Selina Concise Class 10 Physics Solution Chapter No. 2- ‘Work, Energy and Power’ For ICSE Board Students.
Selina Concise Class 10 Physics Chapter 2 Work, Energy and Power Exercise All Questions and Answers by Physics Teacher here in this post.
Exercise 2 (A)
Question: 1
Define work. When is work said to be done by a force?
Solutions: Work is used in a very specific ways in physics. When applied force causes displacement then it is said that work is done by the force.
Question: 2
How is the work done by a force measured when
(i) Force is in direction of displacement,
(ii) Force is at an angle to the direction of displacement?
Solutions:
(i) Work done by force when force is in direction of displacement is product of magnitude of force and displacement caused by that force.
(ii) Work done by force when force is at angle to the direction of displacement is product of magnitude of force and component of displacement in the direction of force(total displacement * cos θ, where θ angle between force and displacement).
Question: 3
A force F acts on a body and displaces it by a distance S in a direction at an angle θ with the direction of force.
(a) Write the expression for the work done by the force.
(b) What should be the angle between force and displacement so that the work done is
(i) Zero,
(ii) Maximum?
Solutions:
(a) Expression for work done in certain angle θ by a force lets call F can be calculated as follows
Let’s consider a force F act along the straight line in a fixed direction XY as shown in the figure and causes displacement in an angle θ. Total displacement is XZ(d) then to find the component of displacement in the direction of force we can draw a perpendicular along the
YZ which make ∆ XYZ a right angle triangle
Hence, the component along direction of force is XY
Work done W = F * XY
As ∆ XYZ is a right angle triangle
Using trigonometric functions
COS θ = base / hypotenuse
COSθ = XY / XZ
XY = COSθ * d ……. (XZ = d)
Therefore,
W = F * d * COS θ
(b) The value of work done is directly proportional to the value of cos θ
Hence, when θ = 90 the value of cos θ is zero hence value of work done also becomes zero
When θ = 0 the value of cos θ is 1 hence value of work done becomes maximum.
Question: 4
A body is acted upon by a force. State two conditions when the work done is zero.
Solutions: as we know the formula for work done is
W = F * d * COS θ
Where F = force applied
d = displacement
COS θ = angle between force and displacement
When the force is applied to have zero work done either displacement must be zero or COSθ (θ = 90) must be zero. These are the two conditions for zero work done for given applied force.
Question: 5
State the condition when the work done by a force is (i) positive, (ii) negative. Explain with the help of examples.
Solutions: (i) for work done to positive θ must be zero i.e. the force and displacement must be same direction. For example when a ball falls to the ground the force acting on the ball in this case gravity is in downward direction and the displacement of the ball is in downward direction hence θ = 0 hence the work done is positive ( COSθ = 1)
(ii) for work done to be negative θ must be 180 i.e. the force and displacement must be in opposite direction. For example, when a body moves on a surface the force of friction between surface and the body is opposite to the displacement hence θ = 180. Therefore, the work done is negative (COS θ = -1).
Question: 6
A body is moved in a direction opposite to the direction of force acting on it. State, whether the work is done by the force or work, is done against the force.
Solutions: As the body moves opposite to the direction of the force value of θ became 180 hence COS θ = -1. Hence, work done is against the force.
Question: 7
When a body moves in a circular path, how much work is done by the body? Give reason.
Solutions: When a body moves in a circular path, work done is zero as the centripetal force on the body acts towards the centre of the circular path and the displacement at any instant is tangential to the circular path. That makes θ = 90 cosθ = 0 hence the work done is zero.
Question: 8
A satellite revolves around the earth in a circular orbit. What is the work done by the satellite? Give reason.
Solutions: when a satellite revolves around the earth in a circular orbit work is zero. When a satellite follows a circular path around the earth the centripetal force is exerted by gravitational force between earth and the satellite which is directed towards the centre of circular motion i.e. towards earth but the displacement of satellite at any given instant is tangential to be the circular path hence θ = 90 and COS θ = 0 hence work done by satellite is zero.
Question: 9
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.
Solutions:
(a) No
(b) No
(c)Yes
Question: 10
A coolie X Carrying a load on his head climbs up a slope and another coolie Y carrying the identical load on his head move the same distance on a frictionless horizontal platform. Who does more work? Explain the reason.
Solutions: Coolie X does more work as he climbing on a slope as a certain angle so forces acting are gravity and friction while coolie Y is moving 90 degree to the force of gravity on a frictionless surface.
Question: 11
The work done by a fielder when he takes a catch in a cricket match, is negative. Explain.
Solutions: To catch a ball fielder has to stop the ball from its direction of displacement as the fielder is applying force opposite to the direction of displacement of the ball the work done is negative.
Question: 12
Give an example when work done by the force of gravity acting on a body is zero even though the body gets displaced from its initial position.
Solutions: when a body is moving to horizontal to the surface of earth the gravity force and displacement are in right angle to each other. The angle θ = 90 which makes COS θ = 0 hence work done is zero even though the body is moved from the initial position.
Question: 13
What are the S.I. and C.G.S. units of work? How are they related? Establish the relationship.
Solutions: S.I. unit of work is joule (J)
C.G.S. unit of work is erg.
Relation between joule and erg.
1 joule = 1N * 1m
But 1N = 105 dyne and 1m = 102 cm
1 joule = 105 dyne * 102 cm = 107 erg.
1 joule = 107 erg.
Question: 14
State and define the S.I. unit of work.
Solutions: S.I. unit of work is joule.
When a force 1 newton force displaces a body through distance of 1m in its own direction 1 joule work is done.
Question: 15
Express joule in terms of erg.
Solutions: 1 joule = 1N * 1m
But 1N = 105 dyne and 1m = 102 cm
1 joule = 105 dyne * 102 cm = 107 erg.
1 joule = 107 erg.
Question: 16
A body of mass m falls through a height h. Obtain an expression for the work done by the force of gravity.
Solutions: Force acting = mass * gravity = mg
Displacement = h
Work done = force applied * displacement
= mg * h
Question: 17
A boy of mass m climbs up a 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?
Solutions: (a) When a boy climbs stairs at certain angle θ then the work done by gravity g
Therefore, force = mass * gravity
Displacement component = h * COS θ
Work done = force * displacement component
= mg * h * COS θ
(b) When boy uses lift to travel same distance force and displacement are opposite to each other. Hence, θ = 180, COS θ = -1
Work done = force * displacement component
= mg * h * COS θ
= – mg * h
Question: 18
Define the term energy and state its S.I unit.
Solutions: Energy contained by body is measured by the amount of work that the body can perform. It is the capacity of the body to perform work. S.I. unit of energy is joule.
Question: 19
What physical quantity does electron volt (eV) measure? How is it related to the S.I. unit of that quantity?
Solutions: The energy related to atomic particles is very small, hence it is measured in electron volt (eV).
1 eV is the energy gained by electron when it is accelerated through potential difference of 1 volt.
1eV = charge on electron * 1 volt
= 1.6 * 10-19 coulomb * 1 volt
= 1.6 *10-19 joule
Question: 20
Complete the following sentence:
(a) 1 J = _____ calorie.
(b) 1 kWh = ______ J.
Solutions: (a) 0.24 calorie
(b) 3.6 * 106 J
Question: 21
Name the physical quantity which is measured in calorie. How is it related to the S.I unit of that quantity?
Solutions: Heat energy is a physical quantity which is measured in calorie. It is related to joule as
1 J = 0.24 calorie or 1 calorie = 4.18 J
Question: 22
Define a kilowatt-hour. How is it related to joule?
Solutions: When energy spend by a source of power 1 kilowatt for an hour it is defined as kilowatt-hour.
1 kilowatt-hour = 3.6 * 10 6 J
Question: 23
Define the term power. State its S.I. unit.
Solutions: the rate of work done is defined as power.
S.I. unit of power is watt
1 watt = 1 joule/ 1 second = 1 Js-1
Question: 24
State two factors on which power spent by a source depends. Explain your answer with examples.
Solutions: According to the formula of power the power spend depends on two factors
One power spend is directly proportional to the amount of work done and two power spend is inversely proportional to the amount of time spend. For example consider a machine doing same work in 5 mins another machine does the same work in 10 mins. Here the work done by both machine is same but as time is doubled the first machine uses double the power of second machine.
Question: 25
Differentiate between work and power.
Solutions:
Work |
Power |
Work done by a force is product of force applied and displacement component caused by the force. | Power is the rate of doing a work. |
Work done is independent of time. | Power is inversely proportional to time. |
S.I. unit of work is joule. | S.I. unit power is watt (joule per second). |
Question: 26
Differentiate between energy and power.
Solutions:
Energy |
Power |
Energy is the capacity of body to do the work. | Power is the rate of doing a work. |
Energy spend is independent of time. | Power is inversely proportional to time. |
S.I. unit of energy is joule. | S.I. unit power is watt (joule per second). |
Question: 27
State and define the S.I unit of power.
Solutions: S.I. unit of power is watt. Work done by a body at the rate of 1joule per second is called 1 watt power
Question: 28
(a) Name the physical quantity measured in terms of horsepower.
(b) How is horsepower related to the S. I. unit of power?
Solutions: (a) Power is measured in terms of horse power.
(b) 1 H.P. = 746 watt= 0.746 kilowatt.
Question: 29
Differentiate between watt and watt-hour.
Solutions:
Watt |
Watt-hour |
It is the unit of power | It is the unit of energy |
It measures the power taken by a device to function | It measures energy consumed by a body over certain period of time. |
Question: 30
Name the quantity which is measured in
(a) kWh
(b) kW
(c) Wh
(d) eV
Solutions:
(a) Energy
(b) Power
(c) Energy
(d) Energy
Question: 31
Is it possible that no transfer of energy takes place even when a force is applied to a body?
Solutions: Yes, when the displacement of body is perpendicular to be force applied which makes θ = 90 i.e. COS θ = 0 hence work done is zero.
Multiple Choice Type:
Question: 1
One horsepower is equal to:
a.) 1000 W
b.) 500 W
c.) 764 W
d.) 746 W
Solutions: (d) 746 W
Question: 2
kWh is the unit of:
a.) Power
b.) Force
c.) Energy
d.) None of these
Solutions: (c) energy
Numerical
Question: 1
A body, when acted upon by a force of 10 kgf, gets displaced by 0.5 m. Calculate the work done by the force, when the displacement is
(i) in the direction of force,
(ii) at an angle of 60owith the force, and
(iii) normal to the force. (g = 10Nkg-1)
Solutions:
(i) Work done when the displacement is in the direction of force
Given, Force = 10 N
Displacement = 0.5 m
θ = 0, COS θ = 1
work done = force * displacement component
= 10 N * 0.5m * 1
= 50 J
(ii) Work done when displacement is angle 60o
Given, θ = 60o, COS 60o = ½
work done = force * displacement component
= 10 N * 0.5 m * ½
= 25 J
(iii) Work done when displacement is normal to the force
Given
θ = 90o, COS 90o = 0
Hence, work done is zero.
Question: 2
A boy of mass 40kg climbs up the stairs and reaches the roof at a height 8m in 5 s. Calculate:
(i) The force of gravity acting on the boy,
(ii)The work done by him against gravity,
(iii)The power spent by the boy.
(Take g= 10ms-2)
Solutions:
(i) The force of gravity acting on the boy.
Given, Mass = 40kg, g = 10 ms-2
Force of gravity = mass * g
= 40 kg * 10 ms-2
= 400 N
(ii) Work done = mass * g * height
= 400 N * 8m
= 3200 J
(iii) Power = work done / time
= 3200J/ 5s
= 640 w
Question: 3
A man spends 6.4 kJ energy in displacing a body by 64 m in the direction in which he applies force, in 2.5 s. Calculate: (i) the force applied and (ii) the power spent (in H.P) by the man.
Solutions:
(i) Given
Work done = 6.4 kJ = 6.4 *103 J
Displacement = 64 m
Work done = force applied * displacement
6.4 *103 J = F * 64 m
F = 6.4 *103 J/ 64
F = 100 N
Force applied is 100 N
(ii) Power = work done / time
Power = 6.4 *103 J/ 2.5 s
Power = 2560 w
But 1 H.P. = 746 w
Therefore
Power in H.P. = 2560 / 746
Power = 3.43 H.P.
Question: 4
A weight lifter lifted a load of 200 kgf to a height of 2.5 m in 5 s. Calculate: (i) the work is done, and (ii) the power developed by him. Take g =10N/kg-1.
Solutions:
(i) Given
Force = mass * gravity = 200 * 10 = 2000N
Displacement = 2.5m
Work done = force * displacement
= 2000 N * 2.5m
= 5000 J
(ii) Power = work done / time
= 5000 J / 5s
= 1000 w
Question: 5
A machine raises a load of 750N through a height of 16m in 5 s. Calculate:
(i) the energy spent by the machine.
(ii) the power of the machine if it is 100% efficient.
Solutions:
(i) Given
Force = 750 N
Displacement = 16m
Work done = force * displacement
= 750 N * 16 m
= 12000 J
(ii) Power = work done / time
= 12000 J / 5s
= 2400 w
Question: 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.
Solutions:
(i) Given
Power = 3kW = 3000 W
Time = 10 hours = 36000s
Energy = power * time
= 3 kW * 10 hours = 30 kWh
Energy = 3000 J * 36000s = 1.08 * 108 J
Question: 7
A water pump raises 50 litres of water through a height of 25m in 5 s. Calculate the power of the pump required.
(Take g= 10N kg-1 and density of water =1000kg m-3).
Solutions: Given
Mass of water = volume of water * density of water
= 50 litre * 1000 kgm-3
= 50000 g= 50 kg
Displacement = 25 m
Time = 5s
Gravity = 10 Nkg-1
Work done = mass * gravity * displacement
= 50 * 10 * 25
= 12500 J
Power = work done / time
= 12500/ 5
= 2500 W
Question: 8
A pump is used to lift 500kg of water from a depth of 80m in 10s. 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= 10m s-2).
Solutions: Given
Mass of water = 500kg
Displacement = 80 m
Time = 10s
Gravity = 10 ms-2
(a) Work done = mass * gravity * displacement
= 500 * 10 * 80
= 4 * 105 J
(b) Power = work done / time
= 400000 / 10
= 4 * 104 W
(c) Efficiency of pump = 40% = 0.4
Power rating = power / efficiency
= 4 * 104 /0.4
= 1 * 105 w = 100 kW
Question: 9
An ox can apply a maximum force of 1000N. It is taking part in a cart race and is able to pull the cart at a constant speed of 30m/s-1 while making its best effort. Calculate the power developed by the ox.
Solutions: Given
Velocity (rate of change of displacement) = 30 m per s
Force = 1000 N
Power = work done / time
Work done = force * displacement
Power = force * displacement / time
Power = force * velocity ……. (velocity = displacement/time)
= 1000 * 30
= 30000 W = 30 kW
Question: 10
The power of a motor is 40kW. At what speed can the motor raise a load of 20,000 N?
Solutions: Given
Power = 40 kW = 40000 W
Force = 20000 N
As we know,
Power = force * velocity
Velocity = Power / force
= 40000 / 20000
= 2 ms-1
Question: 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.
Solutions: Given
Force = 150 N
Velocity = 10 ms-1
As we know,
Power = force * velocity
= 150 * 10
= 1500 W
Question: 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.
Solutions: Given
Force = 350 N
Displacement = 30 * 20 cm = 600 cm = 6 m
Time = 1 min = 60 s
Work done = force * displacement
= 350 N * 6 m
= 2100 J
Power spent = work done / time
= 2100 / 60
= 35 W
Question: 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.
Solutions:
If the mass of persons are same and they travel the same distance then the ratio of work done will be one as the work done will be same
Now for power as work done is same and power is inversely proportional to time
Let us consider power used by person A Pa and power used by person B be Pb .
Hence,
Pa/Pb = 15/20
Pa/Pb = 3/4
Question: 14
A boy of weight 40 kgf climbs up the 30 steps, each 20 cm high in 4 minutes and a girl of weight 30 kgf does the same in 3 minutes. Compare:
(i) the work done, and
(ii) the power developed by them. Take g = 10 N kg-1.
Solutions: Given
Mass of boy = 40 kg
Displacement = 30 * 20 cm = 600 cm = 6 m
Time = 4 minutes = 240 s
Work done by boy = mass * displacement * gravity
= 40 * 6 * 10
= 2400 J
Power spent by boy = work done / time
= 2400 / 240
=10 W
Mass of girl = 30 kg
Time = 3 minutes = 180s
Work done by girl = 30 * 6 * 10
= 1800 J
Power spent = work done / time
= 1800/ 180s
= 10 W
Comparing work done = 2400 J / 1800 J
= 4/3
Comparing power = 10 W / 10 W
= 1/1
Question: 15
A man raises a box of mass 50kg to a height of 2m in 20s, while another man raises the same box to the same height in 50s.
(a) Compare: (i) the work done, and (ii) the power developed by them.
(b) Calculated: (i) the work done, and (ii) the power developed by each man. Take g = 10N kg-1.
Solutions: Given
Force = mass * gravity = 50 * 10 = 500 N
Displacement = 2m
Work done = force * displacement
= 500 * 2
= 1000 J
As for both men the displacement and the force is same the ratio of work done will be 1/1
For power
Power spent by man 1 = 1000 / 20
= 50 W
Power spent by man 2 = 1000 / 50
= 20 W
Comparing power = 50 W / 20 W
= 5/2
Question: 16
A boy takes 3 minutes to lift a 20litre 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.
Solutions: Given
Mass of water = volume * density = 20 * 103= 20000 g = 20 kg
Displacement = 20m
Work done by both = force 8 displacement
= 20 * 9.8 * 20
= 3920 J = 3.92 kJ
As the displacement and force is same ratio of work done is 1/1
Now, for power we know that
Power spent by boy / power spent by father = time taken by father / time taken by boy
Power spent by boy / power spent by father = 120 / 180
= 2/3
Exercise 2 (B)
Question: 1
What are the two forms of mechanical energy?
Solutions: Two forms of mechanical energy are potential energy and kinetic energy.
Question: 2
Name the form of energy which a wound-up watch spring possesses.
Solutions: A wound up watch spring possesses potential energy.
Question: 3
Name the type of energy (kinetic energy K or potential energy U) possessed in the following cases:
(a) A moving cricket ball
(b) A compressed spring
(c) A moving bus
(d) A stretched wire
(e) An arrow shot out of a bow.
(f) A piece of stone placed on the roof.
Solutions:
(a) Kinetic energy
(b) Potential energy
(c) Kinetic energy
(d) Potential energy
(e) Kinetic energy
(f) Potential energy
Question: 4
Define the term potential energy of a body. Name its two forms and give one example of each.
Solutions: The energy of body at position of rest due its position, shape or size is called as potential energy. For example, compressed spring, stretched rubber band etc.
Question: 5
Name the form of energy which a body may possess even when it is not in motion. Give an example to support your answer.
Solutions: A body may possess potential energy even when it is not in motion. For example, compressed spring, stretched rubber band etc.
Question: 6
What is meant by gravitational potential energy? Derive an expression for it for a body placed at a height above the ground.
Solutions: The potential energy contained in a body due to the force of attraction of earth on it, is called its gravitational potential energy.
Expression for a body placed at a height above the ground. Let a body of mass m be placed at a height h from the ground. The least force required to lift the body is equal to the force of gravity. Hence, F = mg
The work done W on the body in lifting it to a height h is
W = force of gravity(mg) * displacement (h)
= mgh
Therefore, gravitational potential energy U = mgh
Question: 7
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.
Solutions: Expression for a body placed at a height above the ground. Let a body of mass m be placed at a height h from the ground. The least force required to lift the body is equal to the force of gravity. Hence, F = mg
The work done W on the body in lifting it to a height h is
W = force of gravity(mg) * displacement (h)
= mgh
Therefore, gravitational potential energy U = mgh
We have to assume that initial gravitational potential energy on the ground is zero then the gravitational potential energy at height h Is mgh.
Question: 8
What do you understand by the kinetic energy of a body?
Solutions: Kinetic energy is the energy possessed by a body due to its motion.
Question: 9
(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.
Solutions:
(a) Suppose a body of mass m is moving with a velocity v. To bring the body to rest we have to apply constant force let that force be F. Let the uniform deacceleration produced by opposing force be a and the body travels distance d before coming to rest.
Then, Deaccelerating force = mass * retardation
F = m * a
Kinetic energy of body = work done by the force F * displacement d
K = F * d
Lets calculate the displacement
Initial velocity (v1) = v
Final velocity (v2) = 0
Deacceleration = -a
As we know, (v2)2 = (v1)2 + 2ad
Substituting,
0 = v2 – 2ad
d = v2 / 2a
therefore, kinetic energy = F * d
K = ma * v2 / 2a
K = ½ mv2
Hence, kinetic energy = ½ * mass * (velocity)2
(b) Based on above formula we derived
Kinetic energy = ½ * mass * (velocity)2
K = ½ * m * v2
Therefore,
Mass = 2K / v2
Question: 10
State the work – energy theorem.
Solutions: The increase in the kinetic energy of a moving body is equal to the work done by a force acting in the direction of the moving body.
Question: 11
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?
Solutions: Let a body of mass m is moving with a uniform velocity u. A force is applied on the body which produces acceleration a due to which its velocity increases from u to v moving up to distance d.
Then, force F = mass * acceleration
F = ma
Work done by the force = force * displacement
W = F * d
As we know, v2= u2 + 2ad
d = (v2 – u2)/ 2a
from above equations,
W = ma * (v2 – u2)/ 2a
W = ½ mv2 – ½ mu2
But initial kinetic energy, K1= ½ mv2
Final kinetic energy, K2 = ½ mu2
Therefore, W = K1 -K2
Hence, work done on the body is increase in kinetic energy,
Question: 12
A light mass and a heavy mass have equal momentum. Which will have more kinetic energy?
Solutions: The relation between momentum and kinetic energy is given by
K = p2/ 2m
Therefore, the mass of the body is inversely proportional to the mass of body if the momentum is constant. Hence, the kinetic energy of light mass is more than the kinetic energy of heavy mass.
Question: 13
Two bodies A and B of masses m and M (M≫ m) have same kinetic energy. Which body will have more momentum?
Solutions: The relation between momentum and kinetic energy is given by
K = p2/ 2m, p = √(2mk)
Therefore, the root mass of the body is directly proportional to the mass of body. If the kinetic energy is constant. Hence, the momentum of heavy mass is more than the momentum of light mass.
Question: 14
Name the three forms of kinetic energy and give one example of each.
Solutions: Three forms of kinetic energy are
(i) Translational kinetic energy. Example of translational kinetic energy is car moving on straight road.
(ii) Rotational kinetic energy. Example of rotational kinetic energy is earth rotating around its own axis.
(iii) Vibrational kinetic energy. Example of vibrational kinetic energy is in a solid, atoms vibrate from there mean position.
Question: 15
State two differences between the potential energy and the kinetic energy.
Solutions:
Potential energy |
Kinetic energy |
The energy possessed by the body at its position of rest is called as potential energy. | The energy possessed by a body due to its motion is called as kinetic energy. |
Potential energy is work done to bring the body to changed state. | Kinetic energy is the work that body can do till it comes to rest. |
Question: 16
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 _______.
Solutions:
(a) Motion
(b) Position
Question: 17
When an arrow is shot from a bow, it has kinetic energy in it. Explain briefly from where does it get its kinetic energy?
Solutions: To shot the arrow we have to pull the string. When the string is pulled the bow contains the potential energy due to change in the shape of bow. That potential energy is turned into kinetic energy to shot the arrow.
Question: 18
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.
Solutions: The energy possessed by compressed spring is potential energy. When the spring is released the ball flies away as the potential energy of the contained in spring gets transferred to the ball as a form of kinetic energy.
Question: 19
A pebble is thrown up. It goes to a height and then comes back on the ground. State the different changes in the form of energy during its motion.
Solutions: The pebble goes with its highest kinetic energy at the start when its released. As the pebble goes up the kinetic energy decreases building up the potential energy of the pebble. The pebble travels upward till the kinetic energy reaches zero and potential energy reaches maximum. After that pebble starts to travel downwards due to force of gravity.
Question: 20
In what way does the temperature of water at the bottom of a waterfall differ from the temperature at the top? Explain the reason.
Solutions: The water at top of waterfall possesses potential energy due to its position. When the water travels downwards then that potential energy is converted into kinetic energy and then into heat. Hence water at the bottom is warmer that water at top.
Question: 21
Name the form of energy in which potential energy can change.
Solutions: Potential energy can change into kinetic energy.
Question: 22
Name the form of mechanical energy, which is put to use.
Solutions: kinetic energy is the form of mechanical energy, which is put to use.
Question: 23
Name six different forms of energy?
Solutions: Solar energy, heat energy, light energy, chemical energy, hydro energy, nuclear energy etc.
Question: 24
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.
Solutions:
(a) The potential energy of spring converted kinetic energy due to unwinding.
(b) Chemical energy is converted into kinetic energy.
(c) Kinetic energy into potential energy.
(d) Light energy into chemical energy.
(e) Electrical energy into chemical energy.
(f) Chemical energy into heat energy.
(g) Chemical energy into heat energy.
(h) Chemical energy into heat, light and sound energy.
Question: 25
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) Biogas 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.
Solutions:
(a) Electrical energy into sound energy
(b) Heat energy into mechanical energy
(c) Sound energy into electrical energy
(d) Electrical energy to mechanical energy
(e) Electrical energy into light energy
(f) Chemical energy to heat energy
(g) Light energy into electrical energy
(h) Chemical energy into heat energy
(i) Chemical energy into electrical energy
(j) Chemical energy to mechanical energy
(k) Electrical energy into heat energy
(I) Light energy into electrical energy
Question: 26
Name the process used for producing electricity from nuclear energy.
Solutions: The process used to produce electricity using nuclear energy is nuclear fission. The energy produced by nuclear fission is used to produce steam which then used to turn turbines to produce electricity.
Question: 27
Is it practically possible to convert a form of energy completely into the other useful form? Explain your answer.
Solutions: No. as the process of conversion is not ideal so some form of the energy is converted into some unwanted form of energy.
Question: 28
What is degraded energy?
Solutions: It is the thermal energy released into surrounding after any process.
Question: 29
What do you mean by degradation of energy? Explain it by taking one example of your daily life.
Solutions: The energy released into surrounding as useless form which reduces the efficiency of the process. When a engine runs on fuel the chemical energy of fuel is not just converted into mechanical energy but sound as well as heat energy which is unwanted.
Question: 30
Complete the following sentence:
The conversion of part of energy into an undesirable form is called ……
Solutions: Degradation of energy.
Multiple Choice Type:
Question: 1
A body at a height possesses:
a.) Kinetic energy
b.) Potential energy
c.) Solar energy
d.) Heat energy
Solutions: (b) Potential energy.
Question: 2
In an electric 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
Solutions: (d) Chemical to electrical
NUMERICAL
Question: 1
Two bodies of equal masses are placed at heights h and 2h. Find the ratio of their gravitational potential energies.
Solutions: The gravitational potential energy= mgh
As m and g are constant
Ratio of gravitational potential energy = ½
Question: 2
Find the gravitational potential energy of 1kg mass kept at a height of 5m above the ground if g =10ms-2.
Solutions: Given
Mass = 1 kg
H = 5m
Gravitational potential energy = mgh
= 1 * 10 * 5
= 50 J
Question: 3
A box of weight 150 kg has gravitational potential energy stored in it equal to 14700 J. Find the height of the box above the ground.
Solutions: Given
Mass = 150 kg
Energy = 14700 J
Gravity = 9.8 ms-2
Gravitational potential energy = mgh
14700 = 150 * 9.8 * h
14700/ 1470 = h
10 = h
Hence, the box is kept at 10 m height.
Question: 4
A body of mass 5 kg falls from a height of 10 m to 4 m. Calculate:
(i) The loss in potential energy of the body,
(ii) The total energy possessed by the body at any instant? (Take g = 10 ms-2).
Solutions: Given
Mass = 5 kg
Initial height = 10 m
Final height = 4 m
Gravity = 10 ms-2
Loss in potential energy = mass * g * difference in height
= 5 * 10 * (10-4)
= 50 * 6
= 300 J
Total energy possessed by the body at any instant is equal to initial potential energy of body
Total energy = mgh
= 5 * 10 * 10
= 500 J
Question: 5
Calculate the height through which a body of mass 0.5 kg is lifted if the energy spent in doing so is 1.0 J. Take g = 10m/s-2.
Solutions: Given
Potential energy = 1 J
Mass = 0.5 kg
Potential energy = mgh
1 = 0.5 * 10 * h
1 / 5 = h
h = 0.2m
body was raised to 0.2 m.
Question: 6
A boy weighing 25 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).
Solutions: Given
Mass = 25
Initial height = 3m
Final height = 9m
Potential energy increase = mass * g * Difference in height
= 25 * 10 * (9-3)
= 250 * 6
= 1500 J
Question: 7
A vessel containing 50 kg of water is placed at a height 15m above the ground. Assuming the gravitational potential energy at the ground to be zero, what will be the gravitational potential energy of water in the vessel? (g = 10ms-2).
Solutions: Given
Mass = 50 kg
h = 15m
gravitational potential energy = mgh
= 50 * 10 * 15
= 7500 J
Question: 8
A man of mass 50 kg climbs up a ladder of height 10m. Calculate: (i) the work done by the man, (ii) the increase in his potential energy.
(g= 9.8m s-2).
Solutions: Given
(i) Mass = 50 kg
Displacement = 10 m
Force = mg = 50 * 9.8 = 490 N
Work done = F * displacement = 490 N * 10 = 4900 J
(ii) Increase in his potential energy = mgh
= 50 * 9.8 * 10 = 4900 J
Question: 9
A block A, whose weight is 100N, is pulled up a slope of length 5m by means of a constant force F (=100N) as illustrated.
(a) What is the work done by the force F in moving the block A, 5m 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.
Solutions: Given
Force = mg = 100 N
(a) Work by the displacement of 5 m
Work done = force * displacement
Work done = 100 * 5 = 500 J
(b)Increase in the potential energy
= mgh
= 100 * 3
= 300 j
(c) Difference in the work done by force and increase in potential
= work done – increase in potential energy
= 500 – 300
= 200 J
This energy is used against friction to move the block.
Question: 10
Find the kinetic energy of a body of mass 1kg moving with a uniform velocity of 10m s-1.
Solutions: Given
Mass = 1 kg
Velocity = 10 ms-1
Kinetic energy = ½ m v2
= ½ * 1 * (10)2
= 100/2
= 50 J
Questions: 11
If the speed of a car is halved, how does its kinetic energy change?
Solutions: The kinetic energy of body is directly proportional to the square of the velocity of body. Hence, if the speed of car is halved then the kinetic energy will be reduced to its ¼ value.
Question: 12
Two bodies of equal masses are moving with uniform velocities v and 2v. Find the ratio of their kinetic energies.
Solutions: Given
Mass is constant
Let the kinetic energy of body with velocity v be k1 and the kinetic energy of body moving with velocity 2v be k2
Hence,
k1/ k2= v2 / (2v)2
k1/ k2= v2 / 4v2
k1/ k2= 1/ 4
hence the ratio of kinetic energy is ¼.
Question: 13
Two bodies have masses in the ratio 5:1 and kinetic energy in the ratio 125:9. Calculate the ration of their velocities.
Solutions: Given, m1/m2 = 5/1
K1/K2 = 125/9
From formula of kinetic energy
K1/K2 = (m1/m2) * (v12/ v22)
K1/ K2= v2 / (2v)2
(v1/v2)2= K1*m2 / K2 * m1
= 125 * 1/ 9 * 5
= 25 / 9
Hence,
v1/ v2 = 5 /3
ratio of velocities is 5:3.
Question: 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.
Solutions: Given
Masses are constant
Therefore, kinetic energies are directly proportional to the square of velocities.
Let K1 be the kinetic energy for vehicle at speed v1 = 15 km per h, and k2 be the kinetic energy of vehicle with speed v2 = 45 km per h.
K1/K2 = v12/ v22
K1 / K2 = 15* 15 / 45 * 45
K1 / K2 = 1 / 9
Hence, ratio of kinetic energy is 1:9.
Question: 15
A ball of mass 0.5 kg slows down from a speed of 5m/s-1 to that of 3m/s-1. Calculate the change in kinetic energy of the ball.
Solutions: Given
Mass = 0.5 kg
Initial velocity v1= 5 m per s
Final velocity v2 = 3 m per s
Kinetic energy at v1 = ½ m * v12
Kinetic energy at v2 = ½ m * v22
Difference in kinetic energy = ½ * m * (v12 – v22)
= ½ * 0.5 * (25 – 9)
= ½ * 0.5 * 16
= 4 J
Change In kinetic energy is 4J decrease.
Question: 16
A canon ball of mass 500g is fired with a speed of 15m/s-1. Find:
(i) its kinetic energy and
(ii) its momentum.
Solutions: Given
Mass = 500 g = 0.5 kg
Speed = 15 m per s
Kinetic energy = ½ m * v2
=½ * 0.5 * 15 * 15
= 56.25 J
Momentum = √(2mK)
= √(2 * 0.5 * 56.25)
= √(56.25)
= 7.5 kg m s-1
Question: 17
A body of mass 10 kg is moving with a velocity 20m 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.
Solutions: Given
Mass = 10 kg
Velocity = 20 m per s
Initial kinetic energy = ½ m * v2
= ½ * 10 * 20 * 20
= 2000 J
Now if the mass is doubled and velocity is halved
Final kinetic energy = ½ m * v2
= ½ * 20 * 10 * 10
= 1000 J
Question: 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)
Solutions: Given
Mass = 1000 kg
Initial velocity, v1 = 36 kmph = 10 m/s
Final velocity, v2= 72 kmph = 20 m/s
As we know,
Work done is change in kinetic energy
Work done = ½ m * (v12 – v22)
= ½ * 1000 * (202 -102)
= ½ * 1000 * 300
= 150000 J
Power = work done / time
Power = 15000 J / 120 s
= 1250 W
Question: 19
A body of mass 60 kg has the momentum 3000 kgm/s-1. Calculate:
(i) The kinetic energy and
(ii) The speed of the body.
Solutions: Given
Mass = 60 kg
Momentum = 3000 kgm/s
As we know,
Momentum = √(2mK)
3000 = √(2mK)
Squaring both sides
9000000 = 2 * 60 * K
K = 9000000 / 120
K = 75000 J
For speed
Velocity = √(2K/m)
= √(2 * 75000 / 60)
= √(2500)
= 50 m/s
Question: 20
How much work is needed to be done on a ball of mass 50g to give it a momentum of 5 kg m s-1?
Solutions: Given
Mass = 50 g = 0.05 kg
Initial velocity = 0
Final momentum = 5 kgm/s
For final velocity,
Final velocity = final momentum / mass = 5 / 0/05 = 100 m/s
Hence, as initial velocity is zero final kinetic energy is work done
Work done = ½ m * v2
= ½ * 0.05 * 100 * 100
= 250 J
Question: 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)
Solutions: Given
Mass = 20 kg
As we know, potential energy = mgh
When the person is in the bus the change in h is zero hence the gain in the energy by box is zero.
When the man runs at speed of 3 m/s
Kinetic energy = ½ m * v2
= ½ * 20 * 3 *3
= 90 J
When the box is raised by 0.5 m
Potential energy = mgh
= 20 * 10 * 0.5
= 100 J
Question: 22
A bullet of mass 50g is moving with a velocity of 500m/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.
Solutions: Given
Mass = 50 g = 0.05 kg
Velocity = 500 m/s
Distance travelled = 10 cm
Kinetic energy = ½ m * v2
= ½ * 0.05 * 500 * 500
= 6250 J
To determine we can use formula of work done as the kinetic energy is changed to 0
Hence, work done = kinetic energy change = 6250 J
Work done = force * displacement
6250 = force * 0.1 m
Force = 6250 / 0.1
Force = 62500 N
Hence, retarding force is 62500 N.
Question: 23
A spring is kept compressed by a small trolley of mass 0.5 kg lying on a smooth horizontal surface as shown. When the trolley is released, it is found to move at a speed of v = 2 m/s-1. What potential energy did the spring possess when compressed?
Solutions: Given
Mass = 0.5 kg
Velocity = 2 m/s
As the spring released the potential energy is converted into kinetic energy
Hence,
Potential energy = kinetic energy = ½ m * v2
= ½ * 0.5 * 2 * 2
= 1 J
Exercise 2 (c)
Question: 1
State the Principle of conservation of energy.
Solutions: According to principle of conservation of energy, energy can neither be created nor be destroyed. It can only be converted into one form to another.
Question: 2
What do you understand by the conservation of mechanical energy? State the condition under which the mechanical energy is conserved.
Solutions: The energy can neither be created nor be destroyed, it can be internally converted from one from to another when forces acting are conservative. Hence, the total mechanical energy of system is conserved.
Question: 3
Name two examples in which the mechanical energy of a system remains constant.
Solutions: The moving car using chemical energy for movement i.e. kinetic energy.
Another example is charging of a battery where electrical energy is converted into chemical energy.
Question: 4
A body is thrown vertically upwards. Its velocity keeps on decreasing. What happens to its kinetic energy as its velocity becomes zero?
Solutions: As the kinetic energy is directly proportional to velocity of the body when velocity becomes zero the kinetic energy also becomes zero.
Question: 5
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.
Solutions:
(a) Potential energy
(b) Kinetic energy and potential energy
(c) Kinetic energy
Question: 6
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.
Solutions: Let a body of mass m be falling freely undergravity from a height h above the ground. As the body falls down, its potential energy changes into kineticenergy. At each point of motion, the sum of potential energy and kinetic energy unchanged.Let us calculate the sum of kinetic energy K and potential energy U at various positions, say at A (at height h above the ground), at B (when it has fallen through a distance d), and at C (on the ground).
At position A (at height & above the ground): Potential energy U= mgh
Initial velocity of body 0 (since body is at rest at A)
Kinetic energy, K=0
Potential energy, U = mgh
Hence total energy = 0 + mgh = mgh
At position B (when it has fallen through a distance. a):
Let v1 be the velocity acquired by the body at B after falling through a distance d.
Then u =0, S=d, a=g
From equation, v2 = u²+2aS
v12= 0+2gd = 2gd
Kinetic energy
K = ½ m v2
= ½ m 2gd
= mgd
Now at B. height of body above the ground = h – d
Potential energy,
U = mg (h-d) Hence, total energy = K+U
= mgd + mg (h-d)
= mgh
At position C (on the ground):
Let the velocity acquired by the body on reaching the
ground be v. Then w=0, S=h, a=g
From equation,
v2 = u²+2aS
v2 = 0+2gh = 2gh
Kinetic energy K = ½ m v2
K = ½ m * 2gh
K = mgh
and potential energy U = 0 (at the ground when h = 0)
Hence total energy,
K+U=mgh+0
K + U =mgh
Question: 7
A pendulum is oscillating on either side of its rest position. Explain the energy changes that take place in the oscillating pendulum. How does the mechanical energy remain constant in it? Draw necessary diagram.
Solutions:
When a pendulum oscillates from its rest position on a fix path its energy changes from maximum potential energy to maximum kinetic energy to maximum potential energy again. When it reaches to its end of path. In the above diagram at position C the pendulum has maximum potential energy and at position A it has maximum kinetic energy and it moves towards B with increasing potential energy. At any time of the path of oscillating the sum of energies is always stays the same.
Question: 9
Name the type of energy possessed by the bob of a simple pendulum when it is at
(a) The extreme position,
(b) The mean position, and
(c) Between the mean and extreme positions.
Solutions:
(a) Potential energy
(b) Kinetic energy
(c) Both kinetic energy and potential energy
Multiple Choice Type:
Question: 1
A ball of mass m is thrown vertically up with an initial velocity so as to reach a height h. The correct statement is:
a.) Potential energy of the ball at the ground is mgh.
b.) Kinetic energy to the ball at the ground is zero.
c.) Kinetic energy of the ball at the highest point is mgh.
d.) The potential energy of the ball at the highest point is mgh.
Solutions: (d) The potential energy of the ball at the highest point is mgh.
Question: 2
A pendulum is oscillating on either side of its rest position. The correct statement is :
(a) It has only the kinetic energy 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 and potential energy remains constant throughout the motion.
Solutions: (d) The sum of its kinetic and potential energy remains constant throughout the motion.
NUMERICAL
Question: 1
A ball of mass 0.20 kg is thrown vertically upwards with an initial velocity of 20m/s-1 Calculate the maximum potential energy it gains as it goes up.
Solutions: Given
Mass = 0.20 kg
Velocity = 20 m/s
Maximum potential energy = initial kinetic energy
= ½ m * v2
= ½ * 0.20 * 20 * 20
= 40 J
Question: 2
A stone of mass 500 g is thrown vertically upwards with a velocity of 15m/s-1. Calculate:
(a) The potential energy at the greatest height,
(b) The kinetic energy on reaching the ground
(c) The total energy at its half waypoint.
Solutions:
Given
Mass = 500g = 0.5 g
Velocity = 15
(a) Potential energy at greatest height (U) = initial kinetic energy (K)
= ½ m * v2
= ½ * 0.5 * 15 * 15
= 56.25 J
(b) Kinetic energy is equal to potential energy as shown in above equation
Hence kinetic energy = 56.25 J
(c) Energy at half way point is = ½ (K + U)
= ½ (56.25 +56.25) = 56.25 J
Hence, total energy at half way point is 56.25 J.
Question: 3
A metal ball of mass 2kg is allowed to fall freely from rest from a height of 5m above the ground.
(a) Taking g = 10m/s2, 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?
Solutions: Given
Mass = 2 kg
Height h = 5 m
Gravity g = 10 ms-2
(a) As we know, potential energy = mgh
Potential energy = 2 * 5 * !0
Potential energy = 100 J
Kinetic energy before the ball just hit the ground is equal to the potential energy
Hence, kinetic energy = 100 J
(b) When the ball hits ground the mechanical energy gets converted sound and hit energy.
Question: 4
The diagram given below shows a ski jump. A skier weighing 60kgf 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 the kinetic energy at B, calculate the speed at which the skier arrives at B.
(Take g = 10 m s-2).
Solutions: Given
Mass = 60 kg
Height at point A = 75 m
Height a point B = 15 m
Therefore,
(a) Potential energy at point A = mgh
= 60 * 10 * 75
= 45000 J
Potential energy at point B = mgh
= 60 * 10 * 15
= 9000 J
Difference in the kinetic energy = 45000 J – 9000 J
= 36000 J
(b) 75 % of total potential energy is converted into kinetic energy then
Kinetic energy = 36000 * 0.75
Kinetic energy = 27000 J
As we know,
Kinetic energy = ½ m * v2
27000 = ½ * 60 * v2
v2 = 27000/ 30
v2 = 900
Velocity = 30 m/s
Question: 5
A hydro-electric power station takes its water from a lake whose water level is 50m 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 1MW.
(g=10 m s-2).
Solutions: Given
Height h = 50 m
Gravity g = 10 m/s2
Potential energy of water = mgh
= m * 10 * 50
= 500 m
As the overall efficiency is 40% the total work = 0.4 * 500 m
Work done = 200m
Now, power produced Is 1 mega watt = 1000000 Watts
Power = work done / time
100000 = 200m / 1
Mass of wate required to flow = 5000 kg
Question: 6
The bob of a simple pendulum is imparted a velocity of 5m 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).
Solutions: Given
Velocity = 5 m/s
Let us consider mass of pendulum be m
The kinetic energy at mean position is
Kinetic energy = ½ m v2
K = ½ m * 5 * 5
K = 12.5 m
As, 60 % of this kinetic energy is lost to overcome friction of air
Potential energy = 0.4 * 12.5m
Potential energy = 5m
Based on the formula of potential energy
E. = mgh
5m = m * 10 * h
Height gained = 5m /10m = 0.5 m
Height gained by the pendulum is 0.5 m.