Class 10 SELINA Solutions Physics Chapter 2  Work, Energy and Power
Work, Energy and Power Exercise Ex. 2A
Solution A.1
(b) W = F.s
In vector form, work done W is written as,
W = F. s
Solution A.2
(a) 0°
For work done to be maximum, the angle between the force and displacement should be 0°.
Because,
W = F.s = Fs cosƟ
And
cos 0° = 1
So, work done will be maximum when the angle between force and displacement is 0°.
Solution A.3
(c) Zero
When a body moves in a circular path in a horizontal plane, the work done is zero because in case of a circular path, the displacement will be zero as the initial and final position is the same.
Solution A.4
(b) mgh
When a ball of mass m is thrown upwards to a height h, the work done by the force of gravity is mgh.
Solution A.5
(a) Joule
The S.I. unit of work is Joule.
Solution A.6
(a) equal to work done by A
The work done does not depend upon the time taken.
W = F × s
The load which is lifted by coolie A and B is the same. And the distance through which it has to be lifted is the same. Thus, the work done by both the coolie will be the same.
Solution A.7
(d) 746 W
1 hp = 746 W
Solution A.8
(c) Energy
The unit kWh is the unit of energy.
Solution A.9
(b) 10^{7}
1 Joule is equivalent to 10^{7}erg.
Solution A.10
(a) Watt
The S.I. unit of power is Watt.
Solution A.11
(b) 4.0 × 10^{19} J
1 eV = 1.6 × 10^{19} J
Thus, the energy of 2.5 eV electron when expressed in Joule will be,
2.5 eV = 2.5 × 1.6 × 10^{19} J = 4 × 10^{19} J
Solution A.12
(a) 600 J
Work done = Force × displacement
W = F × s = 20 × 10 × 3 = 600 J
Solution B.1
(a) When force is at an angle to the direction of displacement, then work done, W= F S cos
(b)
(i)For zero work done, the angle between force and displacement should be 90^{o} as cos 90^{o}=0
W =FScos90^{o}= FSx0=0
(ii)For maximum work done, the angle between force and displacement should be 0^{o} as cos0^{o}=1
Hence, W=FScos 0^{o}=FS
Solution B.2
Work is done against the force.
Solution B.3
(a) No, work is not done since the wall does not move.
(b) No, work is not done since the box on the coolie's head does not move either.
(c) Yes, work is done since the boy moves vertically against gravity while climbing the stairs.
Solution B.4
When a coolie carries a load while moving on a ground, the displacement is in the horizontal direction while the force of gravity acts vertically downward. So the work done by the force of gravity is zero.
Solution B.5
Let a boy of mass m climb up through a vertical height h either through staircase of using a lift. The force of gravity on the boy is F=mg acting vertically downwards and the displacement in the direction opposite to force (i.e., vertical) is S=h. Therefore the work done by the force of gravity on the boy is
W= FS =mgh
or,the work W=mgh is done by the boy against the force of gravity.
Solution B.6
eV measures the energy of atomic particles.
1eV= 1.6 x 10^{19}J
Solution B.7
1 J = 0.24 calorie
1kWh = 3.6× 10^{6} J.
Solution B.8
Calorie measures heat energy.
1calorie = 4.18 J
Solution B.9
(a)Power is measured in terms of horse power.
(b) 1 horse power =746 watt
Solution B.10
(a) Energy is measured in kWh
(b) Power is measure in kW
(c) Energy is measured in Wh
(d) Energy is meaused in eV
Concept insight: Energy has bigger units like kWh (kilowatt hour) and Wh (watt hour). Similarly bigger unit of power is kW (kilo watt).
The energy of atomic particles is very small, and hence, it is measured in eV (electron volt).
Solution B.11
Yes, there is no transfer of energy if the body is acted upon by the force normal to the displacement.
When the body is moving in a circular path, the force is normal to its displacement and the work done is zero.
Thus, there is no transfer of energy.
Solution C.1
Work is said to be done only when the force applied on a body makes the body move. It is a scalar quantity.
Solution C.2
(i)When force is in direction of displacement, then work done , W = FxS
(ii)When force is at an angle to the direction of displacement, then work done, W= F S cos
Solution C.3
Two conditions when the work done is zero are:
(i)When there is no displacement (S=0) and,
(ii)When the displacement is normal to the direction of the force ( =90^{o}).
Solution C.4
When a body moves in a circular path, no work is done since the force on the body is directed towards the centre of circular path (the body is acted upon by the centripetal force), while the displacement at all instants is along the tangent to the circular path, i.e., normal to the direction of force.
Solution C.5
Work done by the force of gravity (which provides the centripetal force) is zero as the force of gravity acting on the satellite is normal to the displacement of the satellite.
Solution C.6
Energy is the capacity to do work and work done is equal to energy spent. Coolie X carrying a load up a slope will do more work as this works involve a change in potential energy, kinetic energy and loss of energy due to friction. Work done in carrying the load in horizontal frictionless surface does not involve change in potential energy and work done by the friction is also zero.
Solution C.7
Force applied by the fielder on the ball is in opposite direction of displacement of ball. So, work done by the fielder on the ball is negative.
Solution C.8
S.I unit of work is Joule.
C.G.S unit of work is erg.
Relation between joule and erg :
1joule= 1N x 1m
But 1N =10^{5}dyne
And 1m=100 cm= 10^{2 }cm
Hence, 1 joule= 10^{5}dyne x 10^{2}cm
=10^{7}dyne x cm=10^{7}erg
Thus, 1 Joule= 10^{7} erg
Solution C.9
S. I unit of work is Joule.
1 joule of work is said to be done when a force of 1 newton displaces a body through a distance of 1 metre in its own direction.
Solution C.10
Relation between joule and erg :
1joule= 1N x 1m
But 1N =10^{5}dyne
And 1m=100 cm= 10^{2 }cm
Hence, 1 joule= 10^{5}dyne x 10^{2}cm
=10^{7}dyne x cm=10^{7}erg
Thus, 1 Joule= 10^{7} erg
Solution C.11
The energy of a body is its capacity to do work. Its S.I unit is Joule (J).
Solution C.12
1kWh is the energy spent (or work done) by a source of power 1kW in 1 h.
1kWh = 3.6 x 10^{6}J
Solution C.13
The rate of doing work is called power. The S.I. unit of power is watt (W).
Solution C.14
Work 
Power 
1.Work done by a force is equal to the product of force and the displacement in the direction of force. 
1.Power of a source is the rate of doing work by it. 
2.Work done does not depend on time. 
2.Power spent depends on the time in which work is done. 
3.S.I unit of work is joule (J). 
3.S.I unit of power is watt (W). 
Solution C.15
Energy 
Power 
1. Energy of a body is its capacity to do work. 
1. Power of a source is the energy spent by it in 1s.

2. Energy spent does not depend on time. 
2. Power spent depends on the time in which energy is spent. 
3. S.I unit of energy is joule (J). 
3. S.I unit of power is watt (W). 
Solution C.16
S.I unit of power is watt (W).
If 1 joule of work is done in 1 second, the power spent is said to be 1 watt.
Solution C.17
Watt (W) is the unit of power, while watt hour (Wh) is the unit of work, since power x time = work.
Solution D.1
(i) If the displacement of the body is in the direction of force, then work done is positive.
Hence, W= F x S
For example: A coolie does work on the load when he raises it up against the force of gravity. The force exerted by coolie (=mg) and displacement, both are in upward direction.
(ii)If the displacement of the body is in the direction opposite to the force, then work done is negative.
Hence, W = F x S
For example: When a body moves on a surface, the force of friction between the body and the surface is in direction opposite to the motion of the body and so the work done by the force of friction is negative.
Solution D.2
Let a body of mass m fall down through a vertical height h either directly or through an inclined plane e.g. a hill, slope or staircase. The force of gravity on the body is F=mg acting vertically downwards and the displacement in the direction of force (i.e., vertical) is S=h. Therefore the work done by the force of gravity is
W= FS =mgh
Solution D.3
Power spent by a source depends on two factors:
(i)The amount of work done by the source, and
(ii)The time taken by the source to do the said work.
Example: If a coolie A takes 1 minute to lift a load to the roof of a bus, while another coolie B takes 2 minutes to lift the same load to the roof of the same bus, the work done by both the coolies is the same, but the power spent by the coolie A is twice the power spent by the coolie B because the coolie A does work at a faster rate.
Solution E.1
Force acting on the body = 20 kgf = 20 x 10 N = 100 N
Displacement, S = 1 m
Work done= force x displacement in the direction of force
(i) When, θ = 0°
W = F x S Cos θ
W = 200 x 1 x 1 = 200 J
(ii) When, θ = 60°
Work = force x displacement in the direction of force
W = F x S cos
W = 200 x 1 cos60^{o}
W = 200 x 1 x 0.5.....(∵ cos60^{o}=0.5)
∴ W =100 J
(iii) Normal to the force:(i.e., θ = 90°)
Work = force x displacement in the direction of force
W = F x S cos
W = 200 x 1 cos 90^{o}
W = 200 x 1 x 0 ...(∵ cos90^{o} =0)
∴ W = 0 J
Solution E.2
Mass of boy=40 kg
Vertical height moved, h=8m
Time taken, t=5s.
(i) Force of gravity on the boy
F= mg =40 x 10 =400N
(ii)While climbing, the boy has to do work against the force of gravity.
Work done by the boy in climbing= Force x distance moved in the direction of force
Or, W = F x S= 400 x 8= 3200 J
(iii) Power spent =
Solution E.3
Given that:
Work = 7.4 kJ = 7400 J
Distance = 74 m
Time = 2.5 s
(i) Work done by the man = Force x distance moved in direction of force
Work, W = F x S
7.4 x 10^{3} = F x 74
(ii) Power spent =
i.e., Power spent (H. P) =
Solution E.4
Force= mg= 200 x 10=2000N
Distance, S= 2.5m
Time , t=5 s
(i)Work done, W= F S
W =2000 x 2.5m= 5000J
(ii)Power developed =
Solution E.5
(i)Energy spent by machine or work done= F S
Work, W =750 x 16= 12000J
(ii)Power spent=
Solution E.6
Energy consumed = power x time
(i)Energy = 3 kW x 10 h=30kWh
(ii)1 kilowatt hour (kWh)= 3.6 x 10^{6 }J
30kWh = 30 x 3.6 x 10^{6 }J
= 1.08 x 10^{8 }J
Solution E.7
Volume of water= 50 L=50 x10^{3} m^{3}
Density of water= 1000kgm^{3}
Mass of water= Volume of water x density of water
= 50 x10^{3} x1000= 50kg
Work done in raising 50kg water to a height of 25m against the force of gravity is:
W = mg x h= mgh
Power P=
Solution E.8
Given that,
Weight, w = 600 kg
Depth, h =75 m
Time, t = 10 s
Efficiency, η = 50%
(a) Work done in raising a 500kg mass to a height of 90m against the force of gravity is:
W = mg × h= mgh
W= 600 ×10 × 75 = 4.5 ×105 J
(b) Now,
Power = Work / time
Power =450000/10
Power = 45000 W = 45 kW
Therefore, the power at which the pump works is 45 kilo Watts.
(c) The power rating of the pump if its efficiency is 40%.
Solution E.9
Given, force = 1000N, velocity=30m/s
Power, P= force x velocity
P = 1000 x 30 = 30,000W = 30kW
Solution E.10
Power =40kW
Force= 20,000N
Power = force x velocity
Velocity =
Solution E.11
Power exerted due to force is
Solution E.12
Total distance covered in 30 steps , S= 30 x 20 cm = 600 cm = 6 m
Work done by the boy in climbing= Force x distance moved in direction of force
Work, W= F x S= 350 x 6 =2100 J
Power developed=
Solution E.13
(i) The work done by persons A and B is independent of time. Hence both A and B will do the same amount of work. Hence,
(ii)Power developed by the person A and B is calculated as follows:
A takes 20 s to climb the stairs while B takes 15 s, to do the same. Hence B does work at a much faster rate than A; more power is spent by B.
Power developed (and amount of work done is same)
Solution E.14
Given that,
m_{1 }= 40 kgf
m_{2} = 30 kgf
h_{1} = h_{2} = h
t_{1} = 4 min = 4× 60
t_{2} = 3 min = 3× 60
Now,
(i) Work done is
(ii) Power developed is
Solution E.15
(a)
(i)Work done to raise the block of mass 50kg is same for both.
(ii) Power= Work done/time. As the time taken by the first man is less therefore power developed is more
(b)
(i)Work done =
(ii) Power developed by first man=1000/20=50W
Power developed by second man=1000/50=20W
Solution E.16
(a)
(i) Work done is same for both as both the people carry the same weight of water to the same height.
(ii) Power developed is
(b) Mass of the water lifted by both is
Hence, the work done is
Work, Energy and Power Exercise Ex. 2B
Solution A.1
Potential energy
Hint: P.E. is the energy possessed by a body by virtue of its position.
Solution A.2
(c) Gravitational potential energy
The energy possessed by a body due to the force of attraction of the earth on it is called its gravitational potential energy.
Solution A.3
(d) Elastic potential energy
The energy possessed by a body in the deformed state due to change in its size or shape is called elastic potential energy.
Solution A.4
(d) work done
According to the workenergy theorem, the increase in the kinetic energy of a moving body is equal to work done by a force acting in the direction of the moving body.
Solution A.5
(c) vibrational kinetic energy
A wire clamped at both the ends is struck in the middle. It will possess vibrational kinetic energy.
Solution A.6
(d) both rotational and translational kinetic energy
The wheel of a moving vehicle possesses both rotational and translational kinetic energy.
Solution A.7
(d) 1/9
Thus, the kinetic energy gets reduced by 1/9.
Solution A.8
(c) 2000 J
A body of mass 10 kg is moving with a velocity of 20 m/s. The mass of the body is doubled and velocity is halved. Its initial kinetic energy is 2000 J
Solution A.9
The kinetic energy K and momentum p are related as
Solution A.10
(b) momentum
On doubling the velocity of a moving body, the quantity which gets doubled is momentum.
Solution A.11
(b) light energy to electrical energy
In a photoelectric cell, the change in energy is from light energy to electrical energy.
Solution A.12
(b) chemical energy to kinetic energy
In a petrol vehicle while in motion, the change in energy is from chemical energy to kinetic energy.
Solution A.13
(c) nuclear energy changes to electrical energy
In the nuclear reactor, when the nuclear fission chain reaction takes place a large amount of nuclear energy is used to boil water into steam. The steam turns the blades of the turbine and drives generators which produce electricity.
Solution A.14
(d) Chemical to electrical
Hint: When current is drawn from an electric cell, the chemical energy stored in it changes into electrical energy.
Solution B.1
Two forms of mechanical energy are:
(i)Kinetic energy
(ii)Potential energy
Solution B.2
Elastic potential energy is possessed by wound up watch spring.
Solution B.3
(a)Kinetic energy (K)
(b)Potential energy (U)
(c)Kinetic energy (K)
(d)Potential energy (U)
(e)Kinetic energy (K)
(f)Potential energy (U)
Solution B.4
(a) Kinetic energy =
(b)
Solution B.5
Kinetic energy is related to momentum and mass as
As the kinetic energy of both bodies are same, momentum is directly proportional to square root of mass.
Now, mass of body B is greater than that of body A.
Hence, body B will have more momentum than body A.
Solution B.6
(a)Motion.
(b)Position.
Solution B.7
Kinetic energy.
Solution B.8
Kinetic energy.
Solution B.9
(a) Potential energy of wound up spring converts into kinetic energy.
(b) Chemical energy of petrol or diesel converts into mechanical energy (kinetic energy)
(c) Kinetic energy to potential energy
(d) Light energy changes into chemical energy
(e) Electrical energy changes into chemical energy
(f) Chemical energy changes into heat energy
(g) Chemical energy changes into heat and light energy
(h) Chemical energy changes into heat, light and sound energy
Solution B.10
(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
(l)electrical energy into mechanical enegy
(m)Electrical energy into magnetic energy.
Solution B.11
The process used to transfer nuclear energy to electrical energy is called nuclear fission.
Solution B.12
No. This is because, whenever there is conversion of energy from one form to another apart of the energy is dissipated in the form of heat which is lost to surroundings.
Solution B.13
Conversion of part of energy into an undesirable form is called dissipation of energy/degradation of energy.
Solution C.1
Potential energy: The energy possessed by a body by virtue of its specific position (or changed configuration) is called the potential energy.
Different forms of P.E. are as listed below:
(i) Gravitational potential energy: The potential energy possessed by a body due to its position relative to the centre of Earth is called its gravitational potential energy.
Example: A stone at a height has gravitational potential energy due to its raised height.
(ii) Elastic potential energy: The potential energy possessed by a body in the deformed state due to change in its configuration is called its elastic potential energy.
Example: A compressed spring has elastic potential energy due to its compressed state.
Solution C.2
Potential energy is possessed by the body even when it is not in motion. For example: a stone at a height has the gravitational potential energy due to its raised position.
Solution C.3
The work done W on the body in lifting it to a height h is
W= force of gravity (mg) x displacement (h)
=mgh
This work is stored in the body when it is at a height h in the form of its gravitational potential energy.
Gravitational potential energy U= mgh
Solution C.4
A body in motion is said to possess the kinetic energy. The energy possessed by a body by virtue of its state of motion is called the kinetic energy.
Solution C.5
According to the workenergy theorem, the work done by a force on a moving body is equal to the increase in its kinetic energy.
Solution C.6
Body of mass m is moving with a uniform velocity u. A force is applied on the body due to which its velocity changes from u to v and produces an acceleration a in moving a distance S.Then,
Work done by the force= force x displacement
W = F x S(i)
From relation : v^{2} = u^{2}+2 a S
Solution C.7
Kinetic energy, where p is the momentum.
Both the masses have same momentum p. The kinetic energy, K is inversely proportional to mass of the body.
Hence light mass body has more kinetic energy because smaller the mass, larger is the kinetic energy.
Solution C.8
The three forms of kinetic energy are:
(i)Translational kinetic energy example: a freely falling body
(ii)Rotational kinetic energyexample: A spinning top.
(iii)Vibrational kinetic energyexample: atoms in a solid vibrating about their mean position.
Solution C.9
Potential energy (U) 
Kinetic energy (K) 
1. The energy possessed by a body by virtue of its specific position or changed configuration is called potential energy. 
1.The energy possessed by a body by virtue of its state of motion is called the kinetic energy. 
2.Two forms of potential energy are gravitational potential energy and elastic potential energy. 
2. Forms of kinetic energy are translational, rotational and vibrational kinetic energy. 
3.Example: A wound up watch spring has potential energy. 
3. For example: a moving car has kinetic energy. 
Solution C.10
The six different forms of energy are:
1.Solar energy
2.Heat energy
3.Light energy
4.Chemical or fuel energy
5.Hydro energy
6.Nuclear energy
Solution C.11
During the transformation of energy from one form to another desired form, some part of the energy gets converted to some undesirable form or a part of it is lost to the surroundings due to the friction or radiations which cannot be used for any productive purpose. This is called dissipation of energy, and this undesirable part of energy is called degraded energy.
Solution C.12
During the transformation of energy from one form to another desired form, some part of energy is converted to some undesirable form or a part of it is lost to the surroundings due to the friction or radiations which cannot be used for any productive purpose. This is called dissipation of energy or degradation of energy.
Example:
When a light bulb glows, a major part of the electrical energy utilised is converted to heat energy while some part is converted to useful light energy.
Solution D.1
Gravitational potential energy is the potential energy possessed by a body due to its position relative to the centre of earth.
For a body placed at a height above the ground, the gravitational potential energy is measured by the amount of work done in lifting it up to that height against the force of gravity.
Let a body of mass m be lifted from the ground to a vertical height h. The least upward force F required to lift the body (without acceleration) must be equal to the force of gravity (=mg) on the body acting vertically downwards. The work done W on the body in lifting it to a height h is
W= force of gravity (mg) x displacement (h)
=mgh
This work is stored in the body when it is at a height h in the form of its gravitational potential energy.
Gravitational potential energyU= mgh
Solution D.2
When the string of a bow is pulled, some work is done which is stored in the deformed state of the bow in the form of its elastic potential energy. On releasing the string to shoot an arrow, the potential energy of the bow changes into the kinetic energy of the arrow which makes it move.
Solution D.3
The compressed spring has elastic potential energy due to its compressed state. When it is released, the potential energy of the spring changes into kinetic energy which does work on the ball if placed on it and changes into kinetic energy of the ball due to which it flies away.
Solution D.4
When the pebble is thrown upwards, the kinetic energy in it is converted to potential energy.
At the top point in its motion, its kinetic energy is completely converted into potential energy.
While coming down, the potential energy is converted into kinetic energy and at the bottom the potential energy is completely converted to kinetic energy.
Solution D.5
When water falls from a height, the potential energy stored in water at a height changes into the kinetic energy of water during the fall. On striking the ground, a part of the kinetic energy of water changes into the heat energy due to which the temperature of water rises.
Solution E.1
Height H_{1}= h
Height H_{2}= 2h
Mass of body 1= m
Mass of body 2= m
Gravitational potential energy of body 1 =mgH_{1}= mgh
Gravitational potential energy of Body 2=mgH_{2}= mg (2h)
Ratio of gravitational potential energies
=
Solution E.2
Given that,
Mass, m=1kg
Height, h=5m
Gravitational potential energy, P. E= mgh
=1 × 10 × 5 = 50J
When the mass just before hits the ground, by conservation of energy, all its potential energy is converted to kinetic energy.
i.e., K.E = P.E = 50 J
Solution E.3
Given that,
Gravitational potential energy = 29400 J
Force of gravity = mg = 300 × 9.8N/kg = 2940N
Gravitational potential energy = mgh
29400 = 2940 × h
h = 10m
Solution E.4
Mass, m = 10 kg
initial height, h_{1} = 20 m
final height, h_{2 }= 8 m
Gravitational acceleration, g = 10 ms^{2}
(i)
Now,
Loss of potential energy of the body, ΔP.E = mg (h_{2}  h_{1})
=10× 10× (20  8)
= 1200 J
(ii)
The total energy possessed by the body at any instant remains constant for freefall.
It is equal to the sum of P.E. and K.E.
∴ At height 20 m, i.e., at topmost point, K .E= 0.
Total energy = P .E. + K.E.
Total energy = mgh_{1} + 0 = 2000 J
Solution E.5
Mass = 0.5 kg
Energy = 1 J
Gravitational potential energy = mgh
1 = 0.5 x 10 x h
1 = 5h
Height, h = 0.2 m
Solution E.6
Force of gravity on boy=mg= 50 x 10 = 500 N
Increase in gravitational potential energy = Mg (h_{2 } h_{1})
= 500 x (9  3)
= 500 x 6 = 3,000 J
Solution E.7
Mass of water, m = 50kg
Height, h = 15m
Gravitational potential energy = mgh
=50 x 10 x 15
=7500 J
Solution E.8
Mass of man= 50 kg
Height of ladder, h_{2}=15 m
(i)Work done by man =mgh_{2}
= 50 x 9.8 x 15 = 7350 J
(ii) Increase in his potential energy:
Initial distance, h1= 0 m
Gravitational potential energy = Mg × Δh
= 50 x 9.8 x (15  0)
= 50 x 9.8 x 15 = 7350 J
Solution E.9
F =150N
a. Work done by the force in moving the block 5m along the slope = Force x displacement in the direction of force =100 x 5 =500 J
b. The potential energy gained by the block
U =mgh where h =3m
=100 x 3 = 300 J
The potential energy gained by the block is 300 J.
c. The difference i.e., 500 – 300 = 200 J energy is used in doing work against friction between the block and the slope, which will appear as heat energy.
Solution E.10
Given that
Mass, m = 5 kg
Velocity, v = 10m/s
Now,
= 250 J
Solution E.11
If the speed is halved (keeping the mass same), the kinetic energy decreases, it becomes onefourth (since kinetic energy is proportional to the square of velocity).
Solution E.12
Given, velocity of first body v_{1}=v
And velocity of second body, v_{2} =2v
Since masses are same, kinetic energy is directly proportional to the square of the velocity ()
Hence, ratio of their kinetic energies is:
Solution E.13
Solution E.14
Solution E.15
Mass of ball= 0.5kg
Initial velocity=5m/s
Initial kinetic energy=
Final velocity of the ball =3m/s
Final kinetic energy of the ball =
Change in the kinetic energy of the ball = 2.25 J  6.25J = 4J
There is a decrease in the kinetic energy of the ball .
Solution E.16
Mass of canon ball= 500g=0.5 kg
Speed, v=15m/s
(a)Kinetic energy of ball =
(b)Momentum of the ball = mass x velocity
=0.5 x15=7.5kgm/s
Solution E.17
Let initial Mass, m_{1}= 10kg and velocity, v_{1 }=20 m/s
Final mass, m_{2}=2 x10=20 kg and velocity, v_{2}=20/2= 10m/s
Initial kinetic energy, K_{1}=
Final kinetic energy, K_{2}=
Solution E.18
u=36 km/h=
and v=72km/h=
mass of the truck =1000 kg
(i)
(ii) Power
Solution E.19
Mass of body = 60kg
Momentum, p=3000kgm/s
(a)Kinetic energy
=7.5 x 10^{4}J
(b)Momentum = mass x velocity
3000 = 60 x velocity
Velocity =50m/s
Solution E.20
Momentum, p= 5 kg m/s
Mass of ball =50 g=0.05kg
(a)Kinetic energy of the ball
Solution E.21
Mass of box=20 kg
(a)Zero work is done as there is no displacement of the man.
(b)Work done, Kinetic energy of man
=
(c) Work done in raising the box, Potential energy = mgh
U= 20 x 10 x0.5=100J
Solution E.22
Mass of bullet =50g = 0.05kg
Velocity=500m/s
Distance penetrated by the bullet=10cm=0.1m
(a)Kinetic energy of the bullet=
(b)Work done by the bullet against the material of the target= resistive force x distance
6250= resistive force x 0.1m
Resistive force=62500N
Solution E.23
Mass of trolley = 0.5 kg
Velocity = 2 m/s
When the compressed spring is released, its potential energy is converted into kinetic energy completely.
Potential energy of compressed spring = kinetic energy of moving trolley
Kinetic energy of trolley =
Hence, potential energy of compressed spring=1.0J
Work, Energy and Power Exercise Ex. 2C
Solution A.1
(d) Total mechanical energy of a system remains constant if no external force is worked on it.
According to the principle of conservation of energy, the total mechanical energy of system remains conserved.
Solution A.2
(d) It remains constant
Assuming there are no external forces acting on the system, the total mechanical energy of the skater remains constant. This is due to the conservation of mechanical energy, where the sum of kinetic energy and potential energy remains unchanged unless acted upon by external forces.
Solution A.3
(d) It remains constant
According to the law of conservation of energy, the total mechanical energy will always be conserved. Thus, the total energy of spring system will remain constant.
Solution A.4
(b) It decreases
When a car moving on a level road is applied with a break it comes to rest and then the mechanical energy will decrease because when the brakes are applied the kinetic energy of the car converts to other forms of energy like heat and sound.
Solution A.5
(d) Potential energy of the ball at the highest point is mgh.
Hint: At the highest point, the ball momentarily comes to rest and thus its kinetic energy becomes zero.
Solution A.6
(d) mechanical energy
A body is released from a height. During its motion in striking the ground, the quantity which does not change is mechanical energy.
Solution A.7
(d) The sum of its kinetic and potential energy remains constant throughout the motion.
Hint: In accordance with the law of conservation of mechanical energy, whenever there is an interchange between the potential energy and kinetic energy, the total mechanical energy remains constant.
Solution A.8
(a) 0.28 J
The total energy of a simple pendulum at any instant will be,
m = 400 g = 0.4 kg
g = 10 m/s^{2}
h = 7 cm = 0.07 m
U = W = mgh
U = 0.4 x 10 x 0.07 = 0.28 J
Solution B.1
Kinetic energy of the body changes to potential energy when it is thrown vertically upwards and its velocity becomes zero.
Solution B.2
(a)Potential energy
(b)Potential energy and kinetic energy
(c)Kinetic energy
Solution C.1
According to the law of conservation of energy, energy can neither be created nor can it be destroyed. It only changes from one form to another.
Solution C.2
According to the law of conservation of mechanical energy, whenever there is an interchange between the potential energy and kinetic energy, the total mechanical energy (i.e., the sum of kinetic energy K and potential energy U) remains constant i.e., K + U = constant when there are no frictional forces.
Mechanical energy is conserved only when there are no frictional forces for a given system (i.e. between body and air). Thus, conservation of mechanical energy is strictly valid only in vacuum, where friction due to air is absent.
Solution C.3
Motion of a simple pendulum and motion of a freely falling body.
Solution C.4
(a)At position A, pendulum has maximum kinetic energy and its potential energy is zero at its resting position. Hence, K=mgh and U= 0.
(b)At B, kinetic energy decreases and potential energy increases. Hence, K= 0 and U=mgh
(c)At C also, kinetic energy K= 0 and potential energy U=mgh.
Solution C.5
a) Extreme position: Potential energy
b) Mean position: Kinetic energy
c) Between mean and extreme: Both kinetic energy and potential energy
Solution D.1
Let a body of mass m be falling freely under gravity from a height h above the ground (i.e., from position A). Let us now 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 x) and at C (on the ground).
(i)At the position A (at height h above the ground):
Initial velocity of body= 0 (since body is at rest at A)
Hence, kinetic energy K =0
Potential energy U = mgh
Hence total energy = K + U= 0 + mgh =mgh(i)
(ii) At the position B (when it has fallen a distance x):
Let v_{1 }be the velocity acquired by the body at B after falling through a distance x. Then u=0, S =x, a=g
From equation v^{2}= u^{2}+2aS
(ii)
(iii) At the position C (on the ground):
Let the velocity acquired by the body on reaching the ground be v. Then u=0, S=h, a=g
From equation: v^{2}= u^{2}+2aS
v^{2}= 0^{2}+2gh
v^{2}= 2gh
And potential energy U=0 (at the ground when h=0)
Hence total energy= K+U= mgh + 0=mgh (iii)
Thus from equation (i), (ii) and (iii), we note that the total mechanical energy i.e., the sum of kinetic energy and potential energy always remain constant at each point of motion and is equal to initial potential energy at height h.
Solution D.2
When the bob swings from A to B, the kinetic energy decreases and the potential energy becomes maximum at B where it is momentarily at rest.
From B to A, the potential energy again changes into the kinetic energy and the process gets repeated again and again.
Thus while swinging, the bob has only the potential energy at the extreme position B or C and only the kinetic energy at the resting position A. At an intermediate position (between A and B or between A and C), the bob has both the kinetic energy and potential energy, and the sum of both the energies (i.e., the total mechanical energy) remains constant throughout the swing.
Solution E.1
Potential energy at the maximum height= initial kinetic energy
Solution E.2
(a)Potential energy at the greatest height = initial kinetic energy
or, mgh
(b)Kinetic energy on reaching the ground= potential energy at the greatest height=56.25 J
(c)Total energy at its half way point==56.25J
Solution E.3
(a)
(i)Potential energy of the ball =mgh
=2 x 10 x 5=100J
(ii)Kinetic energy of the ball just before hitting the ground = Initial potential energy= mgh=2x10x5=100J
(b)Mechanical energy converts into heat and sound energy.
Solution E.4
(a)Mass of skier= 60kg
Loss in potential energy = mg(h_{1} –h_{2})
=60 x 10 x(7515)
= 60 x 10 x60=3.6x10^{4}J
(b)Kinetic energy at B
=2.7x 10^{4}J
Kinetic energy
27000
Solution E.5
Potential energy = mgh
Efficiency = 40 %
Useful work done = 40 % of potential energy
Solution E.6
Total Kinetic energy at mean position=
Solution E.7
Given:
mass of the pendulum bob, m = 200 g = 0.2 kg
height raised at position B, h = 5 m
acceleration due to gravity, g = 10 m/s²
(a) The potential energy of the pendulum at position B can be calculated using the formula for potential energy:
Potential energy at position B, P.E_{B} = mgh
P.E_{B} = 0.2 kg × 10 m/s² × 5 m
P.E_{B} = 10 J
(b)
Total energy at point C, M.E_{C }= Potential energy at its extreme position.
= 10 J
(c) Now,
K.E. at mean position, K.E_{A} = Potential energy at extreme position, P.E_{B}
∴ v_{A} = 10 m/s