Topic covered

`color{red} star` POWER


`➢` The spring force is an example of a variable force which is conservative. Fig. 6.7 shows a block attached to a spring and resting on a smooth horizontal surface.
`➢` The other end of the spring is attached to a rigid wall. The spring is light and may be treated as massless.

`➢` In an ideal spring, the spring force `F_s` is proportional to `x` where x is the displacement of the block from the equilibrium position. The displacement could be either positive [Fig. 6.7(b)] or negative [Fig. 6.7(c)]. This force law for the spring is called Hooke’s law and is mathematically stated as


`=>` The constant `k` is called the spring constant. Its unit is `N m^(-1)`. The spring is said to be stiff if k is large and soft if `k` is small.

`➢` Suppose that we pull the block outwards as in Fig. 6.7(b). If the extension is `x_m`, the work done by the spring force is

`color{purple} {W_s = int_0^(x_m) F_s dx=- int_0^(x_m) kx dx}`

`color {blue}{=- (kx_m^2)/2}`


`➢` This expression may also be obtained by considering the area of the triangle as in Fig. 6.7(d). Note that the work done by the external pulling force F is positive since it overcomes the spring force

`color {blue}{W= + (kx_m^2)/2}`..........(6.16)

`➢` The same is true when the spring is compressed with a displacement `x_c (< 0)`. The spring force does work `W_s=-kx_c^2//2` while the external force `F` does work `+ kx_c^2//2` I
`=>` f the block is moved from an initial displacement `x_i` to a final displacement `x_f` , the work done by the spring force `W_s` is

`color {blue}{W_s= - int_(x_i)^(x_f) kx dx = (kx_i^2)/2 -(k x_f^2)/2}` ...........(6.17)

`➢` Thus the work done by the spring force depends only on the end points. Specifically, if the block is pulled from xi and allowed to return to `x_i` ;

`color {blue}{W_s=- int_(x_i)^(x_f) kx dx= (kx_i^2)/2 - (kx_i^2)/2=0}`


`➢` The work done by the spring force in a cyclic process is zero. We have explicitly demonstrated that the spring force
(i) is position dependent only as first stated by Hooke, `(F_s = − kx)`;
(ii) does work which only depends on the initial and final positions, e.g. Eq. (6.17).

`color{red}☞` Thus, the spring force is a conservative force We define the potential energy `V(x)` of the spring to be zero when block and spring system is in the equilibrium position. For an extension (or compression) `x` the above analysis suggests that

`color {blue}{V(x) =(kx^2)/2}`..............(6.19)

`➢` You may easily verify that `− dV/dx = −k x`, the spring force. If the block of mass m in Fig. 6.7 is extended to `x_m` and released from rest, then its total mechanical energy at any arbitrary point x, where `x` lies between `– x_m` and `+ x_m`, will be given by

`color{orange}{1/2 kx_m^2 =1/2 kx^2 + 1/2 mv^2}`

`=>` where we have invoked the conservation of mechanical energy. This suggests that the speed and the kinetic energy will be maximum at the equilibrium position, `x = 0`, i.e.,

`1/2 kx_m^2+ 1/2 mv^2`

`➢` where we have invoked the conservation of mechanical energy. This suggests that the speed and the kinetic energy will be maximum at the equilibrium position, `x = 0`, i.e.,

`1/2 mv_m^2 = 1/2 kx_m^2` where `v_m` is the maximum speed.

or `v_m = sqrt(k/m) x_m`

`color{red}☞` Note that `k//m` has the dimensions of `[T^(-2)]` and our equation is dimensionally correct. The kinetic energy gets converted to potential energy

and vice versa, however, the total mechanical energy remains constant. This is graphically depicted in Fig. 6.8.

Q 1713680549

To simulate car accidents, auto manufactures 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 18.0 km/h on a smooth road and colliding

with a horizontally mounted spring of spring

constant ` 6.25 xx 10^3 N m^(-1)`.

What is the maximum compression of the spring?


1.25 rn


1 rn


1 .75 rn


2 m


At maximum compression `x_m`, tne potential energy V of

the spring is equal to the kinetic energy K of the moving

car from the principle of conservation of mechanical

energy. `color{purple} {V = 1/2 kx_m^2 = 1/25 xx10^4 J}`

We obtain, `x_m = 2.00 m`

We note that we have idealised the siituation. The spring

is considered to be massless. The surface has been

considered to possess negligible friction.
Correct Answer is `=>` (D) 2 m


`color{red}☞` We conclude this section by making a few remarks on conservative forces.
(i) In the example considered above, we can calculate the compression, but not the time over which the compression occurs. A solution of Newton’s Second Law for this system is required for temporal information.
(ii) Not all forces are conservative. Friction, for example, is a non-conservative force. The principle of conservation of energy will have to be modified in this case. This is illustrated in Example 6.9.
(iii) The zero of the potential energy is arbitrary. It is set according to convenience.
`=>` For the spring force we took V(x) = 0, at x = 0, i.e. the unstretched spring had zero potential energy. For the constant gravitational force `mg,` we took `V = 0` on the earth’s surface.
`➢` In a later chapter we shall see that for the force due to the universal law of gravitation, the zero is best defined at an infinite distance from the gravitational source. However, once the zero of the potential energy is fixed in a given discussion, it must be consistently adhered to throughout the discussion. You cannot change horses in midstream.
Q 1713780649

To simulate car accidents, auto manufactures,

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 18.0 km/h on a smooth road

and colliding with a horizontally mounted spling of

spring constant `6.25 xx 10^3 N m^(-1)` . Taking the

coefficient of friction, `mu` to be 0.5, calculate the

maximum compression of the spring.

Consider Example 6.7 taking the coefficient of friction, μ, to be 0.5 and
calculate the maximum compression of the spring.


1.5 m


1.35 m


2.0 m


2.5 m


In presence ol friction, both the Gpring force and the

frictional force act so as to oppose the compression of

the spring as shown in figure.

We invoke the work-energy theorem, rather than the

conservation of mechanical energy.

The change in kinetic energy is

`color{purple} {Delta K = K_f - K_i = 0 - 1/2 mv^2}`

The work done by the net force is

`W = -1/2 kx_m^2 - mu m g x_m`

Equating, we have `1/2 mv^2 = 1/2 kx_m^2 + mu m gx_m`

` ( : . Delta K = W W.E theorem)`

Now, `mu mg = 0.5 xx 10^3 xx 10 = 5 xx 10^3 N`

(taking `g = 10.0 ms^(-2)).` After rearranging the above

equation, we obtain the following quadratic equation in

the unknown `x_m.`

`color{green} {k x_m^2 + 2 mu m gx_m - mv^2 = 0}`

`x_m = (-mu mg + [ mu^2 m^2 g^2 + m kv^2]^(1//2))/k`

where, we take the positive square root, since `x_m` is

positive. Putting in numerical values, we obtain

`x_m = 1.35 m`

If the two forces on the body consist of a conservative

force `F_c` and a non-conservative force `f_(nc)`, the

conservation of mechanical energy formula will have

to be modified. By the Work-energy theorem,

`color{orange} {(F_c + F_(nc)) Delta x = DeltaK}`

But `F_c Delta x = - Delta V`

`Delta (K + V) = F_(nc) Deltax`

Hence, `DeltaE = F_(nc) Deltax`

where, E is the total mechanical energy. Over the path

this assumes the form

`color{orange} {E_f = E_i = W_(nc)}`

where, `W_(nc)` is the total work clone by the

non-conservative forces over the path. Note that unlike

the conservative force, `W_(nc)` depends on the particular

path i to f.
Correct Answer is `=>` (B) 1.35 m

The Principle of Conservation of Energy

`➢ "We have seen that the total mechanical energy of the system is conserved if the forces doing work on it are conservative."`

`➢` If some of the forces involved are non-conservative, part of the mechanical energy may get transformed into other forms such as heat, light and sound.
`➢` However, the total energy of an isolated system does not change, as long as one accounts for all forms of energy. Energy may be transformed from one form to another but the total energy of an isolated system remains constant.
`➢` `"Energy can neither be created, nor destroyed."`
`➢` Since the universe as a whole may be viewed as an isolated system, the total energy of the universe is constant. If one part of the universe loses energy, another part must gain an equal amount of energy.

`➢` The principle of conservation of energy cannot be proved. However, no violation of this principle has been observed.
`➢` The concept of conservation and transformation of energy into various forms links together various branches of physics, chemistry and life sciences. It provides a unifying, enduring element in our scientific pursuits. From engineering point of view all electronic, communication and mechanical devices rely on some forms of energy transformation.


`➢` In the previous section we have discussed mechanical energy. We have seen that it can be classified into two distinct categories : one based on motion, namely kinetic energy; the other on configuration (position), namely potential energy. Energy comes in many a forms which transform into one another in ways which may not often be clear to us.

`color{red} ► color {blue} bbul" Heat"`

`➢` We have seen that the frictional force is excluded from the category of conservative forces. However, work is associated with the force of friction.

`➢` A block of mass `m` sliding on a rough horizontal surface with speed `v_0` comes to a halt over a distance `x_0`.

`➢` The work done by the force of kinetic friction `f` over `x_0` is `–f x_0`.
`➢` By the work-energy theorem `m v_0^2 //2 = f x_0` . If we confine our scope to mechanics, we would say that the kinetic energy of the block is llost due to the frictional force.

`➢` On examination of the block and the table we would detect a slight increase in their temperatures.
`➢` Thus the work done by friction is this raises the internal energy of the block and the table. In winter, in order to feel warm, we generate heat by vigorously rubbing our palms together.
`➢` We shall see later that the internal energy is associated with the ceaseless, often random, motion of molecules. A quantitative idea of the transfer of heat energy is obtained by noting that `1 kg` of water releases about `42000 J` of energy when it cools by `10 °C`.

`color{red} ► color {blue} bbul" Chemical Energy"`

`➢` As we learnt to rub two flint stones together (mechanical energy), got them to heat up and to ignite a heap of dry leaves (chemical energy), which then provided sustained warmth.
`➢` A matchstick ignites into a bright flame when struck against a specially prepared chemical surface. The lighted matchstick, when applied to a firecracker, results in a spectacular display of sound and light.

`➢` Chemical energy arises from the fact that the molecules participating in the chemical reaction have different binding energies. A stable chemical compound has less energy than the separated parts.
`➢` A chemical reaction is basically a rearrangement of atoms. If the total energy of the reactants is more than the products of the reaction, heat is released and the reaction is said to be an exothermic reaction.
`➢` If the reverse is true, heat is absorbed and the reaction is endothermic. Coal consists of carbon and a kilogram of it when burnt releases `3 xx 10^7 J` of energy.

`color{red}☞` Chemical energy is associated with the forces that give rise to the stability of substances. These forces bind atoms into molecules, molecules into polymeric chains, etc. The chemical energy arising from the combustion of coal, cooking gas, wood and petroleum is indispensable to our daily existence.

`color{red} ► color {blue} bbul " Electrical Energy"`

`➢` The flow of electrical current causes bulbs to glow, fans to rotate and bells to ring. There are laws governing the attraction and repulsion of charges and currents, which we shall learn later. Energy is associated with an electric current. An urban Indian household consumes about 200 J of energy per second on an average.

`color{red} ► color {blue} " The Equivalence of Mass and Energy"`

`➢` Till the end of the nineteenth century, physicists believed that in every physical and chemical process, the mass of an isolated system is conserved. Matter might change its phase, e.g. glacial ice could melt into a gushing stream, but matter is neither created nor destroyed; Albert Einstein (1879-1955) however, showed that mass and energy are equivalent and are related by the relation


where `c`, the speed of light in vacuum is approximately `3 xx 10^8 m s^(–1)`. Thus, a staggering amount of energy is associated with a mere kilogram of matter

`E= 1 xx (3 xx 10^8)^2J = 9 xx 10^16 J`

`=>` This is equivalent to the annual electrical output of a large `(3000 MW)` power generating station.

`color{red} ► color {blue} bbul " Nuclear Energy"`

`=>` The most destructive weapons made by man, the fission and fusion bombs are manifestations of the above equivalence of mass and energy [Eq. (6.20)].

Q 3169178915

Examine Tables 6.1-6.3 and express (a) The energy required to break one bond in DNA in eV; (b) The kinetic energy of an air molecule (`10^(-21) J`) in eV; (c) The daily intake of a human adult in kilocalories.


(a) Energy required to break one bond of DNA is

`(10^(-20) J)/(1.6 xx 10^(-19) J //eV) equiv 0.06 eV`

where the symbol ‘`equiv`’ stands for approximate. Note `0.1 eV = 100 meV` (100 millielectron volt).

(b) The kinetic energy of an air molecule is

`(10^(-21)J)/(1.6 xx 10^(-19) J//eV) equiv 0.0062 eV`

This is the same as `6.2 meV`.

(c) The average human consumption in a day is

`(10^7 J)/(4.2 xx 10^3 J//kcal) equiv 2400 k cal`

We point out a common misconception created by newspapers and magazines. They mention food values in calories and urge us to restrict diet intake to below 2400 calories. What they should be saying is kilocalories (kcal) and not calories. A person consuming 2400 calories a day will soon starve to death! 1 food calorie is 1 kcal.


`➢` Often it is interesting to know not only the work done on an object, but also the rate at which this work is done.
`➢ \ \ "Power"` is defined as the time rate at which work is done or energy is transferred.
`➢` The average power of a force is defined as the ratio of the work, `W`, to the total time `t` taken

`P_(av)= W/t`

`➢` The instantaneous power is defined as the limiting value of the average power as time interval approaches zero,

`color {blue}{P=(dW)/(dt)}`..................(6.21)

`➢` The work `dW` done by a force F for a displacement `dr` is `dW = F*dr`. The instantaneous power can also be expressed as

`P= F * (dr)/(dt)`

`color {blue}{=F*v}`...............(6.22)

`=>` where `v` is the instantaneous velocity when the force is `F`.

`color{red} ►` Power, like work and energy, is a scalar quantity. Its dimensions are `[ML^2T^(–3)]`.
`color{red} ►` In the SI, its unit is called a watt (W). The watt is 1 `J s^(–1)`. The unit of power is named after James Watt, one of the innovators of the steam engine in the eighteenth century. There is another unit of power, namely the horse-power (`hp`)

`1` hp = `746` W

`➢` This unit is still used to describe the output of automobiles, motorbikes, etc. We encounter the unit watt when we buy electrical goods such as bulbs, heaters and refrigerators. A `100` watt bulb which is on for `10` hours uses 1 kilowatt hour (kWh) of energy.

`100` (watt) × `10` (hour)`

`= 1000` watt hour `=1 ` "kilowatt hour (kWh)"`

`= 10^3 (W) xx 3600 (s)`

`color {blue}{= 3.6 xx 10^6 J}`

Our electricity bills carry the energy consumption in units of kWh. Note that kWh is a unit of energy and not of power.
Q 3179178916

An elevator can carry a maximum load of `1800` kg (elevator + passengers) is moving up with a constant speed of `2 m s^(–1)`. The frictional force opposing the motion is `4000` N. Determine the minimum power delivered by the motor to the elevator in watts as well as in horse power.


The downward force on the elevator is

`color{pink} {F = m g + F_f }`

`= (1800 xx 10) + 4000 = 22000 N`

The motor must supply enough power to balance this force. Hence,

`P = F* v = 22000 xx 2 = 44000 W = 59 hp`