# W.P.energy Final Colour

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## Full text

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01. Concept of Energy 02 - 03 02. Work: Definition and Calculation 03 - 12 03. Work Energy Theorem and its Application 13 - 24

04. Power 25 - 31

05. Potential Energy and Energy Conservation 32 - 47

06. More on Conservative forces 48 - 54

07. A Note on Mass and Energy (Optimal) 55 - 59

08. Miscellaneous Examples 60 - 87

## & Energy

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Introduction:

Any body (or an assembly of bodies) represents, in fact, a system of mass points, or particles. If a system changes with time, it is said that its state varies. The state of a system is defined by specifying the positions and velocities of all constituent particles.

Experience shows that if the laws of forces acting on a system and the state of the system at a certain initial moment are known, the motion equations can help predict the subsequent behaviour of the system.

However, an analysis of a system’s behaviour by the use of motion equations requires so much effort (due to the complexity of the system itself), that a comprehensive solution seems to be practically impossible. Moreover, such an approach is absolutely out of the question if the laws of acting forces are not known. Besides, there are some problems in which the accurate consideration of motion of individual particles is meaningless (example: gas).

Under these circumstances the following question naturally comes up: are there any general principles following from Newton’s laws that would help avoid these difficulties by opening up some new approaches to the solution of the problem.

It appears that such principles exist. They are called conservation laws.

As we have already discussed, the state of a system varies in the course of time as that system moves. However, there are some quantities, state functions, which possess the very important and remarkable property of retaining their values with time provided you don’t change your frame. Among these constant quantities, energy, linear

momentum and angular momentum play the most significant role.

The laws of conservation of energy (There used to be a separate conservation law or more until Einstein unified it with energy), momentum and angular momentum fall into the category of the most fundamental principles of physics, (when you will study the law of conservation of linear momentum you will find that if is just a restatement of Newtons’ second law: F =d mv dt( !)/ applied to a system. Angular momentum too, is some thing similar, conservation of energy on the other hand, is a novel concept). These laws have become even more significant since it was discovered that they go beyond the scope of mechanics and represent universal laws of nature. In any case, no phenomena have been observed so far which do not obey these laws. They work reliably in all quarters: in the field of elementary particles, in outer space, in atomic physics and in solid state physics.

Having made possible a new approach to treating various mechanical phenomena, the conservation laws turned into a powerful and efficient tool of research used by physicists. The importance of the conservation principles is due to several reasons:

1. The conservation laws do not depend on either the paths of particles or the nature of acting forces. Consequently, they allow us to draw some general and essential conclusions about the properties of various mechanical processes without resorting to their detailed analysis by means of motion equations. 2. Since the conservation laws do not depend on the nature of the acting forces, they may be applied even when the forces are not known. In these cases the conservation laws are the only tools for research remaining. This is the present trend in the physics of elementary particles.

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3. Even when the forces are known precisely, the conservation laws can help substantially to solve many problems of motion of particles. Although all these problems can be solved with the use of motion equations (and the conservation laws provide no additional information in this case), the utilization of the conservation laws very often allows the solution to be obtained in the most straight-forward and elegant fashion, obviating cumbersome and tedious calculations.

We shall begin examining the conservation laws with the energy conservation law, having introduced the concept of energy and work.

Section - 1 CONCEPT OF ENERGY

Energy : It is possible to give a numerical rating, called energy, to the state of a physical system. The total energy is found by adding up contributions from characteristics of the system such as motion of objects in it, heat content of the objects (though that can also be attributed to mechanical vibrations of particles), and the relative positions of objects that interact via forces. The total energy of a closed system always remains constant. Energy can not be created or destroyed, but only transferred from one system to another.

Energy comes in a variety of forms, and physicists didn’t discover all of them right away. They had to start somewhere, so they picked one form of energy to use as a standard for creating a numerical energy scale. One practical approach is to define an energy unit based on heating of water. The SI unit of energy is the joule, J, named after the British physicist James Joule. One joule is the amount of energy required in order to heat 0.24 g of water by 1°C.

Note that heat, which is a form of energy, is completely different from temperature. In standard, formal terminology, there is another, finer distinction. The word heat is used only to indicate an amount of energy that is transferred, whereas thermal energy indicates an amount of energy contained in an object.

Once a numerical scale of energy bas been established for some form of energy such as heat, it can easily be extended to other types of energy. For instance, the energy stored in one gallon of gasoline can be determined by putting some gasoline and some water in an insulated chamber, igniting the gas, and measuring the rise in the water’s temperature. (The fact that the apparatus is known as a bomb calorimeter will give you some idea of how dangerous these experiments are if you don’t take the right safety precautions). Here are some examples of other types of energy that can be measured using the same units of joules.

Type of energy:

• chemical energy released by burning • energy required to break an object

• energy required to melt a solid substance • chemical energy released by digesting food

• raising a mass against the force of gravity • nuclear energy released in fission, etc. Textbooks often give the impression that a sophisticated physics concept was created by one person who had an inspiration one day, but in reality it is more in the nature of science to rough out an idea and then gradually refine it over many years. The idea of energy was tinkered with from the early 1800’s onwards, and new types of energy kept getting added to the list.

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To establish a new form of energy, a physicist has to

1. show that it could be converted to and from other forms of energy, and

2. show that it is related to some definite measurable property of the object, for example its temperature, motion, position relative to another object, or being in a solid or liquid state.

For example, energy is released when a piece of iron is stored in water, so apparently there is some form of energy already stored in the iron. The release of this energy can also be related to a definite measurable property of the chunk of metal: it turns reddish-orange. There has been a chemical change in its physical state, which we call rusting.

Although the list of types of energy kept getting longer and longer, it was clear that many of the types were just variations on a theme. There is an obvious similarity between the energy needed to melt ice and to melt butter, or between rusting of iron and many other chemical reactions. All the types of energy can be reduced to a very small number by simplifications.

Section - 2 WORK: DEFINITION AND CALCULATION

The concept of work:

The mass contained in closed system is a conserved quantity, but if the system is not closed, we also have ways of measuring the amount of mass that goes in or out.

We often have a system that is not closed, and would like to know how much energy comes in or out. Energy, however, is not a physical substance like water, so energy transfer can not be measured by the same kind of meter which we used to measure flow of water. How can we tell, for instance, how much useful energy a tractor can

put out on one tank of gas?

The law of conservation of energy guarantees that all the chemical energy in the gasoline will reappear in some form, but not necessarily in a form that is useful for doing farm work. Tractors, like cars, are extremely inefficient, and typically 90% of the energy they consume is converted directly into heat, which is carried away by the exhaust and the air flowing over the radiator. We wish to distinguish energy that comes out directly as heat from the energy that serves to accelerate a trailer or to plow a filed, so we defined a technical meaning of the ordinary work to express the distinction.

Definition of work:

Work is the amount of energy transferred into or out of a system, not taking into account energy transferred by heat conduction.

[Based on this definition, is work a vector, or a scalar? What are its units?]

The conduction of heat is to be distinguished from heating by friction. When a hot potato heats up your hands by conduction, the energy transfer occurs without any force, but when friction heats your car’s brake shoes, there is a force involved. The transfer of energy with and without a force are measured by completely different methods, so we wish to include heat transfer by frictional heating under the definition of work, but not heat transfer by conduction. The definition of work could thus be restated as the amount of energy transferred by forces. Work done by a constant force:

Work done by a constant force is defined as product of the force and the component of the displacement

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F

S m

m

fig 5.1.

F

Consider the situation shown in figure 5.1. A constant force F is applied on a block of mass m along the horizontal direction. If the block moves by a distance S on the horizontal surface on which it is placed, as shown in figure, then the work done by the force F is defined as

=

w F S ...(1).

If the force and the displacement are not along the same direction, as shown in figure 5.2, then work done by force F is calculated by multiplying the force and the component of the displacement along the force, as shown in figure 5.3, therefore, for the given case, work done by force F is

F S m m fig. 5.2 F θ S co s θ m m θ S F(const.) fig. 5.3 cos w= ⋅F S θ ⇒ w=F S⋅ ⋅cosθ ...(2)

Here you should note that work done by the force F can also be written as

( cos )

w= F θ ⋅S

and we know that Fcosθθθθθ is the component of F along the displacement, as shown in figure 5.4.

F cos θ m

m θ S

F(const.)

work done by

= force along displacement × displacement

F

fig. 5.4

Hence, work done by a constant force can also be defined as the product of the displacement and the

component of the force along the displacement.

In vector form equations (1) and (2) can be generalized as ⋅

= ! !

w F S ...(3)

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I would like to emphasize that S! is the displacement of the point of application of the force F!. Note: For a constant force F!, in equation (3) you should notice the following:

• When F! ⊥S,! i.e., θ = °90 , wf =0.

• When θ∈(0, 90 ), ° wf >0.

• When θ∈(90 ,180 ), ° ° wf <0.

• When θ =180 , ° wf = − ⋅F S • When θ = °0 , wf = ⋅F S

• In a closed path work done by a constant force is zero. (∵ S! !=0).

(wf denotes work done by the constant force F and S is the magnitude of the displacement S ! ) Equation (3) refers only to the work done on the particle by a particular force F!. The work done on the particle by the other forces must be calculated separately. The total work done on the particle is the sum of the work done by the separate forces.

( = 0°)θ F S F S m fig. 5.5

When θ is zero, as shown in figure 5.5, the work done by F! is simply F × S, in agreement with equation (1). Thus, when a constant horizontal force draws a body horizontally, or when a constant vertical force lifts a body vertically, the work done by the force is the product of the magnitude of the force by the distance moved.

When θ is 90°, as shown in figure 5.6, the force has no component in the direction of motion. That force then does no work on the body. For instance, the vertical force holding a body a fixed distance off the ground does no work on the body, even if the body is moved horizontally over the ground. Also the centripetal force acting on a body in motion does no work on that body because the force is always at right angles

to the direction in which the body is moving.

F S m fig. 5.6 ( = 90°)θ F S

Of course, a force does no work on a body that does not move, for its displacement is then zero.

As I have already mentioned, the work done by a constant force can be calculated in two different ways: Either we multiply the magnitude of the displacement by the component of the force in the direction of the displacement or we multiply the magnitude of the force by the component of the displacement in the direction of the force. These two methods always give the same result.

Work is a scalar, although the two quantities involved in its definition, force and displacement, are vectors. We define the scalar product of two vectors as the scalar quantity that we find when we multiply the magnitude of one vector by the component of a second vector along the direction of the first. Equation (3) shows that work is such a quantity.

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Work can be either positive or negative. If the particle on which a force acts has a component of motion opposite to the direction of the force, the work done by that force (Fig. 5.7) is negative. This corresponds to an obtuse angle, between the force and displacement vectors. For example, when a person lowers an object to the floor, the work done on the object by the upward force of his hand holding the object is negative.

F S m fig. 5.7 ( > 90°)θ θ

Our special definition of the word work does not correspond to the daily usage of the term. This may be confusing. A person holding a heavy weight at rest in the air may say that he is doing hard work- and he may work hard in the physiological sense-but from the point of view of physics we say that he is not doing any work. We say this because the applied force causes no displacement. The word work is used only in the strict sense of equation (3). The unit of work is the work done by a unit force in moving a body a unit distance in direction of the

force. In the mks system the unit of work is 1 newton-meter, called 1 joule. Work done by gravity (near earth surface):

Near the earth’s surface the gravitational force acting on a body due to attraction of the earth can be assumed to be constant. Therefore work done by gravity can be calculated using equation (3), whatever be the path of motion of the body. S cosθ path 2 1 h fig. 5.8 mg! mg! mg! mg! S

### !

S S cosθ mg path 2 1 h fig. 5.9 θ

Consider the situation shown in figure 5.8. A particle of mass m is moved on an arbitrary path in a vertical plane. As the particle is moved from point 1 to point 2, its weight acting on it at different positions during its motion is also shown in figure. From figure it is clear that work done by gravity can be found by multiplying mg (magnitude of the force) by the downward displacement h of the body because the component of the displacement of the body along the force is h only. If you are not satisfied with the discussion above and want to calculate it mathematically, then you can proceed according to the following way:

Work done by gravity when particle moves from point 1to point 2 as shown in figure 5.9 is

mg s = ! !⋅ g w cos mg s θ = ⋅ [∵ mg! is constant] . mg h =

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Hence, we have got the same result as we had predicted earlier.

Therefore, when a particle comes down by a distance h through any path, work done by gravity is given as

(downward motion)

= g

w mgh

...(4)

You should note that this work done is independent of the path of the particle. You should also note that “mgh” is work done by gravity only, not by all forces acting on the particle when it moved from point 1 to point 2. What would be the work done by gravity when the particle moves from point 2 to point 1? In this case the displacement of the particle has a component of length h in the vertical direction but this component is in the opposite direction of mg!, hence, work done by gravity for this case is

(upward motion)

= − g

w mgh

...(5)

You should note that this expression is also independent of the path followed by the particle while moving from point 2 to point 1.

Now, let me propose a common equation for the equations (4) and (5). When a particle of mass m moves near earth surface, work done on it by gravity is given by

= − ∆

g

w mg h ...(6)

where ∆h is the change in height of the particle. If the particle goes up by a height h then h = +h and g

w = −mgh, which is in accordance with equation (5). If the particle comes down by a height h then h = –h and work done by gravity, wg = +mgh, which is in accordance with the equation (4). Hence, for all cases we can use equation (6). The only restriction while using equation (4), (5) or (6) is that ‘mg’ must be uniform over the path for which work done is required.

Note that if the particle moves from 1 to 2 and then from 2 to 1, then work done by gravity is zero. Hence, we can say that work done by gravity in a closed path is zero.

Work done by a variable force:

θ 2

1

d s!

fig 5.10.

When the force varies over the path of motion of a particle (it can vary in magnitude or direction or both), then to calculate work done by the force we divide the path of the particle in many infinitesimally small intervals and calculate the work done in each interval and by adding the work done for all these small intervals we get the work done over the segment of the path under consideration. In figure 5.10, such an interval, d s!, is shown. As the interval is very small, we can assume that in this interval force is constant and hence work done by F! in this interval can be written as

= ⋅! !

dw F ds ...(7)

where d s! is the displacement of the particle for this interval. As the particle moves from point 1 to point 2 on the path shown above, the net work done by the force F! is calculated by adding work done in all subintervals of the path from point 1 to point 2. Therefore, net work done by F! is

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sum of all 's w= dww=

dw ⇒ =

### ∫

2 1 ! ! w F d s ...(8)

You should note that when you apply equation (8) for a constant force you get the same result as given by equation (3).

For a one dimensional case equation (8) can be modified by replacing d s! with dx and replacing F! with F(x). Therefore, we get, ( ) f i x x w=

### ∫

F x dx...(9)

For a 3-dimensional case equation (8) can be expanded as follows: fin in w=

F ds!⋅ !

ˆ ˆ ˆ

. ˆ ˆ ˆ

### )

fin x y z in F i F j F k dxi dyj dzk =

+ + + +

### )

fin x y z in F dx F dy F dz =

### ∫

⋅ + ⋅ + ⋅ ⇒ f f f i i i x y z x y z x y z w=

F dx⋅ +

F dy⋅ +

### ∫

F dz...(10)

If components of F! along x, y and z are constant, then equation (10) reduces to

∆ ∆ ∆

x y z

w F= ⋅ x F+ ⋅ y F+ ⋅ z ...(11)

There is an alternate way of looking at work done by variable forces which would prove to be a powerful method as we will proceed with the topic. Suppose we considerF! as the sum of its components along the tangent and the normal to the path of motion of the particle. In that case equation (8) gives

fin

in

w=

F d s!⋅ !

### )

ˆ and are unit vectors along the tangent ˆ and the normal to the path, respactively, at the point under consideration

ˆ ˆ fin t n in t n F t F n d s     = ⋅ + ⋅ ⋅    

### ∫

!

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ˆ ˆ ( ) ( ) fin t n in F t d s F n d s   =

⋅ ! + ⋅ !

### ∫

fin t in w = F dsnˆ

### ⊥

#d s!and t d sˆ⋅ !=ds ...(12) Hence, work is done only by the tangential component of a force.

Example – 1 WORK DONE BY SPRING FORCE

A light spring of spring constant k is stretched from an initial deformation xi to a final deformation xf quasistatically (i.e., equilibrium is always maintained). Find the work done by

(a) spring force (b) external agent

Solution: In figure 5.11 the spring is shown at some arbitrary deformation x. As the spring always exerts restoring force, at the shown moment it is applying a force in the opposite direction of x. Therefore,

= − sp

F kx (always true) where k is the spring

natural length xi x f x dx F F fig 5.11. ext sp

constant of the spring.

To increase the deformation the external must exert a force in the direction of x and the force exerted by the external agent, Fext, must have a magnitude equal to that of Fsp because equilibrium is always

maintained. Therefore,

= +

ext

F kx

 

true only if equilibrium is maintained

If the spring is further deformed by dx from the shown position then work done by the spring force is

( )

sp

dw = −kx dx

and work done by external agent is

( )

ext

dw = +kx dx

While deforming the spring from xi to xf, we get, total work done by the spring, f i x sp sp x w =

dw = −k

### ∫

x dx⋅ ⇒ wsp = −12kx2f12kxi2   ...(13)

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and total work done by external agent f i x ext ext x w =

dw = +k

### ∫

x dx⋅ ⇒ wext = 12kx2f12kxi2 ...(14)

Initially if the spring has its natural length then substituting xi=0, equation (13) and (14) give

= −1 2 2 sp f w kx ...(15) and = +1 2 2 ext f w kx ...(16)

Note: You must notice the following

• Equations (13) and (15) are true for all cases but equations (14) and (16) are true only if equilibrium is maintained.

• Work done by the spring is independent of path, it depends only upon initial and final position. [See equation (13) and (15)]

• You should note that

### ∫

F x dx( )⋅ denotes area under F(x) curve when it is plotted against x.

• From figure. 5.12, shaded area sp w = fig 5.12. xi xf A1 A2 W – kxi – kxf F x( ) x F x kx F ( )= – = sp sp 1 2 A A = + ( kxi)(xf xi) = − − +1( )( ) 2 −kxf +kxi xfxi 1 2 1 2 2kxf 2kxi   = −  

• If we put xi =xf in equation (13), then we get wsp =0. Therefore, we can say that work done by

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Work Done by Two or More than two forces:

Suppose there are n forces F F! !1, 2,...,F!n acting on a particle of mass m, as shown in figure 5.13. If the particle suffers a displacement ds! in the next time interval dt, then the net work done on the particle is the sum of work done by all the forces acting on it. Hence, we have net work done on the particle

1 2 ... n dw=dw +dw +dw+ +dw 1 2 3 ... n F ds F ds F ds F ds = ⋅ + ⋅ + ⋅ +! ! ! ! ! ! + ⋅! ! F1 F3 F2 Fn m fig. 5.13 ! ! ! ! where dwi= ⋅F ds!i ! is work done by the force F!i.

For a finite displacement, net work done on the particle is

... = 1+ 2+ + n w w w w ...(17) 1 2 ... n F ds F ds F ds =

! ⋅ +!

! ⋅ +! +

! ⋅ !

F1 F2 ... Fn

ds =

### ∫

! + ! + + ! ⋅ ! ⇒ w F= !netds! ...(18)

Therefore, if many forces are acting on a particle, then we have two ways to find out net work done on the particle:

1. find work done by each force individually and then find the sum of work done by all forces;

2. find the net force, sum of all forces, acting on the particle and then find work done by this net force. When you are applying the above approach for a particle then no attention is needed but while applying this concept on a system of particles or on a body you need to be cautious. If points of application of different forces have different displacements, then equation (17) can not be reduced to equation (18). Hence, in such a case you are left with only one option and that is of using equation (17).

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TRY YOURSELF - I Q. 1 No work is done by a force on an object if

(a) the force is always perpendicular to its velocity (b) the force is always perpendicular to its acceleration

(c) the object is stationary but the point of application of the force moves on the object (d) the object moves in such a way that the point of application of the force remains fixed.

Q. 2 Find the work a boy of weight 55 kg has to do against gravity when climbing from the bottom to the top of a 3.0 m high tree. (g = 10 m/s²).

Q. 3 A body is thrown on a rough surface such that friction force acting on it varies linearly with distance traveled as f = ax + b. Find the work done by the friction on the box if before coming to rest the box travels a distance s.

Q. 4 A force is given by 2

F=kx , where x is in meters and k=10Nt m/ 2. What is the work done by this force when it acts from x = 0 to x = 0.1 m?

Q. 5 A body is acted upon by a force which is inversely proportional to the distance covered. The work done will be proportional to :

(a) s (b) s²

(c) s (d) ln s

Q. 6 A force F acting on a particle varies with the position x as shown in figure. Find the work done by this force in displacing the particle from

(a) x= −2m to x=0 10 x m( ) F n( ) –10 –2 2 (b) x=0 to x=2m.

Q. 7 Two unequal masses of 1 kg and 2 kg are attached to the two ends of a light inextensible string passing over a smooth pulley as shown in figure. If the system is released from rest, find the work done by the string on both the blocks in 1 s. (Take g = 10 m/s²)

1 kg 2 kg

Q. 8 A particle moves from a point r!1 =(2 )m iˆ+(3 )m jˆ to another point r!2 =(3 )m iˆ+(2 )m jˆ during which a certain force \$!F =(5 )N iˆ+(5 )N jˆ acts on it. Find the work done by the force on the particle during the displacement.

Q. 9 A block of mass m moving at a speed v compresses a spring through a distance x before its speed is halved. Find the spring constant of the spring.

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Section - 3 WORK ENERGY THEOREM AND ITS APPLICATION

As we have already discussed that the normal component of a force does no work, the work is done only by the tangential component of a force. The tangential component is related to rate of change of speed of the particle. By the Newton’s 2nd law, Ft mdv,

dt

=

### ∑

where v is speed of the particle and Ft denotes the sum of tangential components of all forces acting on the particle. We can now think of speed as a function of the distance s measured along the curve (as shown in figure 5.14) and apply the chain rule for derivatives:

dv dv ds dv v dt = ds dt⋅ = ⋅ds ds dt v= s starting point fig. 5.14t dv F m v ds ∑ = ⋅ ⋅

where we have used the fact that ds

dt is just the speed v. The work done by resultant force is thus

fin fin net t in in dv w F ds mv ds ds =

⋅ =

### ∫

⋅ ⋅ fin fin in in mv dv m v dv =

⋅ =

### ∫

⋅ or 1 2 1 2 2 2 net f i w = mvmv ...(19) The quantity 1 2

2mv has the same unit as that of work and hence it is a form of energy. As it depends upon the

speed of the particle, it is defined as the kinetic energy, k, of the particle. Hence equation (19) can be rewritten as

net f i

w =kk

wnet = ∆k ...(20)

Therefore, net work done on a particle is equal to the change in kinetic energy of the particle. This theorem is known as the work energy theorem.

If positive work is done on a particle then kinetic energy of the particle increases. If work done on a particle is negative then kinetic energy of the particle decreases. If no work is done on a particle then kinetic energy of the particle remains unchanged and hence speed of the particle remains constant. At this juncture you should recall the

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very basic definition of work. We had defined it as energy transferred by forces and here the work energy theorem proves this definition. Now, it is clear that all the work done on a body goes in increasing the kinetic energy of the body.

Using equation (12) and (20), we can get

dw=dk t F ds = ⋅ ⇒ t = dk F ds ...(21)

Therefore, the derivative of kinetic energy with respect to distance covered along the path gives the tangential component of the net force acting on a particle. The normal component of the net force can be found by multiplying the mass of the particle by the centripetal acceleration of the particle. We have already learnt that, in order to change the speed, we need a tangential component of the force (in circular motion). It is in accordance with equation (21). If the tangential force is zero, then no work is done on a particle and hence its kinetic energy remains unchanged or we can say that the speed of the particle remains unchanged. Therefore to change the speed there must be a component of the net force along the tangent to the path and this is exactly what we discussed while covering non uniform circular motion.

Example – 2

A small ball released from rest from a height h on a smooth surface of varying inclination, as shown in figure 5.15. Find the speed of the ball when it reaches the horizontal part of the surface.

smooth

h

fig. 5.15 Solution: Let the speed of the ball when it

reaches the horizontal part of the surface be v0, as shown in figure 5.16(b). fig. 5.16(a) v N mg m v0 fig. 5.16(b)

The position of the ball at some arbitrary point of the inclined part of the surface is shown in figure 5.16(a). During the entire journey there are only two forces acting on the ball:

(1) Its weight; (2) Normal contact force from the surface. Applying the work energy theorem between the moments when it was released from the rest and when it reaches the horizontal surface, we get.

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net w = ∆k ⇒ 0 N g w + w 0 i f k k = −

work done by gravity;

work done by normal contact force

g N w w →       ⇒ 2 0 1 0 0 2 mgh+ = mv − from equation (6), g w mgh    =    ⇒ v0 = 2gh ...(22)

As the normal contact force is always perpendicular to the direction of motion, work done by it is zero.

When a ball falls freely from rest by a distance h, its speed is the same as obtained in equation (22). During free fall the only force doing work is gravity.

Example – 3

In the previous problem suppose that an uncompressed spring is fixed on the horizontal part of the surface, as shown in figure 5.17. Now, if the particle is released from rest from the position shown in the figure, find the maximum compression in the spring if its spring constant

is k.

fig. 5.17 m

k h

Solution: When compression in the spring is maximum, speed of the ball must be zero at that moment, as shown in figure 5.18 (c). If x0 be the maximum compression in the spring then applying work energy theorem on the ball between the instants when it was released from rest and when the compression in the spring is maximum, we get

v N mg m N mg v0 x 0 v=0

fig. 5.18(a) fig. 5.18(b) fig. 5.18(c)

net w = ∆k ⇒ 0 N g w + w 0 f sp w k + = 0 i k

− wsp is work done by spring

⇒ 2 0 1 0 2 mgh kx + − = ⇒ 0 2mgh x k =

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Example – 4

In the previous example suppose the spring is removed from the horizontal part of the surface and this part is made rough. If the coefficient friction between the rough part of the surface and the ball is µ, find the distance covered by the ball on the horizontal part of the surface before it comes to rest.

Solution: Let the ball comes to rest having covered a distance s on the horizontal part of the surface. During the motion of the ball only three forces act on the ball: (1) Gravity; (2) Normal contact force; (3) Frictional

force. v

N

mg Fig. 5.19(a)

Applying work energy theorem between the instants when the ball was released from rest and the ball comes to rest on the horizontal part, we get,

net w = ∆kwg +wN +wfr =kfki (b) v N mg µmg sm ooth rough Fig. 5.19 ⇒ +mgh+ −0 µmg s⋅ = −0 0 ⇒ s= hµ Example – 5

A block of mass m suspended vertically by a light spring of spring constant k is released from rest when the spring is in its natural length state, as shown in figure 5.20. Find the maximum elongation in the spring.

m

k naturallength

“released from rest”

Fig 5.20. Solution: As the block is released from rest it falls downwards

due to its own weight but as it moves downwards its downward acceleration decreases due to the upward force exerted by the spring. Initially the spring force is zero, hence the downward acceleration is maximum

(=g). But as the spring elongates the net downward force decreases and hence the downward acceleration also decreases. After some time the upward spring force becomes equal to the weight of the block. This position is known as equilibrium position because net force in this position is zero. You should note that at this position speed of the block is not zero. In fact, at this position speed of the block is maximum, because before arriving to this position spring force is smaller than the weight of the block and hence the speed of the block is increasing in the downward direction.

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When the block reaches the equilibrium position, the net force on it becomes zero and hence its acceleration also becomes zero. Therefore, speed becomes maximum because it can not increase further. Although acceleration of the block is zero in this position, it will continue moving in the downward direction due to its downward velocity.

mg a = g v=0 x0 mg k kx x a v speed)(increasing kx0 mg vmax kx a x mg v (decreasing speed) 2kx0 mg =2mg maximum elongation position equilibrium position natural length position fig 5.21. a = 0 a = g 2x =0 2mg k =xmax v=0

As the block passes the equilibrium position, the spring force exceeds the weight of the block and hence the block starts decelerating and then after moving some distance with decreasing speed, it momentarily comes to rest. Obviously when the ball comes to rest, elongation in the spring is maximum. If xmax is the maximum elongation in the spring then applying work energy theorem on the block between the moment when it was released and when the elongation is maximum, we get

net w = ∆kwg+wsp =kfki ⇒ 2 max max 1 0 0 0 2 mg x kx + ⋅ − = − = ⇒ max 2mg x k = ⇒ xmax =2x0

where x0 is the elongation in the spring in the equilibrium position. Note: • At x=xmax block momentarily comes to rest.

• At this position velocity of the block is zero but acceleration is not zero. It has an upward acceleration of magnitude g, as shown in figure 5.21.

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• When elongation is maximum in the spring its speed is zero which can be proved very easily if you proceed mathematically rather than analyzing exactly what’s happening there. When elongation in the spring is maximum, distance of block from natural length state, x, as shown in figure 5.21, is also maximum. Hence, we have

0 0.

dx

v

dt = ⇒ =

Example – 6

A small ball of mass m is suspended by means of a light thread of length l. When the ball is hanging vertically it is given a horizontal speed u, as shown in figure 5.22. Find the speed of the ball and tension in the thread supporting the ball when the

thread makes an angle θ with the vertical.

m u

fig. 5.22 O

Solution: At some angular displacement θ, the situation is shown in figure 5.23. Applying work energy theorem on the ball, we get,

all w = ∆kwT+wg =kfki θ mg V T v l 2 l ω θ x s fig. 5.23 cos (1 cos ) x l l l θ = – = − θ O ⇒ 1 2 1 2 0 2 2 g w mv mu + = − T

### ⊥

# V ⇒ 1 2 1 2 2 2 mg h mv mu − ∆ = − ⇒ 1 2 1 2 . 2 2 mg x mv mu − = − ⇒ 2 2 2 v =ugx ⇒ 2 2 2 (1 cos ) v =ugl − θ (i) ⇒ v= u22 (1 cos )gl θ

At the position shown in figure 5.23, the ball is moving in a vertical circle of radius l and has a speed v, therefore, its radial acceleration is v l2/ . Applying Newton’s 2nd law along the radial direction at this moment, we get

net F =maT mgcos mv2 l θ − =

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T mgcos mu2 2mg(1 cos ) l

θ θ

= + − − [using equation (i)]

⇒ 2 3 cos 2 u T mg mg m l θ = − +

* You should notice that we had done this problem in the previous chapter also (CIRCULAR MOTION). That time we solved this problem by analyzing the varying tangential acceleration of the ball.

Example – 7 R m smoo th fig. 5.24

A small ball is released from the top of a smooth hemispherical surface fixed on a horizontal plane as shown in figure 5.24. If m be the mass of the ball and R be the radius of the hemispherical surface, find the speed of the ball and normal contact force between the ball and the hemispherical surface as a function of θ, where θ is the angle between radial position of the ball with respect to the centre of the hemisphere and the vertical direction.

Solution: The position of the ball when it moves through an angle θ is shown in figure 5.25. As the ball is moving on a circular path of radius R, at the shown position it has a centripetal acceleration of magnitude

2 / .

v R Although, I am not discussing about the tangential acceleration of the ball at the shown moment, you should not forget that it is also present.

Applying Work Energy Theorem, we get, wall = ∆kwN+wg =kfkki sm ooth θ fig. 5.25 θ 2 v R x s mg v m N ⇒ 1 2 0 0 2 mg h mv − ⋅∆ = − ⇒ ( ) 1 2 2 mg x mv − − = ⇒ 2 2 v = gx ...(i) ⇒ v=2gx [x=l(1−cos )]θ Applying Newton’s 2nd law along the radial direction, we get.

net

F =ma

2

cos mv 2 (1 cos ) [using ](i)

mg N mg

l

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N =3mgcosθ−2mg

* Notice that at θ =cos (2/3),−1 N becomes zero. What is the physical significance of this θ? What would happen when the ball passes this position?

Example – 8

The kinetic energy of a particle moving along a circle of radius R depends upon the distance covered as 2

, ks

where α is a constant. Find the magnitude of the force acting on the particle as a function of s.

Solution: Let us first find the tangential and normal components of the net force acting upon the particle, then we can find the net force by adding these two components. If Ft be the tangential component of the net force, then 2 ( ) t dk d s F ds ds α = = =2 sα

If Fn be the normal component of the net force, then

2 n v F m R = 2 1 2 2mv R    =    2( s2) [ k s2] R α α = ∵ = 2 2 s R α = ∴ 2 2 net t n F = F +F ( ) 2 2 2 2 2 s s R α α   = +    2 2 s 1 s R α   = +     2 2 2 s R s R α = +

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Example – 9

A particle of mass m starts moving so that its speed varies according to the law vs, where α is a positive constant, and s is the distance covered. Find the total work performed by all the forces which are acting on the particle during the first t seconds after the beginning of motion.

Solution: Using work energy theorem, we have, work done by all forces (change in K.E.) all w = ∆k f i k k = − 2 1 0 [ initially 0 0 0] 2mv s v k = − ∵ = ⇒ = ⇒ = 2 1 2mv = 2 1 2mα s = ...(i)

So, now we have to find s as a function of t. We have

ds v ds s dt = ⇒ dt =α ⇒ ds dt s = ⋅α ⇒ 0 0 s t ds dt s

### ∫

0 0 2 sstt2 st ⇒ 2 2 4 t s=α ...(ii) ∴ 2 2 all m s w = α [Using (i)] 4 2 8 mα t = [Using (ii)]

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Example – 10

A particle of mass m moves along a circle of radius R with a normal acceleration varying with time as ant2, where α is a positive constant. Find the time dependence of the work done by all the forces.

Solution: We have, 2 n atv2 t2 R =α ⇒ 2 2 vRt ...(i)

∴ Work done by all forces, all w = ∆k f i k k = − 2 1 0 2mv = − 2 1 2mv = ⇒ 2 2 all m Rt

w = α [Using equation (i)]

Example – 11

A chain of mass m and length l rests on a rough surfaced table so that one of its ends hangs over the edge. The chain starts sliding off the table all by itself provided the overhanging part equals 1/3of the chain length. What will be the total work performed by the friction forces acting on the chain by the moment it slides completely off the table?

Solution: When the chain has fallen by a distance x its position on the table is shown in figure 5.26. At this instant friction is opposing the motion of the chain and is acting towards the left. If µ be the friction coefficient between the chain and the table and m′ be the mass of the part of the chain on the surface of the table, then friction force is

' fm g 2 3 m l x g l µ    =     l/3 x 2 /3l x fig 5.26

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2 3 mg l x l µ   =  

In the next infinitesimally small time interval dt if the chain slides further by dx, then work done by the frictional force can be given as

dw= − ⋅f dx 2 3 mg l x dx l µ   = − ⋅  

Therefore, till the moment when the chain leaves the table surface completely (i.e., value of x becomes 2l/3), net work done by friction forces is

w=

### ∫

dw 2 /3 0 2 3 x l mg l x dx l µ =   = − − ⋅  

### ( )

2 2 /3 2 2 3 3 2 l mg l l l µ   = −  × −      2 (2 /3) 2 mg l l µ = − 2 9µmgl = −

You should note that µ was not given in the question. You have to find it on your own. (Of course you can find it from the statement of the question).

[Suppose you have to find out the speed of the chain at the moment it just leaves the table, then that can be found by using the result above in the work energy theorem. In this case there are only three forces acting on the chain: gravity, friction and normal contact force from the table. Normal contact force is not doing work so you need work done by friction and gravity only to find the change in K.E. of the chain, which would lead you to the final speed of the chain. And hence in this way you can avoid complicated equations and their solutions which you’d have encountered if you would have chosen methods learnt in chapter NEWTON’S LAWS OF MOTION.]

Example – 12

In the previous example find the work done by gravity (on the chain) for the same duration.

Solution: For the same time interval dt which we considered in the previous example, work done by gravity on the chain is 3 g m l dw x g dx l    = + ⋅     3 mg l x dx l   = + ⋅  

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Hence, net work done on the chain by gravity till the moment it leaves the table completely is g g w =

### ∫

dw 2 /3 0 3 x l mg l x dx l =   = + ⋅  

### ∫

2 2 (2 3) 3 3 2 mg l l l l   = ⋅ +   / l/3 x 2 /3l x fig 5.27 dx 2 2 2 2 9 9 mg l l l   = +   4 9mgl =

Note: In the next topic we will study the concept of CENTRE OF MASS. When you are familiar with that concept, you find the work done by gravity in a much easier way, although the method discussed here is also simple and easy.

If v be the speed of the chain when it just leaves the horizontal surface, then applying work energy theorem on the chain between the moments when it started sliding over the horizontal surface and when it just left the surface, we get

all w = ∆kwg+wf +wN =kfki ⇒ 4 2 1 2 (0) 0 9mglmgl 2mv   + −+ =         ⇒ 4 2 1 2 9mgl−9µmgl=2mv

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Section - 4 POWER Power:

The rate of doing work by a force with respect to time is defined as power developed by that force. We know that when a force does work on a body it transfers energy to that body, therefore, power developed by a force is the rate of energy transfer by that force. Therefore, for power, P, we can write

dw P dt = ...(23)F ds F v [ dw F d s] dt = ⋅! ! = ⋅! ! ∵ = ⋅! ! PP = ⋅F v! ! ...(24)

Therefore, power developed by a force is the scalar product of the force and the velocity of point of application of the force.

The unit of power is Joule/sec which is defined as “watt”.

It is obvious that when force is perpendicular to the velocity, the power developed by the force is zero. If power developed by a force is known it can be used to calculate the work done by that force. We have,

/ P=dw dtdw= ⋅P dt ⇒ =

=

### ∫

. 2 1 t t w dw P dt ...(25)

Here w is the work done by the force which is developing the power P for the time interval [ , ].t t1 2

Note: A common unit of power is horsepower. 1 horsepower = 746 watt

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Example – 13

A body of mass m is thrown at an angle α to the horizontal with an initial velocity v0. Find the mean power developed by gravity over the whole time of motion of the body, and the instantaneous power of gravity as a function of time. Assume that the body is thrown at t = 0.

Solution: The average power developed can be defined as the average rate of doing work. Therefore, the average power developed by gravity for the time interval [0, t] is .

g w P t ∆ = ∆ g F r t ⋅∆ = ∆ ! ! using r mg v v t ∆   = ⋅ = ∆   ! ! ! ! is the component of in the upward direction

y y v v mg v   = − ⋅       !

is the displacement in the vertical direction y y mg t ∆   ∆ = − ⋅ ( ) (0) 0 y t y mg t − = − − ( ) y t mg t = − 2 0 1 sin 2 v t gt mg t α  ⋅ −      = − 0 sin 2 gt mg v α  = − ⋅ −  

The average power developed for the entire duration of flight is zero, because for this duration ∆y is zero. Alternatively this could be explained in the following way: as the displacement, ∆r!, is horizontal for this interval its scalar product with gravity is zero.

The instantaneous power developed by gravity is g

P=F v! !⋅ =mg v! !⋅

is vertically upward component of

y y

mg vv v

= − ⋅ !

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From the equation above it is clear that while rising, the power developed by gravity is negative and while falling, the power developed by gravity is positive. (while rising vy is +ve and while falling vy is –ve.)

Alternate Method: We have g w mg h ∆ = − ∆ ∴ ∴∴ ∴ ∴ g g w P t ∆ = ∆ mg h t ⋅ ∆ = − ∆ h mg t ∆ = − ∆

Now, this result can be used to obtain the desired result. Example – 14

For the situation given in example-10 find the dependence of the power developed by all forces on t and find the average value of this power for the first t seconds of motion.

Solution: We have, 2 n atv2 2 t R =α ⇒ v= αR t ∴ tangential acceleration, t dv a dt = ⇒ at = αR

∴ Normal component of net force,

2

n n

F =ma =m tα

and tangential component of the net force, .

t t

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As we know that work is done by tangential component of the force, power is developed by this component only. We have,

all net P =F! ⋅v! (Ft Fn) v = ! + ! ⋅! ⇒ Pall =F vt

F!t ||v! and F!n

# v!

### )

...(26) =m αR⋅ αR t =m Rtα

Again, work done by all forces for the first t seconds of motion can be obtained by either finding the increment in kinetic energy of the particle or by integrating the power with respect to time. Therefore, for the time interval [0, t], work done by all forces is

0 0 t t all all w =

### ∫

P ⋅ =dt m R t dtα

### ∫

⋅ 2 2 m Rtα =

Therefore, average power developed by all forces for the same interval is

2

### )

/2 work done

length of time interval ( ) av m Rt P t α = = 2 m Rtα =

Alternate-I: We have, Pall =m Rtα , which is a linear function in t, therefore, the average value of Pall final value + initial value

2 av P = ⇒ ( ) 0 2 all av P t P = + 2 m Rtα =

Alternate-II: Using work energy theorem for the time interval [0, t] we get,

all f i w = ∆ =k kk 0 f k = ⇒ 1 2 2 2 2 all m Rt w = mv = α

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∴ Average power developed by all forces

work done by all forces lenth of time interval

= 2 ( 2) ( ) 2 m Rt m Rt t α α = / =

Alternate-III: Average value of f x( ) over the interval [ ,x x1 2] is given as

2 1 2 1 ( ) [ ] x x av f x dx f x x ⋅ = −

### ∫

∴ 0 0 ( ) ( 0) t t av p t dt m R t dt p t t α ⋅ ⋅ = = −

### ∫

2 m Rtα =

Alternate-IV: We have tangential acceleration, at = αR, which is constant with time and we also have

initial 0,

v = therefore, distance traveled by the particle along the path for the first t seconds is

2 2

1 1

2 t 2

s= ⋅ ⋅ =a t αR t

As the tangential force is constant and work done by normal force is always zero, work done by all forces for the same time interval can be written as

### )

1 2 2 all t w = ⋅ =F s m αR ⋅ ⋅ αR t   2 2 m Rtα = Therefore,

### )

2 2 ( ) all m Rt P t α = / 2 m Rtα =

• You should try to understand the physical significance of the equation wall = ⋅F st used here. Just give it a thought, and you are sure to get it.

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TRY YOURSELF - II

Q. 1 A heavy stone is thrown from a cliff of height h with a speed v. The stone will hit the ground with maximum speed if it is thrown

(a) vertically downward (b) vertically upward (c) horizontally

(d) the speed does not depend on the initial direction.

Q. 2 Total work done on a particle is equal to the change in its kinetic energy (a) always

(b) only if the forces acting on it are conservative (c) only if gravitational force alone acts on it (d) only if elastic force alone acts on it.

Q. 3 The kinetic energy of a particle continuously increases with time.

(a) The resultant force on the particle must be parallel to the velocity at all instants (b) The resultant force on the particle must be at an angle less than 90° all the time (c) Its height above the ground level must continuously decrease

(d) The magnitude of its normal acceleration is increasing continuously.

Q. 4 Under the action of a force, a 2 kg body moves such that its position x as a function of time t is given by

3

, 3

=t

x x is in metre and t in second. Calculate the work done by the force in the first 2 second. Q. 5 A block shown in figure slides on a semicircular frictionless

track. If starts from rest at position A, what is its speed at the point marked B?

B A

45º1.0m

Q. 6 An object of mass m is tied to a string of length l and a variable force F is applied on it which brings the string gradually at angle θ with the vertical. Find the work done by the force F.

θ

F l

m

Q. 7 A body of mass m accelerates uniformly from rest to v0 in time t0. As a function of t, the instantaneous power delivered to the body is :

(a) m v t/ (b) 2 / mv t (c) 2 / mvt t (d) 2 2 / mv t .

Q. 8 Show that for the same initial speed v0 the speed v of a projectile will be the same at all points at the same elevation, regardless of the angle of projection.

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Q. 9 A block of mass 10 kg slides down an incline 5 m. long and 3 m high. A man pushes up the block parallel to the incline so that it slides down at constant speed. The coefficient of friction between the block and the incline is 0.1. Find:

(a) the work done by the man on the block (b) the work done by gravity on the block

(c) the work done by the surface on the block (d) the work done by the resultant forces on the block (e) the change in K.E. of the block.

[g = 10 m/s²]

Q. 10 A particle of mass m moves on a straight line with its velocity varying with the distance travelled according to the equation v=a x, where a is a constant. Find the total work done by all the forces during a displacement from x = 0 to x = d.

Q. 11 Power delivered to a body varies as P = 3 t2. Find out the change in kinetic energy of the body from t = 2 to t = 4 sec.

Q. 12 Figure shows a rough horizontal plane which ends in a vertical wall, to which a spring is connected, having a force constant k. Initially spring is in its relaxed state. A block of mass m starts with an initial velocity u towards the spring from a distance l0 from the end of spring, as shown. When block strikes at the end of the spring , it compresses the spring and comes to rest. Find the maximum compression in the spring. The friction coefficient between the block and the floor is µ.

l0

k

m µ

u

Q. 13 A point mass m starts from rest and slides down the surface of a frictionless solid sphere of radius r as in figure. Measure angles from the vertical and potential energy from the top. Find (a) the change in potential energy of the mass with angle; (b) the kinetic energy as a function of angle; (c) the radial and tangential accelerations as a functions of angle (d) the angle at which the mass flies off the sphere. (e) If there is friction between the mass and the sphere, does the mass fly off at a greater or lesser angle than in part (d)?

r m θ

Q. 14 Two disks are connected by a stiff spring. Can one press the upper disk down enough so that when it is released it will spring back and raise the lower disk off the table (see figure)? Can mechanical energy be conserved in such a case?

Q. 15 A smooth sphere of radius R is made to translate in a straight line with a constant acceleration a. A particle kept on the top of the sphere is releases from there at zero speed with respect to the sphere. Find the speed of the particle with respect to the sphere as a function of the angle θ it slides.

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Section - 5 POTENTIAL ENERGY AND ENERGY CONSERVATION

While calculating work done by gravity and an ideal spring, you must have noticed that these were independent of the path of motion of the body. Work done by these forces depends only upon initial and final positions. The same idea leads to the fact that work done by these forces in a closed path is zero. In later chapters you would see that coulomb’s forces exhibit similar behaviour. This feature of these forces allow us to group them together in a different class of forces which are called conservative forces . Therefore a conservative force can be defined as a force whose work done is always independent of path. Alternatively, we can say that work done by a conservative force depends only upon initial and final positions or we can say that work done by a conservative force in a closed path is always zero. You can define a conservative force by any of the three statements. All these three statements have the same physical significance. Therefore if any force behaves like this, that force can be defined as a conservative force. Forces which do not satisfy these conditions can be defined as non conservative forces.

For a while you are urged not to go into conceptual details of this new topic. We will first develop some methods based on the concept of conservative forces. Learn the method and its application initially, and then we will go into the conceptual details of conservative forces.

The region in which a conservative force is acting is defined as the conservative force field of that force. For conservative force fields we associate potential energy with them. Why do we do so? This will be clear to you as you will proceed with this section. In a conservative force field work done by a conservative force is defined as negative of change in potential energy of the system. If potential energy is denoted by U, then, we define

or

= –∆ ∆ = –

cons con

w U U w ...(27)

Therefore, for gravitational field near the earth’s surface, the change in gravitational potential energy,

( )

g g

U w mg h

∆ = − = − − ∆ [using (6)]

Ug= +mg h ...(28)

and for a spring and block system,

2 2 1 1 2 2 sp sp f i U w   kx kx  ∆ = − = − −     [using (13)] ⇒ = 1 21 2 2 2 sp f i U kx kx ...(29)

Now, let us use the definition of change in potential energy in the work energy theorem. We have, all

w = ∆k

wcon+wnoncon= ∆k

wnoncon = ∆ −k wnoncon

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## References

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