Wednesday, February 27, 2019

Chapter 6.20 - Power

In the previous section, we completed a discussion on work and energy'. In this section we will see power
We have seen some basics about power in our high school classes. The following link will give those details

High school physics - Power

It is recommended that the reader get a thorough understanding of those lessons before taking up our present discussion
1. Power is defined as the time rate at which work is done or energy is transferred
• 'time rate' means 'per unit time'
• Also note the terms:
    ♦ 'work is done' or
    ♦ 'energy is transferred'
2. So we can write:
■ In a process where work is done by a machine:
(eg: A motor of a lift used in high rise buildings)
• Power of the machine is the work done by it in unit time
■ In a process where energy is transferred:
(eq: An electric cooker which converts electrical energy into heat energy) 
• Power of the appliance is the energy consumed by it in unit time
3. The unit of power is watt. It's symbol is W
Average power
1. In some cases, the 'work done' or 'energy consumed' may not be uniform
• For example, a machine capable of giving 500 W power,  may not give the same 500 joules of work every second
    ♦ During certain time intervals, it may be 540 J
    ♦ During certain other time intervals, it may be 480 J
2. In such cases, we take the average power
• It is done as follows:
(i) Turn on the machine and a stop-watch at the same instant
(ii) Record the total work 'W' done by the machine during a time interval 't'. 
('t' may be any suitable time interval. Say 25 mins, 40 mins, 1 hr, 3 hr, etc.,)
(iii) Then the average power is given by: pav = W⁄t 
■ Horse power is another unit for power. It's symbol is hp
• 1 hp = 746 W 
• It is mainly used to express the power of automobiles, pumps etc.,
Solved example 6.28
An elevator can lift a maximum load of 1800 kg. (This load includes the mass of the passengers + the mass of the elevator). It is carrying this maximum load and is moving up with a constant velocity of 2 ms-1. The frictional force opposing the motion is 4000 N. Calculate the power of the motor which lifts up the elevator [g = 10 ms-2]
Solution:
1. Power is 'work done in one second'
• Let us first find the work done in a suitable time interval say 10 s
2. 'Work done' is force × displacement
• The force here is the weight of the lift = mg = 1800 × 10 = 18000 N
• displacement is the distance traveled in 10 s
• Since the velocity is uniform, we can write:
displacement = velocity × time = 2 × 10 = 20 m
• So work done during the 10 s = 18000 × 20 = 360000 J
3. In addition to the above, work is done against friction also
• This work is given by:
frictional force × distance = 4000 × 20 = 80000 J
4. So total work done during the 10 s = 360000 + 80000 = 440000 J
5. So Power = work done in 1 s = 440000⁄10 = 44000 W = 44 kW
6. 1 hp = 746 W
• So 1 W = 1⁄746 hp
• So 44000 W = 44000 × 1⁄746 = 58.98 = 59 hp

 Solved example 6.29
A pump at the ground floor of a building can pump up water to fill up a tank of volume 30 m3 in 15 min. If the tank is 40 m above the ground and the efficiency of the pump is 30%, how much electric power is consumed by the pump? [g = 9.8 ms-2]
Solution:
1. Total work done by the pump 
= Work done in completely filling the tank 
= Work done in lifting 30 m3 of water to an elevation of 40 m 
= mgh 
= mass of 30 m3 of water × 9.8 × 40
2. So our next task is to find the ' mass of 30 m3 of water'
• We have: Density = mass⁄volume 
• So mass = volume × density = 30 × 1000 = 30000 kg
(∵ Density of water = 1000 kg m-3)
3. So the result in (1) becomes:
• Total work done by the pump = 30000 × 9.8 × 40 = 11760000 J
4. The efficiency is 30 %
• Let 'I' be the total electrical energy input during the 15 min
• Only 30% of I will be converted into useful work
• The rest will be lost as heat and sound
5. The 'useful work required' in our present case is 11760000 J
• So we can write:
I × 0.3 = 11760000 J
• Thus I = 39200000 J
6. This much electrical energy is consumed in 15 min
So power = $\mathbf\small{\frac{\text{work}}{\text{time}}=\frac{39200000}{15\times 60}=43555.55 \,\,\text{W}=43.55\,\,\text{kW}}$

Solved example 6.30
A family uses 8 kW of power. (a) Direct solar energy is incident on the horizontal surface at an average rate of 200 W per square meter. If 20% of this can be converted into useful electrical energy, how large an area is needed to supply 8 kW. (b) Compare this area to that of the roof of a typical house.
Solution:
1. In this problem, the consumption is given in terms of power. Not in terms of energy
(In fact, consumption is indeed given in terms of power. But for solving numeric problems, we can convert it into energy using suitable time intervals)
• It is given that the family uses 8 kW of power
• We can say: The family uses 8000 J of energy every second
2. The available solar energy is also given in terms of power
• It is given that 1 m2 area receives a power of 200 W 
• We can say: 1 m2 area receives 200 J of energy every second
3. Only 20% of this will be converted into electrical energy
• So we can say: 1 m2 area provide an energy of 200 × 0.2 = 40 J (every second)
4. But the requirement is 8000 J in every second
• This can be achieved by increasing the area
• So area required = Total energy required⁄Energy obtained per unit area = 8000⁄40 = 200 m2
5. A typical house has 9 rooms each of average size 3.6 × 3.6 m
• So area of the roof = 9 × 3.6 × 3.6 = 116.64 m2.
    ♦ This is more than 'half of 200'
• So more than half of the energy requirement can be met using solar energy

 Solved example 6.31
A train moves with a constant velocity of 40 ms-1 on a level road. The friction and air resistance has a total magnitude of 3 × 104 N. What is the power of the engine?
Solution:
1. Power is 'work done in one second'
• Let us first find the work done in a suitable time interval say 10 s
2. 'Work done' is force × displacement
On a level road, no work is done by the engine 'for moving the train' (Details here)
3. But work is done against friction and air resistance
• This work is given by: force × distance = 30000 × (40 ×10) = 12000000 J
[∵ uniform velocity of 40 ms-1 for time10 s will give a distance of (40 ×10) m]
4. So total work done during the 10 s = 12000000 J
5. So Power = work done in 1 s = 12000000⁄10 = 1200000 W = 1200 kW

 Solved example 6.32
A train of mass 2 × 105 kg is moving up a hill with a constant velocity of 20 ms-1. The inclination of the hill is [sin-1 (1⁄50)] to the horizontal. For this motion, the engine works at 8 × 105 W. What is the resistance to motion of the train? [g = 9.8 ms-2]
Solution:
1. The engine is working at 8 × 105 W
• That means, the engine is doing 8 × 105 J of work every second
2. Let us consider a suitable time interval say 10 s
• During this 10 s, the engine will do a work of (8 × 105 × 10) = 8 × 106 J
3. Let the train move from A to B during those 10 s
This is shown in the fig.6.63(a) below:
Fig.6.63
4. Let the level difference between A and B be 'h'
• Then in fig.6.63(b), the altitude BC of the right triangle = 'h' m
5. Distance AB = (velocity × time) = (20 × 10) = 200 m   
• In the right triangle ABC, we have: sin θ = h⁄AB = h⁄200
• But sin θ = 1⁄50.
• So we get: h⁄200 = 1⁄50
• So h = 200⁄50 = 4 m.
6. Work done by the engine during those 10 s is the sum of the following two items:
(i) Work done for raising the train through 4 m
(ii) Work done against 'resistance to motion'
• Work done for raising the train through 4 m = mgh = 2 × 105 × 9.8 × 4 = 78.4 × 105 J.  
7. So we can write:
78.4 × 105 + Work done against 'resistance to motion' = 8 × 106 J 
• So we get: 
Work done against 'resistance to motion' = (8 × 106 - 78.4 × 105) = 1.6 × 105 J
• So during those 10 s, 1.6 × 105 J of work is done against 'resistance to motion'   
8. We have: Work done = force × distance
• So we can write:
1.6 × 105 = (force required to overcome 'resistance to motion') × distance 
⇒ 1.6 × 105 = (force required to overcome 'resistance to motion') × AB
⇒ 1.6 × 105 = (force required to overcome 'resistance to motion') × 200 m
⇒ force required to overcome 'resistance to motion' = 800 N 
By Newton's third law,
9. force required to overcome 'resistance to motion' = 'resistance to motion'
• Thus the required answer is 800 N
In the next section we will see collisions

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Tuesday, February 26, 2019

Chapter 6.19 - Different forms of Energy

In the previous section, we completed a discussion on 'conservation of mechanical energy'. Next we will see the 'law of conservation' in other forms of energy such as heat energy, sound energy, nuclear energy etc.,
■ Total Mechanical energy possessed by a body is the sum of it’s kinetic energy and potential energy
• Kinetic energy is based on motion
• Potential energy is based on position or configuration
    ♦ Gravitational potential energy is based on the position (higher level or lower level)
    ♦ Spring potential energy is based on configuration (stretched or compressed configuration)

• Besides kinetic and potential energies, there are many other forms of energies
• Most of them obey the law of conservation. Let us see some of those energies:

Heat energy

1. Consider a block of mass ‘m’ moving on a horizontal surface with a velocity of ‘v’
• Let μk be the coefficient of kinetic friction between the block and the surface
2. We know that the kinetic energy will be used for overcoming friction
• We saw this:
The distance ‘x’ which the block is able to move is given by the equation: $\mathbf\small{\frac{mv^2}{2}=f_k x}$
$\mathbf\small{\Rightarrow \frac{mv^2}{2}=\mu_k mgx}$
$\mathbf\small{\Rightarrow \frac{v^2}{2}=\mu_k gx}$
$\mathbf\small{\Rightarrow x=\frac{v^2}{2 \mu_k g}}$
3. So when the block travels a distance ‘x’ given by the above equation, all the kinetic energy possessed by the block will be used up.
• We are inclined to say: ‘The kinetic energy possessed by the block is lost’
4. But that is not entirely true. Let us analyze:
(i) Just when the block comes to rest, if we touch the underside of the block and also the top of the horizontal surface, we will find that, the temperature of both the surfaces have increased. 
(ii) That means, the kinetic energy was transformed into heat energy
5. Heat is generated when work is done against friction
• This phenomenon helps us to generate heat (even though in small quantities) by rubbing our hands together. We often do this during cold climates.
6. When two surfaces rub against each other, work is being done to overcome friction
• During the rubbing, the molecules on both the surfaces begins to move with greater velocities
• So the kinetic energies of those molecules increase
• This ‘increase in kinetic energy of molecules’ is the cause of ‘increased heat’. 
• We say that the ‘internal energy of the objects increases’  
• We will see more details in a later chapter

Chemical energy

1. Consider the process of making fire using flint stones
2. When two flint stones are rubbed against each other, we are doing mechanical work
• That is., we are 'spending' mechanical energy to rub the stones
3. This mechanical energy is used to overcome friction between the stones
• As a result, the stones get warmer and sparks are produced
4. The sparks are used to ignite a heap of dry leaves
• The fire is then transferred from the dry leaves to a heap of firewood. Thus early men got sustained heat
5. Now, what makes the dry leaves and firewood give us sustained heat?
• The answer is that, chemical energy is stored in dry leaves and firewood
• When ignited, this chemical energy is converted into heat energy
6. So what is chemical energy?
The answer can be written in the following 5 steps:
(i) Substances like dry leaves, firewood, kerosene etc., are known as fuels
(ii) They are made up of atoms
• Those atoms are bonded together by ‘bond-energy’
(iii) When ignited, reaction takes place between those atoms and oxygen
• As a result of this reaction, new substances such as carbon dioxide, ash, etc., are formed
• That means new bonds are formed
(iv) Now consider the two types of bonds:
(a) Bonds in the original fuel and bonds in oxygen (Reactants)
(b) Bonds in the carbon dioxide and bonds in ash (Products)
(v) Bonds in (b) have lesser bond-energy than the bonds in (a)
• That means, during the chemical reaction, some energy is released
• It is this ‘released energy’ that we receive as heat and light
• This 'released energy' is the chemical energy stored in the fuel
7. The reverse can also sometimes happen. That is.,
• Bonds in (iv.b) have a greater bond-energy than the bonds in (iv.a)
• In such cases, we will not receive any energy
• In fact, in such cases, we will have to supply energy for the reaction to occur
• The energy that we supply is absorbed by the reactants
8. If we receive energy, it is called an exothermic reaction
• If we have to supply energy, it is called an endothermic reaction

Electrical energy

1. We know that 'flow of electric current' is required for operating electrical appliances like fans, bulbs, heater etc., 
2. A flow of current takes place from a point of ‘high electric potential’ to a point of ‘low electric potential’
• Here ‘electric potential’ means ‘electrical energy’
• With out this energy, flow will not occur
3. In a bulb, electrical energy is converted into light and heat energies
• In a fan, electrical energy is converted into mechanical energy
• In an electrical iron, electric energy is converted into heat energy
• We will learn more about electrical energy in a later chapter

The equivalence of mass and energy

1. During physical and chemical processes, we generally do not observe a change in mass.
• Some examples:
(i) Consider the physical process of melting of ice:
If we take ‘m’ kg of ice, we will get the exact ‘m’ kg of water when the ice melts
(ii) Consider a normal chemical reaction:
The ‘sum of the masses of the reactants’ will be exactly equal to the ‘sum of the masses of the products’
(ii) So we are inclined to think that, mass is always conserved
■ But Albert Einstein showed that mass and energy are equivalent and are related by the equation:
$\mathbf\small{E=mc^2}$
• Where c is the speed of light in vacuum which is approximately 3 × 108 ms-1
(iv) According to this equation, even a small quantity of mass can give a large amount of energy
• But the conversion of mass into energy involves expensive processes

Nuclear energy

1. The huge amounts of energy produced in the sun can be explained using the equation: $\mathbf\small{E=mc^2}$
• Let us see the details:
2. Four hydrogen nuclei fuse together to form one helium nucleus
• This is a fusion reaction
3. On the reactant side we have four hydrogen nuclei
• On the product side, we have one helium nucleus
4. But the mass of the product is less than the 'sum of the masses of the reactants'
• We can say that when the reaction takes place, there will be a deficiency of mass on the product side
• This deficiency is indicated as Δm
5. This Δm is converted into energy
The amount of energy released = $\mathbf\small{E=(\Delta m)c^2}$
6. Nuclear energy can be obtained by  fission reaction also
• In this reaction, a heavy nucleus of an 'isotope of uranium' is split into lighter nuclei
7. On the reactant side, we have one uranium nucleus
• On the product side, we have more than one nuclei
8. But 'sum of the masses of the products' is less than the mass of the reactant
• Here also, we can say that, there is a deficiency in the mass of the products
• This deficiency is indicated as Δm
9. This Δm is converted into energy
• The amount of energy released = $\mathbf\small{E=(\Delta m)c^2}$
10. If a fission reaction takes place in a controlled manner, the energy can be supplied to work a power plant. Thus electricity can be produced
• Such power plants are called Nuclear power plants
11. If the fission reaction takes place in an uncontrolled manner, huge amounts of energy will be produced in a very short interval of time
• This property of nuclear fission is used for making nuclear weapons

 Solved example 6.27
A person trying to lose weight (dieter) lifts a 10 kg mass, one thousand times, to a height of 0.5 m each time. Assume that the potential energy lost each time she lowers the mass is dissipated. (a) How much work does she do against the gravitational force? (b) Fat supplies 3.8 × 107 J of energy per kilogram which is converted into mechanical energy with a 20% efficiency rate. How much fat will the dieter use up?
Solution:
• Lifting prescribed weights is an effective way for reducing fat
• When the weight is lifted upwards, the dieter is doing work against gravity
• The energy required for doing that work is obtained from the fat
    ♦ Thus the fat gets used up
• When the dieter lifts up the weight, potential energy get stored in that weight
• When the weight is brought down, no work is needed from the dieter
■ But what happens to the stored energy?
• When the weight comes down, the potential energy gets converted into kinetic energy
• That is why the dieter is able to bring down the weight easily
• But this kinetic energy is dissipated as heat, sound etc., 
• So the dieter has to find ‘new energy’ each time the weight is lifted upwards
• This ‘new energy’ required each time is supplied by the fat
Now we can write the steps:
1. Work done for lifting once = mgh = 10 × 9.8 × 0.5 = 49 J
• So work done for lifting 1000 times = 1000 × 49 = 49000 J
2. Given that, energy supplied by 1 kg of fat = 3.8 × 107 J
• But only 20% of this is available for the ‘mechanical work of weight lifting’
• So energy available from 1 kg = 3.8 × 0.2 × 107 = 0.76 × 107
3. So total fat required = $\mathbf\small{\frac{49000}{0.76 \times 10^{-7}}=0.006447 \,\, \text{kg}}$ = 6.447 grams
In the next section we will see power

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Saturday, February 23, 2019

Chapter 6.18 - More Solved examples on Work and Energy

In the previous section, we have completed a discussion on conservation of mechanical energy. We also saw a number of solved examples. We will see a few more in this section.

 Solved example 6.22
To simulate car accidents, auto manufacturers study the collisions of moving cars with mounted springs of different spring constants. Consider a typical simulation with a car of mass 1000 kg moving with a speed of 18 km h-1 on a smooth road and colliding with a horizontally mounted spring of spring constant 6.25 × 103 Nm-1. 
1. What is the maximum compression of the spring?
2. What will be the maximum compression if friction is also taken into account. Coefficient of kinetic friction is 0.5
Solution:
1. We know that 36 kmh-1 is 10 ms-1
$\mathbf\small{\because 36 \times \left(  \frac{1000}{60 \times 60} \right)=10}$
• 18 is half of 36. So 18 kmh-1 is 5 ms-1  
2. Fig.6.57 below shows the simulation:
Fig.6.57
• The car moving with a certain velocity will be having kinetic energy
• After compression of the spring, the velocity of the car becomes zero
• That means, after compression, the kinetic energy is zero
3. But the initial kinetic energy is not lost. 
• Based on the law of conservation of energy, the initial kinetic energy is transformed into another form of energy: The spring potential energy
4. Now we can write the equations:
• Initial kinetic energy = $\mathbf\small{\frac{mv^2}{2}=\frac{1000 \times 5^2}{2}\,\,J}$
• Let ‘x’ be the compression of the spring
• Then potential energy acquired by the spring = $\mathbf\small{\frac{kx^2}{2}=\frac{6250 \times x^2}{2}\,\,J}$
5. Equating the two energies we get: $\mathbf\small{\frac{1000 \times 25}{2}=\frac{6250 \times x^2}{2}\,\,J}$
• Thus we get x = 2 m (This is the answer for part 1)
6. When friction is also present, we cannot just equate the two energies as in (5)
• The original energy will be the same $\mathbf\small{\frac{1000 \times 5^2}{2}\,\,J}$  
• But some energy will be lost for doing work against friction
• So the car will not be able to make the same compression
7. Let x1 be the new compression
• Work done against friction during the travel through this 'x1' m 
= frictional force × distance = $\mathbf\small{\mu_k \times mg \times x_1=0.5 \times 1000 \times 10 \times x_1=5000x_1 \,\, J}$
8. If we add this energy to the right side, the energies will balance. So we get:
$\mathbf\small{\frac{1000 \times 5^2}{2}=\frac{6250 \times x_1^2}{2}+5000x_1}$
$\mathbf\small{\Longrightarrow 25000=6250x_1^2+10000x_1}$
$\mathbf\small{\Longrightarrow 25=6.250x_1^2+10x_1}$
$\mathbf\small{\Longrightarrow 6.250x_1^2+10x_1-25=0}$
Solving this quadratic equation, we get:
x1= 1.35 m (This is the answer for part 2)

 Another approach using force method:
Part 1:
1. The car makes initial contact with the spring with a velocity of  u = 5 ms-1
• The final velocity of the car v = 0
• The acceleration 'a' experienced by the car = resistive force⁄mass
2. Resistive force is the force exerted by the spring
    ♦ But such a force is not uniform
    ♦ Because it depends on the compression 'x'
3. So we take the average force:
$\mathbf\small{\frac{0+kx}{2}=0.5kx =0.5 \times 6250 \times x=3125x \,\,\text{N} }$
• So a = $\mathbf\small{\frac{3125x}{1000}=3.125x\,\,\text{ms}^{-2}}$
• Note that, this is a negative acceleration
4. We can use the equation: $\mathbf\small{v^2-u^2=2as}$
• Substituting the known values, we get: $\mathbf\small{0^2-5^2=2 \times (-3.125x) \times x}$
$\mathbf\small{\Longrightarrow -25=-6.25x^2}$
• So x = 2.0 m
• This is the same answer that we obtained before
5. Consider the following two instants:
(i) Instant at which the car makes initial contact with the spring
(ii) Instant at which the car comes to a stop
• We want the time interval between the above two instants 
6. Let us use the equation: v = u + at
• Substituting the known values, we get: 0 = 5 + (-3.125x)t
⇒ 0 = 5 - 6.25t
⇒ t = 0.8 seconds
7. While using the 'energy method', we are not able to find 'time'
• We are now able to find 'time' because we applied Newton's second law and the equations of motion

 Part 2:
1. The car makes initial contact with the spring with a velocity of  u = 5 ms-1
• The final velocity of the car v = 0
• The acceleration 'a' experienced by the car = resistive force⁄mass
2. Resistive force = the force exerted by the spring + frictional force
• We have already calculated the force exerted by the spring as 3125x1
• Frictional force = μkmg = 0.5×1000×10 = 5000 N
• So total resistive force = (3125x1+5000) N
• So negative acceleration = Force⁄mass = $\mathbf\small{\frac{3125x_1+5000}{1000}}$
4. We can use the equation: $\mathbf\small{v^2-u^2=2as}$
• Substituting the known values, we get: $\mathbf\small{0^2-5^2=2 \times -1 \times(\frac{3125x_1+5000}{1000}) \times x_1}$
$\mathbf\small{\Longrightarrow -25000=-6250x_1^2-10000x_1}$
$\mathbf\small{\Longrightarrow 6250x_1^2+10000x_1 -25000=0}$
$\mathbf\small{\Longrightarrow 6.25x_1^2+10x_1-25=0}$
• Solving this quadratic equation, we get:
x1= 1.35 m
• This is the same answer that we obtained before
5. Consider the following two instants:
(i) Instant at which the car makes initial contact with the spring
(ii) Instant at which the car comes to a stop
• We want the time interval between the above two instants 
6. Let us use the equation: v = u + at
• Substituting the known values, we get: $\mathbf\small{0=5+-1 \times (\frac{3125x_1+5000}{1000})t_1}$
$\mathbf\small{\Longrightarrow 0=5+-1 \times (\frac{3125 \times 1.35+5000}{1000})t_1}$
$\mathbf\small{\Longrightarrow 0=5+-1 \times (\frac{4218.75+5000}{1000})t_1}$
$\mathbf\small{\Longrightarrow 5000=9218.75t_1}$
So t1 = 0.5423 seconds
• Note that, when 'resistance to motion' increases, the 'time required to bring the car to a stop' decreases 
7. While using the 'energy method', we are not able to find 'time'
• In 'force method', we are able to find 'time' because we apply Newton's second law and the equations of motion

 Solved example 6.23
A block of mass 8 kg is released from the top of an inclined smooth surface. See fig.6.58(a) below:
Fig.6.58
The spring constant of the spring is 200 Nm-1. The block comes to rest after compressing the spring by 1 m. Find the total distance traveled by the block before it comes to rest. [g = 10 ms-2]
Solution:
1. Let 'A' be the point of release of the block
'A' is at a height of 'h1' from the datum. This is shown in fig.b
2. After compressing the spring, the block comes to rest at 'B'
'B' is at a height of 'h2' from the datum
3. By the Law of conservation of energy, the total energy must be the same at those two points
• We can write EA = EB
4. Let us write the various energies:
(i) Total energy EA:
• Kinetic energy = 0
• Gravitational potential energy = $\mathbf\small{mgh_1}$ 
• Spring potential energy = 0
■ So total energy EA = $\mathbf\small{mgh_1}$
(ii) Total energy EB
• Kinetic energy = 0
• Gravitational potential energy = $\mathbf\small{mgh_2}$
• Spring potential energy = $\mathbf\small{\frac{k\,x^2}{2}}$ 
■ So total energy EB = $\mathbf\small{mgh_2+\frac{k\,x^2}{2}}$
5. Equating the two energies we get:
$\mathbf\small{mgh_1=mgh_2+\frac{k\,x^2}{2}}$
$\mathbf\small{\Longrightarrow mg(h_1-h_2)=\frac{k\,x^2}{2}}$
• Substituting the known values, we get: $\mathbf\small{8\times 10(h_1-h_2)=\frac{200 \times \,1^2}{2}}$
• Thus we get: (h1-h2) = 1.25 m
6. Draw a vertical through A
• Also draw a horizontal through B
• These vertical and horizontal will meet at a point. Let us call it 'C'
• A, B and C forms a right triangle as shown in fig.c
• Obviously, (h1-h2) = AC = 1.25 m
• Also, angle at B = 30o
7. In the triangle ABC, sin 30 = 1.25⁄AB
So we get: AB = 1.25⁄sin 30 = 1.25⁄0.5  = 2.5 m

 Solved example 6.24
In fig.6.59(a) below, the spring constant of the spring is 1400 Nm-1. It is compressed by 0.1 m using the block and then released. 
Fig.6.59
What is the total distance traveled by the block if:
1. There is no friction
2. If the coefficient of kinetic friction μk = 0.4
[g = 9.8 ms-2]
Solution:
1. Let 'A' be the point of release of the block
• 'A' is at a height of 'h1' from the datum. This is shown in fig.b
2. After release, the block comes to rest at 'B'
• 'B' is at a height of 'h2' from the datum
3. By the Law of conservation of energy, the total energy must be the same at those two points
• We can write EA = EB
4. Let us write the various energies:
(i) Total energy EA:
• Kinetic energy = 0
• Gravitational potential energy = $\mathbf\small{mgh_1}$ 
• Spring potential energy = $\mathbf\small{\frac{k\,x^2}{2}}$ 
■ So total energy EA = $\mathbf\small{mgh_1+\frac{k\,x^2}{2}}$
(ii) Total energy EB
• Kinetic energy = 0
• Gravitational potential energy = $\mathbf\small{mgh_2}$
• Spring potential energy = 0
■ So total energy EB = $\mathbf\small{mgh_2}$
5. Equating the two energies we get:
$\mathbf\small{mgh_1+\frac{k\,x^2}{2}=mgh_2}$
$\mathbf\small{\Longrightarrow mg(h_2-h_1)=\frac{k\,x^2}{2}}$
• Substituting the known values, we get: $\mathbf\small{0.2\times 9.8(h_2-h_1)=\frac{1400 \times \,0.1^2}{2}}$
• Thus we get: (h2-h1) = 3.571 m
6. Draw a vertical through A
• Also draw a horizontal through B
• These vertical and horizontal will meet at a point. Let us call it 'C'
• A, B and C forms a right triangle as shown in fig.c
• Obviously, (h2-h1) = BC = 3.571 m
• Also, angle at A = 60o
7. In the triangle ABC, sin 60 = 3.571⁄AB
So we get: AB = 3.571⁄sin 60 = 4.124 m (This is the answer for part 1)
8. If there is no friction, EB must be equal to EA
• But since there is friction, EB will be less than EA.
• This is because, some energy is lost for doing work against friction
• If we add this 'lost energy' to EB, the energies will balance. That is:
EA = EB + Energy for doing work against friction
9. So our next task is to find the 'lost energy'
• For that, we have to first find the frictional force
• We have already learnt to find that in the case of inclined planes (Details here)
• So frictional force = μk × FN = (μk × mgcosθ) = (0.4 × 0.2 × 9.8 × cos 60) = 0.392 N 
10. So energy lost = Frictional force × Distance
• This time, the block will not reach up to 'B' which we saw in the fig.6.59(b) above
• Let us call the new point 'B1' This is shown in fig.6.60(a) below:
Fig.6.60
• So the distance traveled is AB1
• Thus 'energy lost' = 0.392AB1
11. Next we write the values of EA and EB
(Note that, EB is now EB1)
(i) The position 'A' has not changed. So EA is the same $\mathbf\small{mgh_1+\frac{k\,x^2}{2}}$ that we obtained in 4(i)
(ii) Total energy EB1
• Kinetic energy = 0
• Gravitational potential energy = $\mathbf\small{mgh_3}$
• Spring potential energy = 0
■ So total energy EB = $\mathbf\small{mgh_3}$
12. Now we can use the result in (8):
EA = EB1 + Energy for doing work against friction
$\mathbf\small{mgh_1+\frac{k\,x^2}{2}=mgh_3+0.392AB_1}$
$\mathbf\small{\Longrightarrow mg(h_3-h_1)=\frac{k\,x^2}{2}-0.392AB_1}$
13. From fig.6.60(b), we have: (h3-h1) = B1C1.  
Also, $\mathbf\small{AB_1=\frac{B_1 C_1}{\sin 60}}$
14. So the result in (12) becomes:
$\mathbf\small{mg(B_1 C_1)=\frac{k\,x^2}{2}-0.392 \times \frac{B_1 C_1}{\sin 60}}$
• Substituting the known values, we get:
$\mathbf\small{0.2 \times 9.8(B_1 C_1)=\frac{1400\times 0.1^2}{2}-0.392 \times \frac{B_1 C_1}{\sin 60}}$
$\mathbf\small{\Longrightarrow 1.96(B_1 C_1)=7-0.392 \times \frac{B_1 C_1}{\sin 60}}$
• Thus we get B1C1 = 2.9 m
• So $\mathbf\small{AB_1=\frac{B_1 C_1}{\sin 60}=\frac{2.9}{\sin 60}=3.348\,\,\text{m}}$

 Solved example 6.25
A 1 kg block situated on a rough incline is connected to a spring of spring constant 100 Nm-1 as shown in fig. 6.61(a) below:
Fig.6.61
The block is released from rest with the spring in the unstretched position. The block moves 10 cm down the incline before coming to rest. Find the coefficient of friction between the block and the incline. Assume that the spring has a negligible mass and the pulley is frictionless. [g = 10 ms-2]

Solution:
1. Let 'A' be the point of release of the block
'A' is at a height of 'h1' from the datum. This is shown in fig.b
2. After stretching the spring, the block comes to rest at 'B'
'B' is at a height of 'h2' from the datum
3. Let us write the various energies:
(i) Total energy EA:
• Kinetic energy = 0
• Gravitational potential energy = $\mathbf\small{mgh_1}$ 
• Spring potential energy = 0
■ So total energy EA = $\mathbf\small{mgh_1}$
(ii) Total energy EB
• Kinetic energy = 0
• Gravitational potential energy = $\mathbf\small{mgh_2}$
• Spring potential energy = $\mathbf\small{\frac{k\,x^2}{2}}$ 
■ So total energy EB = $\mathbf\small{mgh_2+\frac{k\,x^2}{2}}$
4. If there is no friction, EB must be equal to EA
• But since there is friction, EB will be less than EA.
• This is because, some energy is lost for doing work against friction
• If we add this 'lost energy' to EB, the energies will balance. That is:
EA = EB + Energy for doing work against friction
5. So our next task is to find the 'lost energy'
• For that, we have to first find the frictional force
• We have already learnt to find that in the case of inclined planes (Details here)
• So frictional force = μk × FN = (μk × mgcosθ) = (μk × 1 × 10 × cos 37) = 7.986μk N 
• So energy lost = Frictional force × Distance = 7.986μk× 0.1 = 0.7986 μk N
6. Now we can use the expression in (4):
EA = EB + Energy for doing work against friction
That is: $\mathbf\small{mgh_1=mgh_2+\frac{k\,x^2}{2}+0.7986 \mu_k}$
$\mathbf\small{mg(h_1-h_2)=\frac{k\,x^2}{2}+0.7986 \mu_k}$
7. Draw a vertical through A
• Also draw a horizontal through B
• These vertical and horizontal will meet at a point. Let us call it 'C'. This is shown in fig.c
• From the right triangle ABC, we have: (h1-h2) = AC  
• Also, $\mathbf\small{AC=AB\, \sin 37}$
= 0.1 × 0.60182 = 0.0602 
8. So the result in (6) becomes:
$\mathbf\small{mg(0.0602)=\frac{k\,x^2}{2}+0.7986 \mu_k}$
Substituting the known values, we get:
$\mathbf\small{1 \times 10 \times(0.0602)=\frac{100\times 0.1^2}{2}+0.7986 \mu_k}$
$\mathbf\small{\Rightarrow 0.602=0.5+0.7986 \mu_k}$
$\mathbf\small{\Rightarrow \mu_k=0.1277}$

 Solved example 6.26
In the fig.6.62(a) below, the two masses A and B are released from rest. 
Fig.6.62
Find an expression for the velocity of the masses at the instant when the masses have traveled a distance of 'h' m. Use both force method and energy method
Solution:
• When the masses are released, B travels downwards with an acceleration
• Since the string is inextensible, A travels upwards with the same acceleration
• Because of the acceleration, the velocity goes on increasing
• We are asked to find the velocity at the instant when the masses have traveled 'h' m
Force method:
1. Fig.b shows the FBD of A
Taking upward forces as positive and downward forces as negative, we get: T-mAg = mAa 
(Where 'a' is the acceleration with which the masses move)
2. Fig.c shows the FBD of B
From this we get: T-mBg = -mBa
⇒ T = mBg - mBa
Substituting this in (1), we get: mBg - mBa - mAg = mAa
⇒ mBg - mAg = mAa + mBa
⇒ (mB-mA)g = (mB+mA)a
$\mathbf\small{\Rightarrow a=\left(\frac{m_B-m_A}{m_B+m_A}\right)g}$
3. We can use the equation: $\mathbf\small{v^2-u^2=2as}$
• Substituting the known values, we get: $\mathbf\small{v^2-0^2=2ah}$
$\mathbf\small{\Rightarrow v=\sqrt{2ah}}$
Where $\mathbf\small{a=\left(\frac{m_B-m_A}{m_B+m_A}\right)g}$

 Energy method:
1. The initial elevations of A and B are shown in fig.d
Let us write the energies:
(i) Total energy EA1:
• Kinetic energy = 0  
• Gravitational potential energy = $\mathbf\small{m_Agh_{A1}}$ 
■ So total energy EA1 = $\mathbf\small{m_Agh_{A1}}$
(ii) Total energy EB1:
• Kinetic energy = 0
• Gravitational potential energy = $\mathbf\small{m_Bgh_{B1}}$ 
■ So total energy EB1 = $\mathbf\small{m_Bgh_{B1}}$
2. The final elevations (after travelling 'h' m) of A and B are shown in fig.e
Let us write the energies:
(i) Total energy EA2:
• Kinetic energy = $\mathbf\small{\frac{m_A\,v^2}{2}}$
    ♦ Where 'v' is the velocity with which the blocks move when they just pass the elevations shown in fig.e
• Gravitational potential energy = $\mathbf\small{m_Agh_{A2}}$ 
■ So total energy EA2 = $\mathbf\small{\frac{m_A\,v^2}{2}+m_Agh_{A2}}$
(ii) Total energy EB2:
• Kinetic energy = $\mathbf\small{\frac{m_B\,v^2}{2}}$
• Gravitational potential energy = $\mathbf\small{m_Bgh_{B21}}$ 
■ So total energy EB2 = $\mathbf\small{\frac{m_B\,v^2}{2}+m_Bgh_{B2}}$
3. Total initial energy = Ei = (EA1 + EB1) = $\mathbf\small{m_Agh_{A1}+m_Bgh_{B1}}$ 
• Total final energy = Ef = (EA2 + EB2) = $\mathbf\small{\frac{m_A\,v^2}{2}+m_Agh_{A2}+\frac{m_B\,v^2}{2}+m_Agh_{B2}}$
4. Applying the Law of conservation of energy, we have: Ei = Ef.
• Thus we get: $\mathbf\small{m_Agh_{A1}+m_Bgh_{B1}=\frac{m_A\,v^2}{2}+m_Agh_{A2}+\frac{m_B\,v^2}{2}+m_Bgh_{B2}}$
$\mathbf\small{\Rightarrow m_Bgh_{B1}-m_Bgh_{B2}=\frac{m_A\,v^2}{2}+m_Agh_{A2}-m_Agh_{A1}+\frac{m_B\,v^2}{2}}$
$\mathbf\small{\Rightarrow m_Bg(h_{B1}-h_{B2})=\frac{v^2}{2}(m_B+m_A)+m_Ag(h_{A2}-h_{A1})}$
• But (hB1 - hB2) = (hA2 - hA1) = h
• Thus we get: $\mathbf\small{m_Bg(h)=\frac{v^2}{2}(m_B+m_A)+m_Ag(h)}$
$\mathbf\small{(m_B-m_A)gh=\frac{v^2}{2}(m_B+m_A)}$
$\mathbf\small{\Rightarrow v^2=2 \left(\frac{m_B-m_A}{m_B+m_A}\right)gh}$
$\mathbf\small{\Rightarrow v=\sqrt{2ah}}$
• Where $\mathbf\small{a=\left(\frac{m_B-m_A}{m_B+m_A}\right)g}$
• This is the same expression that we obtained by the force method
So we have completed a discussion on conservation of mechanical energy. Next we will see the law of conservation in other forms of energy such as heat energy, sound energy, nuclear energy etc.,   

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