E. = 100 J.
(a) (ii) As the ball falls, its potential energy is converted into kinetic energy. Just before hitting the ground, all the initial potential energy is converted into kinetic energy.
Kinetic energy just before hitting the ground = Initial Potential Energy
=> K.E. = 100 J.
(b) After the ball hits the ground and comes to rest, its mechanical energy (kinetic and potential) becomes zero. This mechanical energy is converted into other forms of energy, primarily heat energy (due to the deformation of the ball and ground) and sound energy.
4. A person swings on a rope from a cliff that is 20 m high. How fast is the person moving at the lowest point of the swing? (take g = 10 m s⁻²)
Answer:
At the highest point (the cliff), the person has only potential energy. At the lowest point of the swing, this potential energy is converted into kinetic energy.
Given:
Height (h) = 20 m
Acceleration due to gravity (g) = 10 m s⁻²
According to the principle of conservation of energy:
Potential Energy at the top = Kinetic Energy at the bottom
mgh = (1/2)mv²
We can cancel ‘m’ from both sides:
gh = (1/2)v²
=> v² = 2gh
=> v = √(2gh)
=> v = √(2 × 10 m s⁻² × 20 m)
=> v = √(400) m/s
=> v = 20 m s⁻¹.
5. The diagram given alongside shows a ski jump. A skier weighing 60 kgf stands at A at the top of ski jump. He moves from A and takes off for his jump at B.
(a) Calculate the change in the gravitational potential energy of the skier between A and B.
(b) If 75% of the energy in part (a) becomes kinetic energy at B, calculate the speed at which the skier arrives at B. (Take g = 10 m s⁻²)
Answer:
Given:
Weight of the skier = 60 kgf, so mass (m) = 60 kg
Initial height at A (h₁) = 75 m
Final height at B (h₂) = 15 m
Acceleration due to gravity (g) = 10 m s⁻²
(a) The change in height (Δh) = h₁ – h₂ = 75 m – 15 m = 60 m.
The change in gravitational potential energy (ΔP.E.) is the loss in P.E. as the skier moves from A to B.
ΔP.E. = mg(Δh)
=> ΔP.E. = 60 kg × 10 m s⁻² × 60 m
=> ΔP.E. = 36000 J or 3.6 × 10⁴ J.
(b) The kinetic energy at B is 75% of the change in potential energy.
K.E. at B = 75% of ΔP.E.
=> K.E. = (75/100) × 36000 J
=> K.E. = 0.75 × 36000 J
=> K.E. = 27000 J
Now, we use the formula for kinetic energy to find the speed (v).
K.E. = (1/2)mv²
=> 27000 J = (1/2) × 60 kg × v²
=> 27000 = 30 × v²
=> v² = 27000 / 30
=> v² = 900
=> v = √(900) m/s
=> v = 30 m s⁻¹.
6. A hydroelectric power station takes its water from a lake whose water level is at a height of 50 m above the turbine. Assuming an overall efficiency of 40%, calculate the mass of water which must flow through the turbine each second to produce power output of 1 MW. (Take g = 10 m s⁻²).
Answer:
Given:
Height (h) = 50 m
Efficiency (η) = 40% = 0.40
Power output (P_out) = 1 MW = 1 × 10⁶ W
Acceleration due to gravity (g) = 10 m s⁻²
The efficiency of the power station relates the power output to the power input.
Efficiency (η) = Power output / Power input
=> Power input (P_in) = Power output / Efficiency
=> P_in = (1 × 10⁶ W) / 0.40
=> P_in = 2.5 × 10⁶ W
The power input is the rate at which the potential energy of the water is supplied to the turbine.
P_in = (Potential Energy) / time = (mgh) / t
P_in = (m/t) × g × h
Here, (m/t) is the mass of water flowing per second.
Let’s find (m/t):
2.5 × 10⁶ W = (m/t) × 10 m s⁻² × 50 m
=> 2.5 × 10⁶ = (m/t) × 500
=> (m/t) = (2.5 × 10⁶) / 500
=> (m/t) = 5000 kg s⁻¹.
7. The bob of a simple pendulum is imparted a velocity of 5 m s⁻¹ when it is at its mean position. To what maximum vertical height will it rise on reaching at its extreme position if 60% of its energy is lost in overcoming the friction of air ? (Take g = 10 m s⁻²).
Answer:
Given:
Initial velocity at mean position (v) = 5 m s⁻¹
Energy lost to friction = 60%
Acceleration due to gravity (g) = 10 m s⁻²
The initial energy of the bob is its kinetic energy at the mean position.
Initial K.E. = (1/2)mv²
Since 60% of this energy is lost, the remaining energy that gets converted into potential energy is 100% – 60% = 40%.
Potential energy at extreme position (P.E.) = 40% of Initial K.E.
mgh = (40/100) × (1/2)mv²
We can cancel ‘m’ from both sides:
gh = 0.40 × (1/2)v²
gh = 0.2 × v²
=> h = (0.2 × v²) / g
=> h = (0.2 × (5 m s⁻¹)²) / (10 m s⁻²)
=> h = (0.2 × 25) / 10
=> h = 5 / 10
=> h = 0.5 m.
8. The figure alongside shows a simple pendulum of mass 200 g. It is displaced from the mean position A to the extreme position B. The potential energy at position A is zero. At position B the bob is raised by 5 m.
(a) What is the potential energy of the pendulum at position B ?
(b) What is the total mechanical energy at point C?
(c) What is the speed of the bob at position A when released from position B ? (Take g = 10 m s⁻²)
Answer:
Given:
Mass of the bob (m) = 200 g = 0.2 kg
Vertical height at B (h) = 5 m
Potential energy at A = 0
Acceleration due to gravity (g) = 10 m s⁻²
(a) The potential energy at position B is calculated as:
P.E. at B = mgh
=> P.E. at B = 0.2 kg × 10 m s⁻² × 5 m
=> P.E. at B = 10 J.
(b) Assuming no energy loss to friction, the total mechanical energy of the pendulum is conserved. The total energy is equal to the energy at the extreme position B. Point C is an intermediate point, so its total energy will be the same.
Total mechanical energy at C = Total mechanical energy at B = P.E. at B
=> Total energy at C = 10 J.
(c) When the bob is released from B and reaches position A, all its potential energy from position B is converted into kinetic energy.
Kinetic energy at A = Potential energy at B
(1/2)mv² = 10 J
=> (1/2) × 0.2 kg × v² = 10 J
=> 0.1 × v² = 10
=> v² = 10 / 0.1
=> v² = 100
=> v = √(100) m/s
=> v = 10 m s⁻¹.
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