CBSE Class 9 Science Work and Energy Exam Notes

CBSE Class 9 Science Work and Energy Exam Notes. Please refer to the examination notes which you can use for preparing and revising for exams. These notes will help you to revise the concepts quickly and get good marks.

Work, power and energy are the basic ideas of mechanics in Class 9 Science. These exam notes cover the definition and unit of work, work done by and against a force, work done at an angle, power and its units, kinetic energy and potential energy with their formulae, the law of conservation of energy, the transformation of energy, and solved examples, followed by Exercise 1 (very short, short and long answer questions) and Exercise 2 (37 objective questions) with answers.

WORK

Definition: In our daily life "work" implies an activity resulting in muscular or mental exertion. However, in physics the term 'work' is used in a specific sense and involves the displacement of a particle or body under the action of a force. Work is said to be done when the point of application of a force moves.
Work done in moving a body is equal to the product of force exerted on the body and the distance moved by the body in the direction of force.
Work = Force × Distance moved in the direction of force.

The work done by a force on a body depends on two factors:
(i) Magnitude of the force, and
(ii) Distance through which the body moves (in the direction of force)

Unit of Work:
When a force of 1 newton moves a body through a distance of 1 metre in its own direction, then the work done is known as 1 joule.

[Figure: A block shown at its initial position A, with a force F acting on it towards the right, and at its final position B after a displacement S]

Work = Force × Displacement
1 joule = 1 N × 1 m
or 1 J = 1 Nm (In SI unit)

Ex.1 How much work is done by a force of 10N in moving an object through a distance of 1 m in the direction of the force ?

Answer:
1. The work done is calculated by using the formula W = F × S.
2. Here, Force F = 10 N and Distance S = 1 m.
3. Work done, W = 10 × 1 J = 10 J
4. Thus, the work done is 10 joules.

Teacher's Note:
a) Work = force × displacement along the force.
b) Always write the unit: N × m = joule.

Ex.2 Find the work done by a force of 10 N in moving an object through a distance of 2 m.

Answer:
1. Work done = Force × Distance moved
2. Here, Force = 10 N and Distance moved = 2 m.
3. Work done, W = 10 N × 2 m = 20 Joule = 20 J

Teacher's Note:
a) Doubling the distance doubles the work for the same force.
b) Check the units: N × m = J.

WORK DONE ANALYSIS

Work done when force and displacement are along same line.

Work done by a force: Work is said to be done by a force if the direction of displacement is the same as the direction of the applied force.
Work done against the force: Work is said to be done against a force if the direction of the displacement is opposite to that of the force.
Work done against Gravity: To lift an object, an applied force has to be equal and opposite to the force of gravity acting on the object. If 'm' is the mass of the object and 'h' is the height through which it is raised, then the upward force (F) = force of gravity = mg.
If 'W' stands for work done, then W = F × h = mg × h. Thus W = mgh.
Therefore we can say that, "The amount of work done is equal to the product of weight of the body and the vertical distance through which the body is lifted."

Ex.3 Calculate the work done in pushing a cart, through a distance of 100 m against the force of friction equal to 120 N.

Answer:
1. Force, F = 120 N and Distance, s = 100 m.
2. Using the formula, W = Fs = 120 N × 100 m = 12,000 J

Teacher's Note:
a) Work is done against friction, so the pushing force equals the friction force.
b) Write the answer with the unit joule.

Ex.4 A body of mass 5 kg is displaced through a distance of 4m under an acceleration of 3 m/s². Calculate the work done.

Answer:
1. Given: mass m = 5 kg and acceleration a = 3 m/s².
2. Force acting on the body is given by F = ma = 5 × 3 = 15 N
3. Now, work done is given by W = Fs = 15 N × 4 m = 60 J

Teacher's Note:
a) Find the force from Newton's second law first.
b) Then use work = force × distance.

Ex.5 Calculate the work done in raising a bucket full of water and weighing 200 kg through a height of 5 m. (Take g = 9.8 ms^-2).

Answer:
1. Force of gravity mg = 200 × 9.8 = 1960 N and h = 5 m.
2. Work done, W = mgh
3. W = 1960 × 5 = 9800 J

Teacher's Note:
a) Here the 200 kg is a mass, so multiply by g to get the weight in newton.
b) Work against gravity depends only on the height raised.

Work done when force and displacement are inclined (Oblique case):
Consider a force 'F' acting at angle θ to the direction of displacement 's' as shown in the figure. The component of the force along the displacement is F cos θ and the perpendicular component is F sin θ.
Work done, W = F × s × cos θ

[Figure: A block moving from A to B through displacement S, with a force F acting at angle θ to the displacement; the components F cos θ along the displacement and F sin θ perpendicular to it are marked. The relation w = f × s × cos θ is written beside it]

Work done when force is perpendicular to Displacement:
θ = 90°
W = F × S × cos 90° = F × S × 0 = 0
Thus no work is done when a force acts at right angle to the displacement.

Special Examples:
1. When a bob attached to a string is whirled along a circular horizontal path, the force acting on the bob acts towards the centre of the circle and is called as the centripetal force. Since the bob is always displaced perpendicular to this force, thus no work is done in this case.
2. Earth revolves around the sun. A satellite moves around the earth. In all these cases, the direction of displacement is always perpendicular to the direction of force (centripetal force) and hence no work is done.
3. A person walking on a road with a load on his head actually does no work because the weight of the load (force of gravity) acts vertically downwards, while the motion is horizontal that is perpendicular to the direction of force resulting in no work done. Here, one can ask that if no work is done, then why the person gets tired. It is because the person has to do work in moving his muscles or to work against friction and air resistance.

Ex.6 A boy pulls a toy cart with a force of 100 N by a string which makes an angle of 60° with the horizontal so as to move the toy cart by a distance of 3 m horizontally. Calculate the work done.

Answer:
1. Given F = 100 N, s = 3 m, θ = 60°.
2. Work done is given by W = Fs cos θ = 100 × 3 × cos 60°
3. Since cos 60° = \( \frac{1}{2} \), W = 100 × 3 × \( \frac{1}{2} \) = 150 J

Teacher's Note:
a) The printed question leaves out the distance and the printed working has a stray factor 2; the given data s = 3 m and the final answer 150 J are used here.
b) Only the horizontal component F cos θ does work.

Ex.7 An engine does 64,000 J of work by exerting a force of 8,000 N. Calculate the displacement in the direction of force.

Answer:
1. Given W = 64,000 J and F = 8,000 N.
2. Work done is given by W = Fs, so 64000 = 8000 × s
3. s = 8 m

Teacher's Note:
a) Rearrange W = Fs to s = W ÷ F.
b) Check: 8000 × 8 = 64000.

POWER

Definition: Power is defined as the rate of doing work.
Power = \( \frac{Work done}{Time taken} \) ⇒ P = \( \frac{W}{t} \)
In other words, power is the work done per unit time. Power is a scalar quantity.
Since W = F × S, therefore P = \( \frac{W}{t} \) = \( \frac{F × S}{t} \) = F × V = force × velocity

Unit of power: The S.I. unit of power is watt and it is the rate of doing work at 1 joule per second.
1 watt = 1 joule per 1 second
1 kilowatt = 1 kW = 1000 W
1 Horse power = 1 H.P. = 746 W

Ex.8 A machine raises a load of 750 N through a height of 15 m in 5s. Calculate :
(i) the work done by the machine.
(ii) the power at which the machine works.

Answer:
1. (i) Work done is given by W = F × s. Here F = 750 N and s = 15 m.
2. W = 750 × 15 = 11250 J = 11.250 kJ
3. (ii) Power of the machine is given by P = \( \frac{W}{t} \). Here W = 11250 J and t = 5 s.
4. P = \( \frac{11250 J}{5 s} \) = 2250 W = 2.250 kW

Teacher's Note:
a) For a load given in newton, the force is already the weight; do not multiply by g.
b) Power = work ÷ time.

Ex.9 A weight lifter lifted a load of 100 kg to a height of 3 m in 10 s. Calculate the following:
(i) amount of work done
(ii) power developed by him

Answer:
1. (i) Work done is given by W = F × s. Here F = mg = 100 × 10 = 1000 N (taking g = 10 m/s²).
2. W = 1000 N × 3 m = 3000 joule
3. (ii) P = \( \frac{W}{t} \), where W = 3000 J and t = 10 s.
4. P = \( \frac{3000 J}{10 s} \) = 300 W

Teacher's Note:
a) The value g = 10 m/s² is used in this example.
b) Divide the work by the time taken to get power.

Ex.10 A water pump raises 60 liters of water through a height of 20 m in 5 s. Calculate the power of the pump. (Given: g = 10 m/s², density of water = 1000 kg/m³)

Answer:
1. Work done, W = F × s ... (1) and F = mg ... (2)
2. Mass = volume × density. Volume = 60 liters = 60 × 10^-3 m³ and density = 1000 kg/m³.
3. Mass, m = (60 × 10^-3 m³) × (1000 kg/m³) = 60 kg
4. F = 60 kg × 10 m/s² = 600 N
5. W = F × s = 600 N × 20 m = 12000 J
6. Power = \( \frac{W}{t} \) = \( \frac{12000 J}{5 s} \) = 2400 W

Teacher's Note:
a) 1 liter = 10^-3 m³, so 60 liters of water has a mass of 60 kg.
b) Weight first, then work, then power.

Ex.11 A woman pulls a bucket of water of total mass 5 kg from a well which is 10 m deep in 10 s. Calculate the power used by her.

Answer:
1. Given that m = 5 kg, h = 10 m, t = 10 s and g = 10 m/s².
2. P = \( \frac{W}{t} \) = \( \frac{mgh}{t} \) = \( \frac{5 × 10 × 10}{10} \) = 50 W

Teacher's Note:
a) Work done in lifting the bucket = mgh = 500 J.
b) Power is 500 J in 10 s, that is 50 W.

ENERGY

Definition: Energy is the ability to do work. The amount of energy possessed by a body is equal to the amount of work it can do when its energy is released. Thus, energy is defined as the capacity of doing work. Energy is a scalar quantity and it exists in various forms.

Units of energy: The units of energy are the same as that of work. In SI system, the unit of energy is joule (J). In CGS system, the unit of energy is erg.
1 Joule = 10⁷ ergs
Other units of energy in common use are watt-hour and kilowatt hour.
1 watt-hour = 1 watt × 1 hour = 1 watt × 60 × 60 sec = 3600 J
1 kilowatt-hour (kWh) = 3.6 × 10⁶ Joule
Heat energy is usually measured in calorie or kilocalorie such that 1 calorie = 4.18 J
A very small unit of energy is electron volt (eV). 1 eV = 1.6 × 10^-19 J

MECHANICAL ENERGY

The energy possessed by a body due to its state of rest or state of motion is called mechanical energy.
Mechanical energy is of two types - (A) Kinetic Energy (B) Potential Energy.

Kinetic Energy: The energy of a body due to its motion is called kinetic energy. In other words, the ability of a body to do work by virtue of its motion is called its kinetic energy.

Expression for Kinetic Energy: The kinetic energy of a body is measured in terms of the amount of work done by an opposing force that brings the body to rest from its present state of motion.

[Figure: A body moving with velocity v from A to B through a distance S]

Suppose a body of mass m is moving with a velocity v and is brought to rest by an opposing force F.
Now retarding force is given by F = ma ...(1)
Now using the equation of motion, v² - u² = 2as, we get 0² = v² - 2as
s = \( \frac{v²}{2a} \) ...(2)
Kinetic energy of the body = work done by the retarding force
or Kinetic energy = force × displacement = F × s ...(3)
Substituting the value of F from equation (1) and the value of s from equation (2) in equation (3), we get
K.E. = ma × \( \frac{v²}{2a} \) = \( \frac{1}{2} \) mv²
Thus, a body of mass m and moving with a velocity v has the capacity of doing work equal to \( \frac{1}{2} \) mv² before it stops.

Ex.12 A bullet of mass 100 gm is fired with a velocity 50 m/s from a gun. Calculate the kinetic energy of the bullet.

Answer:
1. Kinetic energy is given by K.E. = \( \frac{1}{2} \) mv²
2. Here m = 100 gm = 0.1 kg and v = 50 m/s.
3. K.E. = \( \frac{1}{2} \) × 0.1 × (50)² = \( \frac{1}{2} \) × 0.1 × 50 × 50 = 125 J

Teacher's Note:
a) Convert grams to kilograms before using the formula.
b) The printed working shows v = 500 m/s at one step, which is a slip; the given velocity is 50 m/s and 125 J is correct.

Ex.13 A 4 kg body is dropped from the top of a building of height 2.5 m. With what velocity will it strike the ground ? What is its kinetic energy when it strikes the ground ? (Takes g = 9.8 m/s²)

Answer:
1. Velocity with which the body strikes the ground is found from v² = u² + 2gh.
2. Here u = 0, g = 9.8 m/s² and h = 2.5 m, so v² = 0² + 2 × 9.8 × 2.5 = 49
3. v = 7 m/s. Thus, the speed of the body with which it strikes the ground = 7 m/s.
4. Kinetic energy = \( \frac{1}{2} \) mv² = \( \frac{1}{2} \) × 4 × 49 = 98 J

Teacher's Note:
a) The printed working stops at the velocity; the kinetic energy asked in the question is 98 J.
b) Check: loss of potential energy = mgh = 4 × 9.8 × 2.5 = 98 J.
c) A body dropped from rest has u = 0.

Ex.14 Calculate the velocity of 4 kg mass with kinetic energy of 128 J.

Answer:
1. The formula for kinetic energy is K.E. = \( \frac{1}{2} \) mv²
2. Here K.E. = 128 J and m = 4 kg, so 128 = \( \frac{1}{2} \) × 4 × v²
3. v² = 64, so v = 8 m/s

Teacher's Note:
a) Rearrange to v = √(2 × K.E. ÷ m).
b) Take the positive root as speed.

Ex.15 Which would have a greater effect on the kinetic energy of an object, doubling the mass or doubling the velocity ?

Answer:
1. (i) The kinetic energy of a body is directly proportional to its "mass" (m). So, if we double the mass (so that it becomes 2m), then the kinetic energy will also get doubled.
2. (ii) On the other hand, kinetic energy of a body is directly proportional to the "square of its velocity" (v²). So, if we double the velocity (so that it becomes 2v), then the kinetic energy will become four times. This is because (2v)² = 4v².
3. It is clear from the above discussion that doubling the velocity has a greater effect on the kinetic energy of an object.

Teacher's Note:
a) K.E. depends on v², so velocity changes have a bigger effect than mass changes.
b) This is why high speed is dangerous in road accidents.

Potential Energy:
The energy possessed by a body by virtue of its position or change in shape is known as potential energy. It is obvious that a body may possess energy even when it is not in motion.

Expression for Potential Energy:
Suppose a body of mass m be lifted from the ground to a vertical height h, then the minimum force required to lift the body is equal to the force of gravity, i.e. F = mg.
This force of gravity acts on the body vertically downwards.
Now, work done in lifting the body to a height h will be Work = force × distance = mgh.
This work done is stored as potential energy in the body such that Potential energy, U = mgh, i.e. gain in potential energy of the body and the earth.

Ex.16 What will be the potential energy of a body of mass 2 kg kept at a height of 10 m ?

Answer:
1. The potential energy is given by U = mgh.
2. Here m = 2 kg, g = 10 m/s² and h = 10 m.
3. U = 2 × 10 × 10 = 200 J

Teacher's Note:
a) Potential energy depends on mass, g and height.

Ex.17 In lifting a mass of 25 kg to a certain height 1250 J energy is utilized. Calculate to what height it has been lifted ? (Take g = 10 m/s²)

Answer:
1. In lifting a mass through a height h the work done is given by U = mgh.
2. Here U = 1250 J, g = 10 m/s² and m = 25 kg, so 1250 = 25 × 10 × h
3. h = 5 m

Teacher's Note:
a) Rearrange U = mgh to h = U ÷ (mg).
b) Check: 25 × 10 × 5 = 1250.

LAW OF CONSERVATION OF ENERGY

Energy can neither be created nor be destroyed, it can only be changed from one form to another. Appearing amount of energy in one form is always equal to the disappearing amount of energy in some other form. The total energy thus remains constant.

INTER CONVERSION OF POTENTIAL AND KINETIC ENERGY

Mechanical Energy of a Freely Falling Body:
Assume, a body of mass m is at rest at a height h from the earth's surface, as it starts falling, its velocity after travelling a distance x (point B) becomes v and its velocity on the earth's surface is v'.

[Figure: A vertical line with point A at height h above the ground, point B below A at a distance x, and point C near the ground at a height (h - x) below B marked as the remaining distance; the ground is shown at the bottom]

Mechanical energy of the body at point A:
E_A = Kinetic energy + Potential energy
E_A = \( \frac{1}{2} \) m(0)² + mgh
E_A = mgh ......... (i)

Mechanical energy of the body at point B:
E_B = \( \frac{1}{2} \) mv² + mg (h - x) ....... (ii)

Mechanical energy of the body at point C:
E_C = \( \frac{1}{2} \) m (v')² + mg × 0
E_C = \( \frac{1}{2} \) m (v')² ........ (iii)

Use: E_A = E_B = E_C
Hence, when a body falls freely, its mechanical energy will be constant. That means, the total energy of the body during free fall, remains constant at all positions. However, the form of energy keeps on changing at all points during the motion.

[Figure: Graph of energy against position from A to C - the total energy T.E. is a horizontal dashed line, the potential energy P.E. decreases and the kinetic energy K.E. increases from A to C]

TRANSFORMATION OF ENERGY

Definition: The change of one form of energy into another form of energy is known as transformation of energy.

Different Forms of Energy:
Heat energy: Burning of fuels like diesel or petrol in vehicles provides heat energy to do work.
Electrical energy: Electric motors are used in home, industry and even for driving electric trains.
Light energy: When light energy falls on light-meter used in photography, it causes its pointer to move across a scale.
Sound energy: Sound energy causes a thin plate of microphone diaphragm to vibrate.
Chemical energy: Chemical energy is the source of energy in our food and it provides us energy to move the various objects.
Nuclear Energy: The energy in the nucleus of an atom is used to produce heat energy which in turn is used to generate electrical power.

Energy transformation

A swinging simple pendulum is an example of conservation of energy:
This is because a swinging simple pendulum is a body whose energy can either be potential or kinetic, or a mixture of potential and kinetic, but its total energy at any instant of time remains the same.

[Figure: A simple pendulum hanging by a thread from a rigid support, with the bob shown at the extreme position C (only P.E.), the centre position A (only K.E.) and the other extreme position B (only P.E.); positions between them are labelled (PE + KE)]

1. When the pendulum bob is at position B, it has only potential energy (but no kinetic energy).
2. As the bob starts moving down from position B to position A, its potential energy goes on decreasing but its kinetic energy goes on increasing.
3. When the bob reaches the centre position A, it has only kinetic energy (but no potential energy).
4. As the bob goes from position A towards position C, its kinetic energy goes on decreasing but its potential energy goes on increasing.
5. On reaching the extreme position C, the bob stops for a very small instant of time. So at position C, the bob has only potential energy (but no kinetic energy).

MISCELLANEOUS EXAMPLES

Ex.18 A car weighing 1200 kg and travelling at a speed of 20 m/s stops at a distance of 40 m retarding uniformly. Calculate the force exerted by the brakes. Also calculate the work done by the brakes.

Answer:
1. In order to calculate the force applied by the brakes, first calculate the retardation. Initial speed u = 20 m/s, final speed v = 0 and distance covered s = 40 m.
2. Using the equation v² = u² + 2as, we get 0² = (20)² + 2 × a × 40
3. 80a = -400, so a = -5 m/s²
4. Force exerted by the brakes is given by F = ma = 1200 × (-5) = -6000 N. The negative sign shows that it is a retarding force.
5. Work done by the brakes is W = Fs = 6000 × 40 J = 240000 J = 2.4 × 10⁵ J
6. Work done by the brakes = 2.4 × 10⁵ J

Teacher's Note:
a) The printed working writes s = 90 m once; the correct distance, 40 m, is used in the calculation.
b) Check: loss of kinetic energy = \( \frac{1}{2} \) × 1200 × 400 = 2.4 × 10⁵ J.
c) The brake force is opposite to the motion, so the work done by it is negative in sign.

Ex.19 A horse applying a force of 800 N in pulling a cart with a constant speed of 20 m/s. Calculate the power at which horse is working.

Answer:
1. Power P is given by force × velocity, that is P = F × v.
2. Here F = 800 N and v = 20 m/s.
3. P = 800 × 20 = 16000 watt = 16 kW

Teacher's Note:
a) Use P = Fv for a body moving at constant speed.
b) 1 kW = 1000 W.

Ex.20 A boy keeps on his palm a mass of 0.5 kg. He lifts the palm vertically by a distance of 0.5 m. Calculate the amount of work done. Use g = 9.8 m/s².

Answer:
1. Work done, W = F × s
2. Force F of gravity applied to lift the mass is F = mg = (0.5 kg) × (9.8 m/s²) = 4.9 N and s = 0.5 m.
3. Therefore, W = (4.9) × (0.5 m) = 2.45 J

Teacher's Note:
a) The lifting force equals the weight of the mass.
b) Check: mgh = 0.5 × 9.8 × 0.5 = 2.45 J.

Ex.21 A truck of mass 2500 kg is stopped by a force of 1000 N. It stops at a distance of 320 m. What is the amount of work done ? Is the work done by the force or against the force?

Answer:
1. Here the force F = 1000 N and displacement s = 320 m.
2. Work done, W = F × s = (1000 N) × (320 m) = 320000 J
3. In this case, the force acts opposite to the direction of displacement. So the work is done against the force.

Teacher's Note:
a) The mass 2500 kg is not needed in this calculation.
b) Work against the force carries a negative sign: W = -320000 J.

Ex.22 Two bodies of equal masses move with uniform velocity v and 3v respectively. Find the ratio of their kinetic energies.

Answer:
1. The masses of the bodies are equal, so let the mass of each body be m.
2. (i) Mass of first body = m and velocity of first body = v, so K.E. of first body = \( \frac{1}{2} \) mv² ...(1)
3. (ii) Mass of second body = m and velocity of second body = 3v, so K.E. of second body = \( \frac{1}{2} \) m (3v)² = \( \frac{1}{2} \) m × 9v² = \( \frac{9}{2} \) mv² ...(2)
4. Dividing equation (1) by equation (2): \( \frac{K.E. of first body}{K.E. of second body} \) = \( \frac{1}{9} \) ...(3)
5. Thus, the ratio of the kinetic energies is 1 : 9. We can also write that K.E. of second body = 9 × K.E. of first body.
6. It is clear from this example that when the velocity (or speed) of a body is "tripled" (from v to 3v), then its kinetic energy becomes "nine times".

Teacher's Note:
a) Kinetic energy is proportional to the square of velocity.
b) Cancel the common factor \( \frac{1}{2} \) m before dividing.

EXERCISE - 1

A Very Short Answer Type Questions

Q.1 Is work a scalar or a vector quantity ?

Answer: Work is a scalar quantity.

Teacher's Note:
a) Work has magnitude but no direction, although force and displacement are vectors.

Q.2 What name is given to the product of force and distance ?

Answer: The product of force and distance (moved in the direction of the force) is called work.

Teacher's Note:
a) Work = force × distance moved along the force.

Q.3 Give the units of work in SI system and in CGS system.

Answer: The unit of work in the SI system is the joule (J) and in the CGS system it is the erg. 1 joule = 10⁷ erg.

Teacher's Note:
a) 1 joule = 1 newton × 1 metre; 1 erg = 1 dyne × 1 centimetre.

Q.4 What is the work done, when the displacement of a body is perpendicular to the direction of force acting on it ?

Answer: The work done is zero, because W = Fs cos 90° = 0.

Teacher's Note:
a) cos 90° = 0, so a force perpendicular to the motion does no work.

Q.5 Give the SI unit of power.

Answer: The SI unit of power is the watt (W), where 1 watt = 1 joule per second.

Teacher's Note:
a) Power is the rate of doing work.

Q.6 What is the relationship between watt and horse power ?

Answer: 1 horse power (H.P.) = 746 watt.

Teacher's Note:
a) The printed key writes this the wrong way round as "1 watt = 746 H.P"; the correct relation is 1 H.P. = 746 W.
b) A horse power is the larger unit.

Q.7 What are the units of work and energy ?

Answer: The unit of both work and energy is the joule (J) in the SI system and the erg in the CGS system.

Teacher's Note:
a) Work and energy have the same unit because energy is the capacity to do work.

Q.8 A cell converts one form of energy into another. Name the two forms.

Answer: A cell converts chemical energy into electrical energy.

Teacher's Note:
a) Chemical reactions inside the cell produce the electric current.

Q.9 Name the device which converts electrical energy into mechanical energy.

Answer: An electric motor converts electrical energy into mechanical energy.

Teacher's Note:
a) An electric fan is a common household use of an electric motor.

Q.10 Is energy a scalar quantity or a vector quantity ?

Answer: Energy is a scalar quantity.

Teacher's Note:
a) Energy has only magnitude and no direction.

Q.11 What are the two different forms of mechanical energy ?

Answer: The two forms of mechanical energy are kinetic energy and potential energy.

Teacher's Note:
a) Mechanical energy = kinetic energy + potential energy.

Q.12 How much work is done when a body of mass m is raised to a height h above the ground ?

Answer: The work done is W = mgh.

Teacher's Note:
a) The lifting force equals the weight mg and the displacement is h.

Q.13 How much work is done when a force of 1 N moves a body through a distance of 1 m in its own direction ?

Answer: The work done is 1 joule.

Teacher's Note:
a) This is the definition of the joule.

Q.14 What is the power of a body which is doing work at the rate of one joule per second ?

Answer: The power is 1 watt.

Teacher's Note:
a) 1 watt = 1 J/s.

B Short Answer Type Questions

Q.15 Write the formula for the work done on a body when the force is applied at an angle with the direction of motion of the body.

Answer: W = F s cos θ, where F is the force, s is the displacement and θ is the angle between the force and the displacement.

Teacher's Note:
a) F cos θ is the component of the force along the displacement.
b) If θ = 0°, W = Fs.

Q.16 A satellite revolves around the earth in a circular orbit. Calculate the work done by the force of gravity ?

Answer: The work done is zero. The force of gravity acts towards the centre of the earth, while the displacement of the satellite at every instant is along the tangent to the orbit. They are perpendicular, so W = Fs cos 90° = 0.

Teacher's Note:
a) Centripetal force never changes the speed, so no work is done.
b) The kinetic energy of the satellite stays constant.

Q.17 In which of the following case the work done by a force will be maximum : when the angle between the direction of force and direction of motion is 0° or 90° ?

Answer: The work done is maximum when the angle is 0°, because W = Fs cos 0° = Fs, whereas for 90° the work done is zero.

Teacher's Note:
a) cos θ is largest (equal to 1) at θ = 0°.

Q.18 State two situations in which a body moves with uniform speed and force acts on the body but work done on the body by the force acting is zero.

Answer: (i) A bob tied to a string and whirled in a horizontal circle with uniform speed: the centripetal force is always perpendicular to the displacement, so the work done is zero.
(ii) A satellite moving around the earth (or the earth around the sun) with uniform speed: the gravitational force is perpendicular to the displacement, so the work done is zero.

Teacher's Note:
a) The common condition is that the force is perpendicular to the displacement.
b) A person carrying a load on the head along a level road is another case.

Q.19 What do you understand by the kinetic energy of a body ? Deduce the formula for kinetic energy.

Answer:
1. The energy possessed by a body due to its motion is called its kinetic energy. It is the ability of a body to do work by virtue of its motion.
2. Let a body of mass m move with velocity v and be brought to rest by an opposing force F, so the retardation is a, and F = ma.
3. Using v² - u² = 2as with final velocity 0, 0 = v² - 2as, so the distance moved is s = \( \frac{v²}{2a} \).
4. Kinetic energy = work done by the retarding force = F × s = ma × \( \frac{v²}{2a} \).
5. Hence kinetic energy K.E. = \( \frac{1}{2} \) mv².

Teacher's Note:
a) Kinetic energy is measured by the work needed to stop the body.
b) The "a" cancels in the last step, which is why the formula has only m and v.

Q.20 On what factors does the kinetic energy of a body depend ?

Answer: The kinetic energy of a body depends on its mass (K.E. is directly proportional to m) and its velocity (K.E. is directly proportional to v²).

Teacher's Note:
a) Doubling the velocity makes the kinetic energy four times.

Q.21 What is the difference between potential energy and kinetic energy ?

Answer:

Teacher's Note:
a) Always compare the two on the basis of definition, formula and example.
b) A body can have both at the same time, as a flying aeroplane does.

Q.22 When a ball is thrown vertically upwards, its velocity goes on decreasing. What happens to its potential energy as its velocity becomes zero ?

Answer: As the velocity decreases, the kinetic energy changes into potential energy. When the velocity becomes zero at the highest point, the kinetic energy is zero and the potential energy becomes maximum.

Teacher's Note:
a) Total mechanical energy remains the same throughout the motion.

Q.23 State whether the following objects possess potential energy, kinetic energy or both ?
(i) A flying aeroplane
(ii) A stretched spring
(iii) A rotating ceiling fan
(iv) A man climbing upstairs
(v) A stone placed on the roof
(vi) A running car
(vii) Water stored in a dam

Answer:

Teacher's Note:
a) Motion gives kinetic energy; position or change of shape gives potential energy.
b) A body that both moves and is above the ground has both kinds.

Q.24 What do you understand by the term "transformation of energy" ? Explain with an example.

Answer: The change of one form of energy into another form of energy is called transformation of energy. For example, in an electric fan electrical energy is transformed into mechanical (kinetic) energy; in a steam engine, heat energy is transformed into mechanical energy.

Teacher's Note:
a) Total energy remains constant during every transformation.
b) Name the initial and final forms in the example.

Q.25 A car of mass 1000 kg moving with a speed of 10 m/s stops after moving a distance of 8 m after applying the brakes. Calculate the force applied by the brakes and work done by the brakes.

Answer:
1. Mass m = 1000 kg, initial speed u = 10 m/s, final speed v = 0 and distance s = 8 m.
2. Using v² = u² + 2as: 0 = 100 + 2 × a × 8, so a = -6.25 m/s².
3. Force applied by the brakes F = ma = 1000 × 6.25 = 6250 N (opposite to the motion).
4. Work done by the brakes W = F × s = 6250 × 8 = 50000 J = 5 × 10⁴ J

Teacher's Note:
a) The printed answer key gives 5 × 10⁶ J; the correct value is 5 × 10⁴ J.
b) Check: initial kinetic energy = \( \frac{1}{2} \) × 1000 × 10² = 50000 J, which equals the work done by the brakes.
c) The force is 6250 N in the direction opposite to the motion.

Q.26 A car is being driven by a force of 2.5 × 10¹⁰N. Travelling at a constant speed of 5 m/s, it takes 2 minutes to reach a certain place. Calculate the work done.

Answer:
1. Speed v = 5 m/s and time t = 2 minutes = 120 s.
2. Distance travelled s = v × t = 5 × 120 = 600 m
3. Work done W = F × s = 2.5 × 10¹⁰ × 600 = 1.5 × 10¹³ J

Teacher's Note:
a) The printed answer key gives 15 × 10¹⁴ J; the correct value is 1.5 × 10¹³ J.
b) Convert minutes to seconds before using the speed.

Q.27 How much is the mass of a man if he has to do 2500 joules of work in climbing a tree 5 m tall ?

Answer:
1. Work done W = mgh = 2500 J, with h = 5 m and g = 10 m/s².
2. m = \( \frac{2500}{10 × 5} \) = 50 kg

Teacher's Note:
a) With g = 9.8 m/s² the mass would be about 51 kg; the key uses g = 10 m/s².
b) Rearrange W = mgh to m = W ÷ (gh).

Q.28 A man weighing 500 N carried a load of 100 N up a flight of stairs 4 m high in 5 seconds. What is the power ?

Answer:
1. Total weight lifted = 500 N + 100 N = 600 N and height = 4 m.
2. Work done = 600 × 4 = 2400 J
3. Power = \( \frac{2400}{5} \) = 480 W

Teacher's Note:
a) The man lifts his own weight along with the load.
b) Power = work done ÷ time taken.

Q.29 An athlete weighing 60 kg makes a high jump of 1.8 m. Determine the following :
(i) kinetic energy at the highest point.
(ii) potential energy at the highest point.

Answer:
1. (i) At the highest point the athlete is momentarily at rest (vertical velocity is zero), so the kinetic energy is zero.
2. (ii) Potential energy at the highest point = mgh = 60 × 10 × 1.8 = 1080 J (taking g = 10 m/s²).

Teacher's Note:
a) At the top of the jump all the kinetic energy has changed into potential energy.
b) With g = 9.8 m/s² the value would be 1058.4 J.

Q.30 If an electric bulb of 100 W is light up for 2 hrs, how much electrical energy will be consumed ?

Answer:
1. Power = 100 W = 0.1 kW and time = 2 h.
2. Energy consumed = power × time = 0.1 kW × 2 h = 0.2 kWh (that is, 0.2 × 3.6 × 10⁶ = 7.2 × 10⁵ J)

Teacher's Note:
a) Electrical energy is billed in kilowatt hour (kWh).
b) 1 kWh = 3.6 × 10⁶ J.

Q.31 A person weighing 800 N carries a packet from the base camp B to point A of a hill at a height of 1200 m. The weight of the packet is 200 N. Calculate the following :
(i) How much work he does against gravity ?
(ii) What is the potential energy of the packet at A if it assumed to be zero at B ?

Answer:
1. (i) Total weight lifted = 800 + 200 = 1000 N and height = 1200 m.
2. Work done against gravity = 1000 × 1200 = 1.2 × 10⁶ J = 12 × 10⁵ J
3. (ii) Potential energy of the packet at A = weight of packet × height = 200 × 1200 = 2.4 × 10⁵ J

Teacher's Note:
a) Part (i) includes the person's own weight; part (ii) is only for the packet.
b) Potential energy is measured from the reference level B.

Q.32 A man weighing 600 N carries a load of 100 N up a flight of stairs 4 m high in 5 s. Calculate the power.

Answer:
1. Total weight lifted = 600 + 100 = 700 N and height = 4 m.
2. Work done = 700 × 4 = 2800 J
3. Power = \( \frac{2800}{5} \) = 560 W

Teacher's Note:
a) Add the weight of the man and the load before finding work.
b) Power = work ÷ time.

Q.33 Water is falling on the blades of a turbine at the rate of 6 × 10³ kg/min. The height of the fall is 10 m. Calculate the power of the motor to be used.

Answer:
1. Mass of water per second = \( \frac{6 × 103}{60} \) = 100 kg/s
2. Work done per second = mgh = 100 × 10 × 10 = 10000 J (taking g = 10 m/s²)
3. Power = 10000 W = 10 kW

Teacher's Note:
a) Power is the work done per second, so convert kg per minute to kg per second.
b) The key uses g = 10 m/s².

Q.34 An electric motor drives a machine which lifts a mass of 2 kg through a height of 6 m, in 4 s at a constant speed. Assume g = 9.8 N kg^-1 and calculate (i) the amount of work done and (ii) the power of the machine to lift the mass of 2 kg.

Answer:
1. (i) Work done = mgh = 2 × 9.8 × 6 = 117.6 J
2. (ii) Power = \( \frac{W}{t} \) = \( \frac{117.6}{4} \) = 29.4 W

Teacher's Note:
a) At constant speed the lifting force equals the weight mg.
b) Power = work ÷ time.

C Long Answer Type Questions

Q.35 Define the term work. What are the quantities on which the amount of work done depends ? How are they related to work ? What is the condition for a force to do work on a body?

Answer:
1. Work: In physics, work is said to be done when a force acts on a body and the body moves in the direction of the force. Work done is the product of force and the distance moved by the body in the direction of the force.
2. The work done depends on two quantities: (i) the magnitude of the force F and (ii) the distance s through which the body moves in the direction of the force.
3. The relation is W = F × s. Work is directly proportional to the force and directly proportional to the displacement.
4. Condition for work: the force must act on the body and the body must be displaced in the direction of the force (or against it). If the body is not displaced, or the displacement is perpendicular to the force, no work is done.

Teacher's Note:
a) A person pushing a wall does no work, because the displacement is zero.
b) The unit of work is the joule: 1 J = 1 N × 1 m.
c) Work is a scalar quantity.

Q.36 Write the formula for work done on a body when the body moves at an angle to the direction of force. Give the meaning of each symbol used. What will happen to the work done if angle between the direction of force and motion of the body is increased gradually? Will it increase, decrease or remain constant ?

Answer:
1. Formula: W = F s cos θ
2. Meaning of symbols: W is the work done, F is the magnitude of the force, s is the magnitude of the displacement of the body and θ is the angle between the direction of the force and the direction of the displacement.
3. As θ is increased gradually from 0° to 90°, cos θ decreases from 1 to 0, so the work done decreases.
4. At θ = 90° the work done becomes zero. Beyond 90° (up to 180°) it becomes negative.

Teacher's Note:
a) Work is maximum at θ = 0° and zero at θ = 90°.
b) Do not say that the work is constant; it changes with cos θ.

Q.37 Write an expression for the kinetic energy of a body of mass m moving with a velocity v. Explain by an example what is meant by potential energy. Write down the expression for gravitational potential energy of a body of mass m placed at a height h above the surface of the earth.

Answer:
1. Kinetic energy of a body of mass m moving with velocity v: K.E. = \( \frac{1}{2} \) mv².
2. Potential energy is the energy possessed by a body due to its position or change in shape. Example: water stored in an overhead tank or a dam has potential energy because of its height, and it can do work (turn a turbine) when it falls. A stretched rubber band also has potential energy.
3. Gravitational potential energy of a body of mass m at a height h above the earth's surface: U = mgh, where g is the acceleration due to gravity.

Teacher's Note:
a) The potential energy equals the work done in lifting the body to the height h.
b) Quote the formula along with the meaning of each symbol.

Q.38 How can you explain the oscillation of a simple pendulum on the basis of conservation of energy ?

Answer:
1. A swinging simple pendulum is a body whose energy is either potential or kinetic or a mixture of the two, but its total energy at any instant remains the same.
2. At the extreme position (B or C) the bob stops for an instant, so it has only potential energy and no kinetic energy.
3. As the bob moves from the extreme position towards the centre position A, its potential energy decreases and its kinetic energy increases.
4. At the centre position A the bob has only kinetic energy (maximum speed) and no potential energy.
5. As the bob moves from A towards the other extreme position, its kinetic energy decreases and its potential energy increases. The total energy stays constant, which is the law of conservation of energy.

Teacher's Note:
a) Energy keeps changing form between potential and kinetic during every swing.
b) In real life the swing slowly dies down because some energy is lost as heat due to air resistance.

EXERCISE - 2

Single Correct Answer Type Questions

Q.1 Work done upon a body is-
(A) a vector quantity
(B) a scalar quantity
(C) (A) and (B) both are correct
(D) none of these

Answer: (B) a scalar quantity

1. Work has magnitude only and no direction.
2. Hence work is a scalar quantity.

Teacher's Note:
a) Force and displacement are vectors, but their product W = Fs cos θ is a scalar.

Q.2 Work done -
(A) is always positive
(B) is always nagative
(C) can be positive, negative or zero
(D) none of these

Answer: (C) can be positive, negative or zero

1. Work is positive when the force and displacement are in the same direction, negative when they are opposite and zero when they are perpendicular.
2. Hence work can be positive, negative or zero.

Teacher's Note:
a) W = Fs cos θ has the sign of cos θ.
b) Be careful with "always" statements in options.

Q.3 No work is done when -
(A) a nail is plugged in a wooden board
(B) a box is pushed along a horizontal floor
(C) there is no component of force parallel to the direction of motion
(D) there is no component of force perpendicular to the direction of motion

Answer: (C) there is no component of force parallel to the direction of motion

1. Work depends on the component of force along the displacement.
2. If there is no component of force parallel to the direction of motion, then W = 0.

Teacher's Note:
a) A force perpendicular to the motion does no work.
b) Option (D) is the opposite and is wrong.

Q.4 A body at rest can have :
(A) speed
(B) velocity
(C) momentum
(D) energy

Answer: (D) energy

1. A body at rest has zero speed, zero velocity and zero momentum.
2. It can still have energy, for example potential energy due to its position.

Teacher's Note:
a) A stone on a roof is at rest but has potential energy.

Q.5 Types of mechanical energy are:
(A) kinetic energy only
(B) potential energy only
(C) kinetic energy and potential energy both
(D) neither kinetic energy nor potential energy

Answer: (C) kinetic energy and potential energy both

1. Mechanical energy is the energy due to the state of rest or motion of a body.
2. It is of two types, kinetic energy and potential energy.

Teacher's Note:
a) Both forms together make the mechanical energy.

Q.6 Work means:
(A) effort
(B) interview
(C) achievement
(D) get-together

Answer: (A) effort

1. In the everyday sense, work means an activity that needs physical or mental effort.
2. Interview and get-together are not work, and achievement is the result of work, not the work itself.
3. Hence, the correct option is (A).

Teacher's Note:
a) The printed key shows (C) achievement, but the meaning of work is effort; (A) is the correct answer.
b) In physics, work has a specific meaning: force × displacement.

Q.7 Work is done on a body when :
(A) force acts on the body but the body is not displaced
(B) force does not act on the body but it is displaced
(C) force acts on the body in a direction perpendicular to the direction of the displacement of the body
(D) force acts on the body and the body is either displaced in the direction of force or opposite to the direction of force

Answer: (D) force acts on the body and the body is either displaced in the direction of force or opposite to the direction of force

1. Work needs both a force and a displacement along (or against) the force.
2. Only option (D) satisfies this.

Teacher's Note:
a) No displacement, no force or a perpendicular force all mean zero work.

Q.8 Force F acts on a body such that force F makes an angle θ with the horizontal direction and the body is also displaced through a distance S in the horizontal direction, then the work done by the force is -
(A) FS
(B) FS cos θ
(C) FS sin θ
(D) zero

Answer: (B) FS cos θ

1. The component of F along the horizontal displacement is F cos θ.
2. Work done = F cos θ × S = FS cos θ.

Teacher's Note:
a) The sin θ component is perpendicular to the displacement and does no work.

Q.9 In tug of war work done by winning team is :
(A) zero
(B) positive
(C) negative
(D) none of these

Answer: (B) positive

1. The winning team pulls the rope in the direction of its force and the rope moves in that direction.
2. So the work done by the winning team is positive.

Teacher's Note:
a) Force and displacement in the same direction give positive work.

Q.10 In tug of war work done by loosing team is :
(A) zero
(B) positive
(C) negative
(D) none of these

Answer: (C) negative

1. The losing team pulls the rope towards itself, but the rope moves in the opposite direction.
2. Force and displacement are opposite, so the work done by the losing team is negative.

Teacher's Note:
a) It is the same rope, but the sign of work depends on the direction of each team's force.

Q.11 Work done by the force of gravity, when a body is lifted to height h above the ground is :
(A) zero
(B) positive
(C) negative
(D) none of these

Answer: (C) negative

1. Gravity acts downwards while the body moves upwards, so force and displacement are opposite.
2. The work done by gravity is negative, equal to -mgh.

Teacher's Note:
a) The lifting force does positive work; gravity does negative work.

Q.12 When work is done on a body:
(A) it gains energy
(B) it looses energy
(C) its energy remains constant
(D) none of these

Answer: (A) it gains energy

1. Work done on a body is stored in it as energy.
2. So the body gains energy.

Teacher's Note:
a) Work done by a body decreases its energy.

Q.13 Choose correct relation :
(A) 1 J = 10⁵ erg
(B) 1 J = 10⁷ erg
(C) 1 J = 10³ erg
(D) none of these

Answer: (B) 1 J = 10⁷ erg

1. 1 J = 1 N × 1 m = 10⁵ dyne × 10² cm = 10⁷ erg.

Teacher's Note:
a) 1 erg = 1 dyne × 1 cm; learn 1 J = 10⁷ erg.

Q.14 The kinetic energy of an object is K. If its velocity is doubled than its kinetic energy will be -
(A) K
(B) 2K
(C) \( \frac{K}{2} \)
(D) 4K

Answer: (D) 4K

1. K.E. = \( \frac{1}{2} \) mv² is proportional to v².
2. If v becomes 2v, K.E. becomes (2)² = 4 times, that is 4K.

Teacher's Note:
a) Doubling the velocity makes the kinetic energy four times, not two times.

Q.15 Two bodies of mass 1 kg and 4 kg possess equal momentum. The ratio of their K.E.
(A) 4 : 1
(B) 1 : 4
(C) 2 : 1
(D) 1 : 2

Answer: (A) 4 : 1

1. K.E. = \( \frac{p²}{2m} \), so for equal momentum p, K.E. is inversely proportional to mass.
2. K.E.₁ : K.E.₂ = 4 : 1

Teacher's Note:
a) Use K.E. = p² ÷ 2m when the momentum is equal.
b) The lighter body has the greater kinetic energy.

Q.16 Which one is not the unit of energy?
(A) kilocalorie
(B) kWh
(C) erg
(D) watt

Answer: (D) watt

1. Kilocalorie, kWh and erg are all units of energy.
2. Watt is the unit of power.

Teacher's Note:
a) 1 watt = 1 J/s, a rate, not an amount of energy.

Q.17 1 kg mass has K.E. of 1 J when its speed is-
(A) 0.45 ms^-1
(B) 1 ms^-1
(C) 1.4 ms^-1
(D) 4.4 ms^-1

Answer: (C) 1.4 ms^-1

1. K.E. = \( \frac{1}{2} \) mv², so 1 = \( \frac{1}{2} \) × 1 × v², which gives v² = 2.
2. v = √2 ≈ 1.4 m/s

Teacher's Note:
a) Rearrange to v = √(2K.E. ÷ m).
b) √2 = 1.414.

Q.18 When you compress a spring you do work on it. The elastic potential energy of the spring-
(A) increases
(B) decreases
(C) disappears
(D) remains constant

Answer: (A) increases

1. The work done in compressing the spring is stored in it as elastic potential energy.
2. So the elastic potential energy increases.

Teacher's Note:
a) The same is true when a spring is stretched.

Q.19 When a ball is thrown upward, its total energy-
(A) increases
(B) decreases
(C) remains same
(D) none of these

Answer: (C) remains same

1. While the ball rises, kinetic energy changes into potential energy.
2. The total energy remains the same (neglecting air resistance).

Teacher's Note:
a) This is the law of conservation of energy.

Q.20 If a stone of mass 'm' falls a vertical distance 'd' the decrease in gravitational potential energy is-
(A) \( \frac{Mg}{d} \)
(B) \( \frac{Mg²}{2} \)
(C) mgd
(D) \( \frac{Mg}{d²} \)

Answer: (C) mgd

1. Decrease in gravitational potential energy = mg × (distance fallen) = mgd.

Teacher's Note:
a) U = mgh, with h replaced by the distance d fallen.

Q.21 An object of mass 10 kg falls from height 10 m.. Kinetic energy gained by the body will be approximately equal to-
(A) 1000 J
(B) 500 J
(C) 100 J
(D) None of these

Answer: (A) 1000 J

1. Kinetic energy gained = loss of potential energy = mgh.
2. = 10 × 9.8 × 10 = 980 J, which is approximately 1000 J.

Teacher's Note:
a) Using g = 10 m/s² gives exactly 1000 J.
b) Energy gained equals energy lost.

Q.22 A spring is stretched. The potential energy in stretching the spring-
(A) remains the same
(B) increases
(C) decreases
(D) becomes zero

Answer: (B) increases

1. The work done in stretching the spring is stored as elastic potential energy.
2. So the potential energy increases.

Teacher's Note:
a) More stretch stores more energy.

Q.23 The potential energy of a boy is maximum when he is-
(A) standing
(B) sleeping on the ground
(C) sitting on the ground
(D) sitting on chair

Answer: (A) standing

1. Potential energy is mgh, so it is greatest when the centre of mass is highest.
2. The centre of the boy's mass is highest when he is standing.

Teacher's Note:
a) Compare the height of the centre of mass in each case.

Q.24 The potential energy of a freely falling object decreases continuously. What happens to the loss of potential energy ?
(A) It is continuously converted into sound energy
(B) it is continuously converted into kinetic energy
(C) it is continuously destroyed
(D) None of these

Answer: (B) it is continuously converted into kinetic energy

1. Energy cannot be destroyed.
2. The lost potential energy is converted into kinetic energy as the speed of the object increases.

Teacher's Note:
a) Total mechanical energy of a freely falling body is constant.

Q.25 A device which converts mechanical energy into electrical energy is known as-
(A) electric motor
(B) lever
(C) generator
(D) microphone

Answer: (C) generator

1. A generator (dynamo) rotates a coil in a magnetic field, converting mechanical energy into electrical energy.

Teacher's Note:
a) An electric motor does the opposite.
b) A microphone converts sound energy into electrical energy.

Q.26 The value of g on moon is 1/6th of the value of g on the earth. A man can jump 1.5 m high on the earth. On moon he can jump up to a height of-
(A) 9 m
(B) 7.5 m
(C) 6 m
(D) 4.5 m

Answer: (A) 9 m

1. The man gives the same energy in the jump on both, so mgh is the same: h is inversely proportional to g.
2. On the moon h = 6 × 1.5 = 9 m.

Teacher's Note:
a) With g one-sixth, the height is six times.

Q.27 A raised hammer possess-
(A) kinetic energy only
(B) gravitational potential energy
(C) electrical energy
(D) sound energy

Answer: (B) gravitational potential energy

1. A raised hammer is at a height above the ground and at rest.
2. So it possesses gravitational potential energy.

Teacher's Note:
a) Potential energy is due to position.

Q.28 An object of mass 1 kg has a P.E. of 1 J relative to the ground when it is at a height of : (g = 9.8 m/s²).
(A) 0.10 m
(B) 10 m
(C) 9.8 m
(D) 32 m

Answer: (A) 0.10 m

1. P.E. = mgh, so 1 = 1 × 9.8 × h.
2. h = \( \frac{1}{9.8} \) ≈ 0.10 m

Teacher's Note:
a) Rearrange to h = U ÷ (mg).

Q.29 To lift a 5 kg mass to a certain height, amount of energy spent is 245 J. The mass was raised to a height of-
(A) 15 m
(B) 10 m
(C) 7.5 m
(D) 5 m

Answer: (D) 5 m

1. Energy spent = mgh, so 245 = 5 × 9.8 × h = 49 h.
2. h = 5 m

Teacher's Note:
a) Check: 5 × 9.8 × 5 = 245.

Q.30 Chlorophyll in the plants convert the light energy into-
(A) heat energy
(B) chemical energy
(C) mechanical energy
(D) electrical energy

Answer: (B) chemical energy

1. In photosynthesis, chlorophyll helps plants convert light energy into chemical energy stored in glucose.

Teacher's Note:
a) The chemical energy is stored in food made by the plant.

Q.31 Kilowatt is the unit of-
(A) energy
(B) power
(C) force
(D) momentum

Answer: (B) power

1. 1 kilowatt = 1000 watt, and watt is the unit of power.

Teacher's Note:
a) Kilowatt hour (kWh) is the unit of energy; kilowatt is power.

Q.32 Work is product of time and-
(A) energy
(B) power
(C) force
(D) distance

Answer: (B) power

1. Power = \( \frac{Work}{Time} \), so Work = Power × Time.
2. Hence work is the product of time and power.

Teacher's Note:
a) Rearranging the definition of power gives the answer.

Q.33 A young son work quickly for two hours and prepares 16 items in a day. His old father works slowly for eight hours and prepare 24 items a day.
(A) son has more power
(B) son has more energy
(C) both have equal power
(D) both have equal energy

Answer: (A) son has more power

1. Rate of work of the son = \( \frac{16}{2} \) = 8 items per hour.
2. Rate of work of the father = \( \frac{24}{8} \) = 3 items per hour.
3. Power is the rate of doing work, so the son has more power.

Teacher's Note:
a) Power depends on how fast the work is done.
b) Compare the work done per hour.

Q.34 One horse power is
(A) 746 W
(B) 550 W
(C) 980 W
(D) 32 W

Answer: (A) 746 W

1. 1 horse power (H.P.) = 746 watt.

Teacher's Note:
a) 550 is the value in foot-pound per second, not in watt.

Q.35 Power of a moving body is stored in the form of-
(A) Work and distance
(B) force and distance
(C) force and velocity
(D) force and time

Answer: (C) force and velocity

1. Power = \( \frac{W}{t} \) = \( \frac{Fs}{t} \) = F × v.
2. So the power of a moving body is the product of force and velocity.

Teacher's Note:
a) Force × velocity gives power directly.

Q.36 A weight lifter lifts 240 kg from the ground to a height of 2.5 m in 3 second his power is-
(A) 1960 W
(B) 19.6 W
(C) 1.96 W
(D) 196 W

Answer: (A) 1960 W

1. Work done = mgh = 240 × 9.8 × 2.5 = 5880 J
2. Power = \( \frac{5880}{3} \) = 1960 W

Teacher's Note:
a) Multiply the mass by g to get the force.
b) Check the unit: W = J/s.

Q.37 Which of the following is not the unit of power ?
(A) J/s
(B) Watt
(C) kJ/h
(D) kWh

Answer: (D) kWh

1. J/s, watt and kJ/h are all work per unit time, so they are units of power.
2. kWh is work, a unit of energy.

Teacher's Note:
a) kWh = kW × h is an energy unit.

W = Fs cos q = 100 × 2 × cos 60º

= 100 × 3 × 1/2

1 = 150 J (Q cos 60º = 1/2 )

Power = Work done /T ime taken

 P = W/t

 Mass , m = (60 × 10–3 m3) × (1000 kg/m3) = 60 kg

 Equation (2) becomes