Home Lakhmir Singh Solution

Question 1 How much work is done when a body of mass m is raised to a height h above the ground?

Solution 1

Mass = m

Height above the ground = h

Work done = Potential energy acquired by the body = m x g x h

where g is acceleration due to gravity

Question 2 State the SI unit of work.

Solution 2 SI unit of work is Joule (J).

Question 3 Is work a scalar or a vector quantity?

Solution 3 Work is a scalar quantity.

Work, as a physical quantity, requires only magnitude to be represented. Hence, it is scalar quantity.

Question 4 Define 1 joule of work.

Solution 4 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.

Question 5 What is the condition for a force to do work on a body?

Solution 5 The condition for a force to do work on a body is that it should produce motion in the body.

Question 6 Is energy a vector quantity?

Solution 6 Energy is a scalar quantity. It has only magnitude but no direction.

Question 7 What are the units of (a) work and ( b) energy?

Solution 7

a) Unit of work is joule.

b) Unit of energy is joule.

Question 8 What is the work done against gravity when a body is moved horizontally along a frictionless surface?

Solution 8 The work done against gravity is zero when a body is moved horizontally along a frictionless surface because force of gravity acts perpendicular to the direction of motion.

Question 9 By how much will the kinetic energy of a body increase if its speed is doubled?

Solution 9Kinetic energy will become four times when the speed is doubled because kinetic energy is directly proportional to square of speed of the body.

K.E. ∝ v^{2}

Question 10 Write an expression for the kinetic energy of a body of mass m moving with a velocity v.

Solution 10

Mass = m

Velocity = v

K.E.=1⁄2 mv^{2}

Question 11 If the speed of a body is halved, what will be the change in its kinetic energy?

Solution 11 Kinetic energy will become one-fourth when the speed is halved because kinetic energy is directly proportional to square of speed of the body

K.E. ∝ v^{2}

Question 12 On what factors does the kinetic energy of a body depend?

Solution 12 The kinetic energy of a body depends on

a) Mass of the body, m

b) Square of the velocity of the body, v^{2}

Question 13 Which would have a greater effect on the kinetic energy of an object: doubling the mass or doubling the velocity?

Solution 13 Doubling the velocity would have a greater effect on kinetic energy.

Question 14 How fast should a man of 50kg run so that his kinetic energy is 625 J?

Solution 14

m=50 Kg

K.E.=625 J

Question 15 State whether the following objects possess kinetic energy, potential energy, or both:

a) A man climbing a hill

b) A flying aeroplane

c) A bird running on the ground

d) A ceiling fan in the off position

e) A stretched spring lying on the ground

Solution 15

a) Both kinetic and potential energy

b) Both kinetic and potential energy

c) Only kinetic energy

d) Only potential energy

e) Only potential energy

Note: In all the above cases we take ground as reference level where potential energy is zero.

Question 16 Two bodies A and B of equal masses are kept at heights of h and 2h respectively. What will be the ratio of their potential energies?

Solution 16 Let masses of body A and B be m

Height of body A = h

Height of body B = 2h

Potential energy for body A, PE_{A} = m x g x h

Potential energy for body B, PE_{B} = m x g x 2h

Question 17 What is the kinetic energy of a body of mass 1kg moving with a speed of 2m/s?

Solution 17

Mass = 1 kg

Velocity = 2 m/s

Question 18 Is potential energy a vector or a scalar quantity?

Solution 18 Potential energy is a scalar quantity as it has magnitude only and it does not require any specification of direction.

Question 19 A load of 100kg is pulled up by 5m. Calculate the work done. (g=9.8m/s^{2})

Solution 19

Mass = 100 kg

Height = 5 m

g = 9.8 m/s^{2}

P.E. = m x g x h = 100 x 5 x 9.8 = 4900 J

Work done is equal to PE acquired by the body.

Question 20 State whether the following statement is true or false:

The potential energy of a body of mass 1kg kept at a height of 1m is 1 J.

Solution 20vFalse.

PE = m x g x h

= 1 x 9.8 x 1 = 9.8 J

Question 21 What happens to the potential energy of a body when its height is doubled?

Solution 21 The potential energy is doubled when the height is doubled since potential energy is directly proportional to height, h to which body is raised.

Question 22 What kind of energy is possessed by the following?

a) A stone kept on roof-top.

b) A running car.

c) Water stored in the reservoir of a dam.

d) A stretched rubber band.

Solution 22

a) Potential energy

b) Kinetic energy

c) Potential energy

d) Potential energy

e) Potential energy

Question 23 Fill in the following blanks suitable words:

a)Work is measured as a product of ........... and ...........

b)The work done on a body moving in a circular path is ..............

c)Joule is the work done when a force of one ................ moves an through a distance of one ........... in the direction of ...........

d)The ability of a body to do work is called ............ The ability of a body to do work because of its motion is called .........

e)The sum of the potential and kinetic energies of a body is called ............... energy.

Solution 23

a)Force ; distance

b)Zero

c)newton; metre; force

d)Energy; kinetic energy

e)mechanical

Question 24 What are the quantities on which the amount of work done depends? How are they related to work?

Solution 24 The work done by a force on a body depends on two factors

a) magnitude of force

b) distance through which the body moves

Work done is directly proportional to the force applied and the distance through which the body moves.

W = F x s

where W is work done, F is force applied and s is distance through which the body moves.

Question 25 It is possible that a force is acting on a body but still the work done is zero? Explain giving one example.

Solution 25 Yes, it is possible that a force is acting on a body but still work done is zero. For example, in the case of a man pushing a wall, the work done is zero despite of non-zero force, since there is no displacement of the wall.

Question 26 A boy throws a rubber ball vertically upwards. What type of work, positive or negative, is done:

a) by the force applied by the boy?

b) by the gravitational force of earth?

Solution 26 a) Work done by force applied by the boy is positive because this force is in the direction of motion of the body.

b) Work done by the gravitational force is negative because this force is against the direction of motion of the body.

Question 27 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

Solution 27

W=Fcosθ x s

F is the force applied

θ is the angle between the direction of motion of the body.

s is the distance moved by the body

Question 28 How does the kinetic energy of a moving body depend on its (i) speed, and (ii) mass?

Solution 28

Kinetic energy is directly proportional to the mass of the body, m.

KE ∝ m

Kinetic energy is directly proportional to square of speed of the body, v

KE ∝ v^{2}

Question 29 Give one example each in which a force does (a) positive work (b) negative work (c) zero work.

Solution 29

(a) Positive work: Work done by the force applied by a person on a ball that is thrown upwards.

(b) Negative work: Work done by gravitational force of earth on a ball thrown upwards.

(c) Zero work: Work done by gravitational force of earth on a box that is sliding horizontally on the ground.

Question 30 A ball of mass 200g falls from a height of 5metres. What is its kinetic energy it just reaches the ground? (g=9.8m/s^{2})

Solution 30

Question 31 Find the momentum of a body of mass 100g having a kinetic energy of 20 J.

Solution 31

KE = 20 J

Mass = 100 g = 0.1 kg

KE = 1/2 mv^{2}

20 = 1/2 x 0.1 x v^{2}

v^{2} = 400

v = 20 m/s

Momentum = m x v = 0.1 x 20 = 2 kg.m/s

Question 32 Two objects having equal masses are moving with uniform velocities of 2 m/s and 6m/s respectively. Calculate the ratio of their kinetic energies.

Solution 32

Question 33 A boy of 2kg falls from rest. What will be its kinetic energy during the fall at the end of 2 s? (Assume g=10m/s^{2})

Solution 33

Mass of body = 2 kg

Initial velocity u = 0

Time taken = 2 s

Acceleration due to gravity, g = 10 m/s^{2}

Final velocity v

Using first equation of motion

v = u + gt

= 0 + 10 x 2 = 20 m/s

K.E.=1/2 x mv^{2}

=1/2 x 2 x 20^{2}

=400J

Question 34 On a level road, a scooterist applies brakes to slow down from a speed of 10 m/s to 5 m/s. if the mass of the scooter and the scooterist be 150 kg, calculate the work done by the brakes.(Neglect air resistance and friction)

Solution 34

Mass of scooter+scooterist = 150 kg

Initial velocity u = 10 m/s

Final velocity v = 5 m/s

Retardation = a

Distance covered= s

Using third equation of motion

v^{2} - u^{2} = 2as

5^{2} - 10^{2} = 2as

as = -75/2 ------(i)

Work done W = F x s

But F = m x a

So, W= m x a x s

Put the value of 'as' from eq(i)

W = 150 x (-75/2) = -5625 J

Neagtive sign implies that force of brakes acts opposite to the direction of motion.

Question 35 A man drops a 10 kg rock from the top of a 5 m ladder. What is its speed just before it hits the ground? What is its kinetic energy when it reaches the ground?(g=10 m/s^{2})

Solution 35

Mass of rock = 10 kg

Height of ladder, h = 5 m

Initial velocity of rock, u = 0

Final velocity v

g = 10 m/s^{2}

using third equation of motion

v^{2} - u^{2} = 2gh

v^{2} - 02 = 2 x 10 x 5

v = 10 m/s

K.E.=(1/2)x10x10^{2}

=500J

Question 36 Calculate the work done by the brakes of a car of mass 1000 kg when its speed is reduced from 20 m/s to 10 m/s?

Solution 36

Mass of car = 1000 kg

Initial velocity u = 20 m/s

Final velocity v = 10 m/s

Retardation = a

Distance covered = s

Using third equation of motion

v^{2} -u^{2} = 2as

10^{2}- 20^{2} = 2as

as = -150 ------(i)

Work done W = F x s

But F = m x a

So, W= m x a x s

Put the value of 'as' from equation (i)

W = 1000 x -150 = -150000 = -150 kJ

Neagtive sign implies that force of brakes acts opposite to the direction of motion.

Question 37 A body of mass 100 kg is lifted up by 10 m. Find

i) the amount of work done

ii) potential energy of the body at that height?(value of g = 10 m/s^{2})

Solution 37

Mass of body, m = 100 kg

Height , h= 10 m

Acceleration due to gravity, g = 10 m/s^{2}

i) Work done, W = m x g x h = 100 x 10 x 10 = 10000 = 10 kJ

ii) Potential energy of the body = work done = 10 kJ

Question 38 A boy weighing 50 kg climbs up a vertical height of 1000m. Calculate the amount of work done by him. How much potential energy does he gain? (g = 9.8 m/s^{2})

Solution 38

Mass of the boy, m = 50 kg

Height, h = 100 m

Acceleration due to gravity, g = 9.8m/s^{2}

Work done by the boy, W = m x g x h = 50 x 9.8 x 100 = 49000 J = 49 kJ

Potential energy gained by the boy = work done by the boy = 49 kJ

Question 39 When is the work done by a force on a body

i) positive

ii) negative

iii) zero?

Solution 39 Work done by a force applied on a body is

i) positive when the force acts in the direction of motion of the body.

ii) negative when the force acts in the direction opposite to the direction of motion of the body.

iii) zero when the force acts at right angle to the direction of motion of the body.

Question 40 To what height should a box of mass 150 kg be lifted, so that its potential energy may become 7350 J? (g = 9.8 m/s^{2})

Solution 40

Mass of the box, m = 150 kg

PE = 7350 J

Acceleration due to gravity, g = 9.8m/s^{2}

PE = m x g x h

7350 = 150 x 9.8 x h

Question 41 A body of mass 2 kg is thrown vertically upwards with an initial velocity of 20 m/s. what will be its potential energy at the end of 2 s? (g = 10 m/s^{2})

Solution 41

Mass of the body, m = 2 kg

Initial velocity, u = 20 m/s

Acceleration due to gravity, g = 10 m/s^{2}

Height reached = h

Time, t = 2 s

Using second equation of motion

Question 42 How much work is done when a force of 1 N moves a body through a distance of 1 m in its own direction?

Solution 42

Force, F = 1 N

Distance, s = 1 m

Work done W = F x s

= 1 x 1 = 1 J

Question 43 A car is being driven by a force of 2.5 x 10^{10} N. travelling at a conctant speed of 5 m/s, it takes 2 minutes to reach a certain place. Calculate the work done.

Solution 43

Force, F = 2.5 x 1010 N

Velocity, v = 5 m/s

Time, t = 2 minutes = 120 s

Distance, s = v x t = 5 x 120 = 600 m

Work done, W = F x s = 2.5 x 10^{10} x 600 = 15 x 10^{12} J

Question 44 Explain by an example that a body may possess energy even when it is not in motion.

Solution 44 A stretched rubber band is an example of a body possessing energy while it is not in motion. The rubber band contains potential energy due to the change in its shape or configuration.

Question 45

a) On what factors does gravitational potential energy of a body depend?

b) Give one example each of a body possessing

a. Kinetic energy

b. Potential energy

Solution 45

a) Gravitational potential energy of a body depends on:

i) mass of the body, m

ii) height to which the body is lifted, h

iii) acceleration due to gravity, g

b)

i. A moving cricket ball has kinetic energy

ii. A stretched rubber band has potential energy

Question 46 Give two examples where a body possesses both, kinetic energy as well as potential energy.

Solution 46 Two examples where a body possesses both kinetic energy as well as potential energy are

i) a man climbing up a hill

ii) a flying aeroplane

Question 47 How much is the mass of a man if he has to do 2500 J of work in climbing a tree 5 m tall? (g = 10 m/s^{2})

Solution 47

Mass of man, m

Height of tree, h = 5 m

Work done, W = 2500 J

Acceleration due to gravity, g = 10 m/s^{2}

W = m x g x h

2500 = m x 10 x 5

Question 48 If the work done by a force moving an object through a distance of 20 cm is 24.2 J, what is the magnitude of the force?

Solution 48

Work done, W =24.2 J

Distance, s = 20 cm = 0.2 m

Force, F

W = F x s

24.2 = F x 0.2

F = 24.2/0.2 = 121 N

Question 49 A boy weighing 40 kg makes a high jump of 1.5 m

i) What is his kinetic energy at the highest point?

ii) What is his potential energy at the highest point?(g = 10 m/s^{2})

Solution 49

Mass of boy, m = 40 kg

Height, h = 1.5 m

Acceleration due to gravity, g = 10 m/s^{2}

i) At highest point, velocity, v = 0

Therefore KE = 0

ii) PE = m x g x h = 40 x 10 x 1.5 = 600 J

Question 50 What type of energy is possessed by

i) By the stretched rubber strings of catapult?

ii) By the piece of stone which is thrown away on releasing the stretched rubber strings of catapult?

Solution 50

a) Potential energy

b) Both potential and kinetic energy

Question 51 A weightlifter is lifting weights of mass 200 kg upto a height of 2 m, if g = 9.8 m/s^{2},calculate:

i) Potential energy acquired by the weights

ii) Work done by the weightlifter

Solution 51

Mass, m = 200 kg

Height, h = 2 m

Acceleration due to gravity, g = 9.8 m/s^{2}

i) Potential energy = m x g x h = 200 x 9.8 x 2 = 3920 J

ii) Work done is against gravity = Potential energy gained by the weights, therefore

Work done W = m x g x h = 200 x 9.8 x 2 = 3920 J

Question 52 a) Define the term work. Write the formula for the work done on a when a force acts on the body in the direction of its displacement. Give the meaning of each symbol which occurs in the formula

b) A person of mass 50 kg climbs a tower of height 72 meters. Calculate the work done(g = 9.8 m/s^{2})

Solution 52

(a) Work is done when a force applied on a body produces motion in it.

Formula for work done:

W = F x s

where W is the work done

F is force applied

S is the displacement of the body in the direction of force

(b) Mass of the person, m = 50 kg

Height of tower, h = 72 m

Acceleration due to gravity = 9.8 m/s^{2}

Work done W = m x g x h

= 50 x 9.8 x 72 = 35280 J

Question 53 a) When do we say that work is done? Write the formula for the work done by a body in moving up against gravity. Give the meaning of each symbol which occurs in it.

b) How much work is done when a force of 2 N moves a body through a distance of 10 cm in the direction of force?

Solution 53

(a) Work is said to be done when the force applied on a body produces motion in it.

Work done by a body in moving up is given by

W = m x g x h

where W is the work done against the gravity

m = mass of the body

g = acceleration due to gravity

h = height through which the body is lifted above the ground

(b) Force, F = 2 N

Distance, s = 10 cm = 0.1 m

Work done W = F x s = 2 x 0.1 = 0.2 J

Question 54 a) What happens to the work done when the displacement of a body is at right angles to the direction of force acting on it? Explain your answer.

b) A force of 50 N acts on a body and moves it a distance of 4 m on a horizontal surface. Calculate the work done if the direction of force is at angle of 60^{o} to the horizontal surface.

Solution 54 (a) When the displacement of a body is at right angles to the direction of foce acting on it, then work done is zero.

(b) Force, F = 50 N

Distance, s = 4 m

Work And Energy = 60^{o}

W = F cos θ

when θ=90^{o}, cos90^{o}=0

W=0xs=0

(b) Force F= 50

Distance s= 4m

angle between direction of force and direction of motion, θ=60^{o}

work done, W = F cosθ x s

=50 x cos60^{o} x 4 = 50 x 0.5 x 4= 100 J

Question 55 a) Define the term energy of a body. What is the SI unit of energy?

b) What are the various forms of energy?

c) Two bodies having equal masses are moving with uniform speeds of v and 2v respectively. Find the ratio of their kinetic energies.

Solution 55

(a) Energy is the ability to do work. SI unit of energy is Joule.

(b) Various forms of energy are:

1. Kinetic energy

2. Potential energy

3. Chemical energy

4. Heat energy

5. Light energy

6. Sound energy

7. Electrical energy

8. Nuclear energy

(c) Let masses of bodies be m

Velocity of one body, v_{1}= v

Velocity of another body, v_{2} = 2v

Question 56 a) What do you understand by kinetic energy of a body?

b) A body is thrown vertically upwards. Its velocity goes on decreasing. What happens to its kinetic energy as its velocity becomes zero?

c) A horse and a dog are running with the same speed. If the weight of the horse is ten times that of the dog, what is the ratio of their kinetic energies?

Solution 56

(a) The energy of a body due to its motion is called kinetic energy

(b) When the velocity becomes zero, the kinetic energy also becomes zero since kinetic energy is directly proportional to square of the velocity.

(c)

Question 57 a) 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.

b) What is the difference between kinetic energy and potential energy?

c) A ball of mass 0.5 kg slows down from a speed of 5 m/s to that of 3 m/s. Calculate the change in kinetic energy of the ball. state your answer giving proper units.

Solution 57

(a) The energy of a body due to its position or change in its shape is known as its potential energy. E.g. a stretched rubber has potential energy due to change in its shape and water in the overhead tank has potential energy due to its height above the ground.

PE = m x g x h

where, PE is the potential energy of the body,

m is the mass of the body,

g is the acceleration due to gravity,

h is the height above the surface ofearth.

(b) Kinetic energy of a body is due to motion of the body while potential energy is due to position or change in shape of the body.

Kinetic energy is zero for a still body, while potential energy may or may not be zero for a still body.

Kinetic energy of a body is directly proportional to its speed while potential energy is directly

proportional to the height to which the body is above the ground.(c) Mass of ball, m = 0.5 kg

Speed v_{1} = 5 m/s

Speed v_{2} = 3 m/s

Question 58 a) What is the difference between gravitational potential energy and elastic potential energy? Give one example of body having gravitational potential energy and another having elastic potential energy

b) If 784 J of work was done in lifting a 20 kg mass, calculate the height through which it was lifted? (g = 9.8 m/s^{2})

Solution 58(a) The potential energy due to the position of the body above the ground is gravitational potential energy and the potential energy due to change in shape and size of the body is elastic potential energy. E.g. a stretched rubber has elastic potential energy due to change in its shape while water in the overhead tank has gravitational potential energy due to its height above the ground.

(b) Work done, W = 784 J

Mass, m = 20 kg

g = 9.8 m/s^{2}

W = m x g x h

784 = 20 x 9.8 x h

Question 71 A boy tries to push a truck parked on the roadside. The truck does not move at all. Another boy pushes a bicycle. The bicycle moves through a certain distance. in which case was the work done more: on the truck or on the bicycle? Give a reason to support your answer.

Solution 71 Work done is more on the bicycle because the truck does not move at all and the bicycle moves through a certain distance. And work is said to be done only when applied force produces motion in the body.

Question 72 The work done by force acting obliquely is given by the formula: W = Fcosθ x s. what will happen to the work done if the angle θbetween the direction of fore and motion of the body is increased gradually? Will it increase, decrease or remain constant?

Solution 72 The work done will decrease as the angle between the direction of force and direction of motion is increased gradually because W = Fcosθ x s

And cosθ decreases as the angle θ is increased.

Question 73 What should be the angle between direction of fore and the direction of motion of a body so that the work done is zero?

Solution 73 The work done will be zero when angle between the direction of force and direction of motion is 90^{o} because

W = Fcosθ

and cos90^{o}=0

Question 74 In which of the following cases, is the work done by a force will be maximum: when angle between the direction of force and direction of motion is 0^{o} or 90^{o}?

Solution 74 Work done will be maximum when angle between the direction of force and direction of motion is 0^{o} because

W = Fcosθ

cos0^{o}=1 and cos90^{o}=0

Question 75 How much work is done by the gravitational force of earth acting on a satellite moving around it in a circular path? Give reason for your answer.

Solution 75 The work done is zero because the gravitational force acts along the radius of the circular path, at right angles i.e. 90^{o} to the motion of satellite.

Question 76 A man is instructed to carry a package from the base camp at B to summit A of a hill at a height of 1200 meters. The man weighs 800 N and the package weighs 200 N. (g= 10 m/s^{2})

i) How much work does the man do against gravity?

ii) What is the potential energy of the package at A if it isassumed to be zero at B?

Solution 76

Weight of man = Mg = 800 N

Weight of package = mg = 200 N

Total weight of man and package = Mg + mg = (M+m)g = 1000 N

Height of the summit, h = 1200 m

i) Work done = (M+m) x g x h = 1000 x 1200 = 12 x 10^{5} J

ii) Potential energy of the package = m x g x h = 200 x 1200 = 2.4 x 10^{5} J

Question 77 When a ball is thrown vertically upwards, its velocity goes on decreasing. What happens to its potential energy as its velocity becomes zero?

Solution 77Potential energy becomes maximum.

Question 78 A man X goes to the top of the building by a vertical spiral staircase. Another man Y of the same mass goes to the top of the same building by a slanting ladder. Which of the two man does more work against gravity and why?

Solution 78 The work done by both X and Y are equal because irrespective of whether they reach the top of building by using a spiral or slanted ladder, the vertical distance moved by them against the gravity is same.Question 79 When a ball is thrown inside a moving bus, does its kinetic energy depend on the speed of the bus? Explain.

Solution 79 Yes, the kinetic energy of the ball thrown inside a moving bus depends on the speed of the bus because the speed of the bus adds up to the speed with which the ball is thrown inside the moving bus.

Question 80 A bullet of mass 15 g has a speed of 400 m/s. what is its kinetic energy? If the bullet strikes a thick target and is brought to rest in 2 cm, calculate the average net force acting on the bullet. What happens to the kinetic energy originally in the bullet?

Solution 80

Question 1 Name the commercial unit of measurement of energy.

Solution 1 The commercial unit of energy is kilowatt-hour.

Question 2 Define one kilowatt-hour.

Solution 2 One kilowatt-hour is the amount of electrical energy consumed when an electrical appliance having power of 1 kilowatt is used for 1 hour.

Question 3 Name two units of power bigger than watt.

Solution 3 Megawatt and kilowatt are the units of power bigger than watt.

Question 4 Define the term watt.

Solution 4 1 watt is the power of an appliance which does work at the rate of 1 joule per second.

Question 5 How many watts equal one horse power?

Solution 5 1 horse power = 746 watt

Question 6 Name the physical quantity whose unit is watt.

Solution 6 Power has watt as its physical unit

Question 7 What is the power of a body which is doing work at the rate of one joule per second?

Solution 7 1 watt

Question 8 A body does 1200 J of work in 2 minutes. Calculate its power.

Solution 8 Work done = 1200 J

Time taken = 2 minutes = 2 x 60 = 120 s

Question 9 How many joules are there in one kilowatt-hour?

Solution 9 One kilowatt = 3.6 x 10^{6} J

Question 10 Name the quantity whose unit is

a) Kilowatt

b) Kilowatt-hour

Solution 10

a) Power

b) Electrical Energy

Question 11 What is the common name for 1 kWh of electrical energy?

Solution 11 1 kW-h of electrical energy is commonly known as unit of electricity.

Question 12 A cell converts one form of energy into another form. Name the two forms.

Solution 12 A cell converts chemical energy into electrical energy.

Question 13 Name the device which converts electrical energy into mechanical energy.

Solution 13 Electric motor

Question 14 Name the devices or machines which convert

a) Mechanical energy into electrical energy

b) Chemical energy into electrical energy

c) Electrical energy into heat energy

d) Light energy into electrical energy

e) Electrical energy into light energy

Solution 14

a) Electric generator

b) Cell

c) Electric iron

d) Solar cell

e) Electric bulb

Question 15 Name the devices or machines which convert:

a) Electrical energy into sound energy

b) Heat energy into kinetic (or mechanical energy)

c) Chemical energy into kinetic energy (or mechanical energy)

d) Chemical energy into heat energy

e) Light energy into heat energy

Solution 15

a) Speaker

b) Steam engine

c) Car engine

d) Gas stove

e) Solar water heater

Question 16 Fill in the blanks with suitable words

a) Power is the rate of doing__________

b) 1 watt is a rate of working of one_______ per ________

c) The electricity meter installed in our home measures electric energy in the units of __________

d) The principle of ________ of energy says that energy can be _______ from one form to another, but it cannot be _________ or ________

e) When a ball is thrown upwards,______ energy is transformed into ________ energy

Solution 16

a) Work

b) Joule ; second

c) kWh

d) conservation; transformed; created; destroyed

e) kinetic; potential

Question 17 A trolley is pushed along a road with a force of 400 N through a distance of 60 m in 1 minute. Calculate the power developed

Solution 17 Force, F = 400 N

Distance, s = 60 m

Time taken, t = 1 minute = 60 s

Work done, W = F x s = 400 x 60

Question 18 What kind of energy transformations takes place at a hydroelectric power station?

Solution 18 At a hydroelectric power station, the potential energy of water is transformed into kinetic energy and then into electrical energy.

Question 19 What kind of energy transformations takes place at a coal-based thermal power station?

Solution 19 At a coal-based thermal power station, the chemical energy of coal is transformed into heat energy, whhich is further converted into kinetic energy and electrical energy.

Question 20 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?

Solution 20

Weight of the man = 500 N

Weight of the load = 100 N

Total weight = 600 N

Height of stairs = 4 m

Time taken = 5 s

Work done = mg x h = weight X h = 600 x 4

Question 21 The power output of an engine is 3 kW. How much work does the engine do in 20 s?

Solution 21

Power = 3 kW

Time = 20 s

Work done = power x time = 3 x 20 kWs = 60 kJ

Question 22 An electric heater uses 600 kJ of electrical energy in 5 minutes. Calculate its power rating

Solution 22

Energy consumed = 600 kJ

Time taken = 5 minutes = 300 s

Question 23 How much electrical energy in joules does a 100 watt lamp consume

a)in 1 second?

b)in 1 minute?

Solution 23

Power = 100 W

a)time = 1 s

energy = power x time = 100 J

b)time = 1 minute = 60 s

energy = power x time= 100 x 60 = 6 kJ

Question 24 Five electric fans of 120 W each are used for 4 hours. Calculate the electrical energy consumed in kilowatt-hours.

Solution 24

Power of 1 fan = 120 W

Power of 5 fans = 5 x 120 = 600 W = 0.6 kW

Time = 4 hours

Electrical energy = 0.6 x 4 = 2.4 kWh

Question 25 Describe the energy changes which take place in a radio.

Solution 25 A radio first converts electrical energy into kinetic energy and then into sound energy

Question 26 Write the energy transformations which take place in an electric bulb. (or electric lamp).

Solution 26 In an electric bulb, electrical energy is first converted into heat energy and then into light energy

Question 2 7Name five appliances or machines which use electric motor.

Solution 27 Fan, washing machine, mixer grinder, water pump, hair dryer use electric motor

Question 28 A bulb lights up when connected to a battery. State the energy change which takes place:

i) in the battery

ii) in the bulb

Solution 28

i) chemical energy to electrical energy

ii) electrical energy to heat and light energy

Question 29 The hanging bob of a simple pendulum is displaced to one extreme position B and then released. It swings towards centre position A and then to the other extreme position C. in which position does the bob have:

i) maximum potential energy

ii) maximum kinetic energy

Give reason for your answer.

Solution 29

i) Maximum potential energy is present in the bob at point C as at point C bob is at maximum height.

ii) Maximum kinetic energy is present in the bob at point A as at point A bob is at maximum speed

Question 30 A car of weight 20000 N, climbs up a hill at a steady speed of 8 m/s, gaining a height of 120 m in 100 s, calculate :

a) Work done by the car

b) Power of engine of car.

Solution 30

Weight of the car = 20000 N= 20 kN

Speed = 8 m/s

Distance s= 120 m

Time = 100 s

a) Work done W = f x s = 20 x 120 = 2400 kJ

Question 31 a) What do you understand by the term transformation of energy? Explain with an example

b) Explain the transformation of energy in the following cases:

a. A ball thrown upwards

b. A stone dropped from the roof of a building.

Solution 31

a) The change of one form of energy into another form of energy is known as transformation of energy, e.g. in a cell chemical energy is transformed into electrical energy

b)

a. When a ball is thrown upwards its kinetic energy gradually converts into potential energy and potential energy becomes maximum at the maximum height attained by the ball

b. When a stone is dropped from the roof of the building its potential energy gradually converts into kinetic energy and kinetic energy becomes maximum when the stone is just above the ground

Question 32 a)State and explain the law of conservation of energy with an example.

b)Explain how, the total energy of the swinging pendulum at any instant of time remains conserved. Illustrate your answer with the help of alabeled diagram.

Solution 32 a)Law of conservation of energy states that whenever energy changes from one form to another form, the total amount of energy remains constant. Energy can never be created nor destroyed, it transforms from one form to another. For example, when electrical energy is converted into light energy in an electric bulb, then some energy is wasted as heat during conversion but the total energy remains the same.

b)Initially the pendulum is at rest. The bob is pulled to one side to position B to give it potential energy due to higher position of B and then released, the bob starts swinging

i.When the bob is at position B, it has potential energy but no kinetic energy

ii.As the bob starts moving down from position B to A, its potential energy starts decreasing and kinetic energy starts increasing

iii.When the bob is at position A, it has maximum kinetic energy and zero potential energy

iv.As the bob starts moving down from position A to C, its kinetic energy starts decreasing and potential energy starts increasing

v.On reaching the extreme position C, the bob stops for a very small instant of time and bob maximum potential energy and zero kinetic energy.

Therefore at extreme positions B and C bob has only potential energy and at A it has only kinetic energy and at other intermediate positions bob has both kinetic and potential energy. Thus the total energy of the pendulum is same for any instant of time (conserved)

Question 33 a) What is the meaning of symbol kWh? What quantity does it represent?

b) How much electrical energy in kWh is consumed by an electrical appliance of 1000 W when it is switched on 60 minutes?

Solution 33

a) The unit kWh stands for kilowatt-hour. One kilowatt-hour is the amount of electrical energy consumed when an electrical appliance having power of 1 kilowatt is used for 1 hour. It represents the amount of electrical energy consumed in 1 hour.

b) Power = 1000 W = 1 kW

Time = 60 minutes = 1 hour

Energy(kWh) = 1 x 1 = 1 kWh

Question 34

a) Derive the relation between commercial unit of energy (kWh) and SI unit of energy (joule).

b) A certain household consumes 650 units of electricity in a month. How much is this electricity in joule?

Solution 34

a) 1 kilowatt-hour = 1 kW for 1 hour

= 1000 W for 1 hour

Question 35 a) Define power. Give the SI unit of power.

b) A boy weighing 40 kg carries a box weighing 20 kg to the top of a building 15 m high in 25 s. Calculate the power(g = 10 m/s^{2}).

Solution 35

a) Power is the rate of doing work.

Question 46 The following data was obtained for a body of mass 1 kg dropped from a height of 5 meters:

Show by calculations that the above data verifies the law of conservation of energy (Neglect air resistance) (g = 10 m/s^{2})

Solution 46

Total energy = PE + KE = 50 J

The total energy in all three cases is constant. This proves the law of conservation of energy

Question 47 A ball falls to the ground as shown below

A Potential energy= 80J & kinetic energy=0

B Kinetic energy=48 J

C Potential energy=0

(a)what is the kinetic energy of ball when it hits the ground?

(b)what is the Potential energy of ball at B?

(c)which law you have made use of in answering this question?

Solution 47

Since from point A total energy = PE + KE = 80 J

And according to the law of conservation of energy Total energy remains constant

a) PE = 0

Total energy = PE + KE = 80 J

KE = 80 J PE = 80 J

b) At point B KE = 48 J

Total energy = PE + KE = 80 J

PE = 80 J KE = 80 – 48 = 32 J

c) Law of conservation of energy

Question 48 If an experiment to measure his power, a student records the time taken by him in running up a flight of steps on a staircase. Use the following data to calculate the power of the student:

Number of steps = 28

Height of each step = 20 cm

Time taken = 5.4 s

Mass of student = 55 kg

Acceleration = 9.8 m/s^{2} due to gravity

Solution 48

No. of steps = 28

Height of each step = 28 cm

Total height = 20 x 28 = 560 cm = 5.6 m

Mass of student = 55 kg

g = 9.8 m/s^{2}

time = 5.4 s

Work done = m x g x h = 55 x 9.8 x 5.6 = 3018.4 J

Question 49 In loading a truck, man lifts boxes of 100 N each through a height of 1.5 m

a) How much work does he do in lifting the box?

b) How much energy is transferred when one box is lifted?

c) If the man lifts 4 boxes per minute, at what rate is he working?(g=10 ms-2)

Solution 49

Weight of box = 100 N

Height = 1.5 m

i. work done = m x g x h = 100 x 1.5 = 150 J

ii. potential energy = m x g x h= 100 x 1.5 = 150 J

iii. weight of 4 boxes = 400 N

time = 1 minute = 60 s

work done = 400 x 1.5 = 600 J

Question 50 Name the energy transfers which occur when:

a) an electrical bell rings

b) someone speaks into a microphone

c) there is a picture on a television screen

d) a torch is on

Solution 50

a) Electrical energy to sound energy

b) Sound energy to electrical energy

c) Electrical energy to light (and heat) energy

d) Chemical energy to electrical energy to light energy(and heat energy)

Question 65:A car is travelling at 20 m/s along a road. A child runs out into the road 50 m ahead and the car driver steps on the brake pedal. What must the car’s deceleration be if the car is to stop just before it reaches the child ?

Solution :

Initial velocity, u=20m/s

Final velocity, v=0m/s

Distance, s=50m