Physics THE CONCEPT OF POTENTIAL ENERGY AND THE CONSERVATION OF MECHANICAL ENERGY
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CLASS 11 CHAPTER 6 - WORK, ENERGY AND POWER

THE CONCEPT OF POTENTIAL ENERGY AND THE CONSERVATION OF MECHANICAL ENERGY

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`color{red} star` THE CONCEPT OF POTENTIAL ENERGY
`color{red} star` THE CONSERVATION OF MECHANICAL ENERGY

THE CONCEPT OF POTENTIAL ENERGY

`➢` The word potential suggests possibility or capacity for action.
`➢` When it is released, the arrow flies off at a great speed. The earth’s crust is not uniform, but has discontinuities and dislocations that are called fault lines. These fault lines in the earth’s crust are like ‘compressed springs’. They possess a large amount of potential energy.

`➢` Thus, potential energy is the stored energy’ by virtue of the position or configuration of a body.
`➢` The body left to itself releases this stored energy in the form of kinetic energy.

`➢` Let us make our notion of potential energy more concrete. The gravitational force on a ball of mass `m` is `mg` . `g` may be treated as a constant near the earth surface. By ‘near’ we imply that the height `h` of the ball above the earth’s surface is very small compared to the earth’s radius `R_E (h < < R_E)` so that we can ignore the variation of `g` near the earth’s surface*. In what follows we have taken the upward direction to be positive. Let us raise the ball up to a height `h`.

`➢` The work done by the external agency against the gravitational force is `mgh`. This work gets stored as potential energy. Gravitational potential energy of an object, as a function of the height `h`, is denoted by `V(h)` and it is the negative of work done by the gravitational force in raising the object to that height.

`V (h) = mgh`

`➢` If `h` is taken as a variable, it is easily seen that the gravitational force `F` equals the negative of the derivative of `V(h)` with respect to h. Thus,

`color{orange}{F=- d/(dh) V(h) =-mg}`



`➢` The negative sign indicates that the gravitational force is downward. When released, the ball comes down with an increasing speed.
Just before it hits the ground, its speed is given by the kinematic relation,

`v^2 = 2gh`

This equation can be written as `color{blue}{1/2 mv^2 =mgh}` which shows that the gravitational potential energy of the object at height `h`, when the object is released, manifests itself as kinetic energy of the object on reaching the ground.

`➢` Physically, the notion of potential energy is applicable only to the class of forces where work done against the force gets ‘stored up’ as energy. When external constraints are removed, it manifests itself as kinetic energy. Mathematically, (for simplicity, in one dimension) the potential energy `V(x)` is defined if the force `F(x)` can be written as

`F(x) =- (dV)/(dx)`

This implies that

`color{blue}bb(int_(x_i)^(x_f) F(x) dx =- int_(v_i)^(v_f) dV =V_t -V_f}`



`➢"The work done by a conservative force such as gravity depends on the initial and final positions only. "`

`➢` If an object of mass `m` is released from rest, from the top of a smooth (frictionless) inclined plane of height `h`, its speed at the bottom is `sqrt(2gh)` irrespective of the angle of inclination. Thus, at the bottom of the inclined plane it acquires a kinetic energy, `mgh`.
`➢` If the work done or the kinetic energy did depend on other factors such as the velocity or the particular path taken by the object, the force would be called non conservative.

`color{red}☞` The dimensions of potential energy are `[ML^2T ^(–2)]` and the unit is joule `(J)`, the same as kinetic energy or work. To reiterate, the change in potential energy, for a conservative force, `ΔV` is equal to the negative of the work done by the force

`color {blue}{Delta V=-F(x) Delta x}`.............(6.9)

`➢` In the example of the falling ball considered in this section we saw how potential energy was converted to kinetic energy. This hints at an important principle of conservation in mechanics, which we now proceed to examine.

THE CONSERVATION OF MECHANICAL ENERGY

`➢` For simplicity we demonstrate this important principle for one-dimensional motion.
`=>` Suppose that a body undergoes displacement Δx under the action of a conservative force `F`. Then from the `WE` theorem we have,

`color{orange}{ΔK = F(x) Δx}`

`➢` If the force is conservative, the potential energy function `V(x)` can be defined such that

`− ΔV = F(x) Δx`



`➢` The above equations imply that

`ΔK + ΔV = 0`

`Δ(K + V ) = 0`



which means that `K + V`, the sum of the kinetic and potential energies of the body is a constant. Over the whole path, `x_i` to `x_f`, this means that

`color {blue}{K_i + V(x_i ) = K_f + V(x_f )}` ....................(6.11)

`➢` The quantity `K +V(x)`, is called the total mechanical energy of the system. Individually the kinetic energy `K` and the potential energy `V(x)` may vary from point to point, but the sum is a constant. The aptness of the term conservative force’ is now clear.


`color{red} ►` Let us consider some of the definitions of a conservative force.

✍️ A force F(x) is conservative if it can be derived from a scalar quantity `V(x)` by the relation given by Eq. (6.9). The three-dimensional
generalisation requires the use of a vector derivative, which is outside the scope of this book.

✍️ The work done by the conservative force depends only on the end points. This can be seen from the relation,

`color{blue}{W = K_f – K_i = V (x_i ) – V(x_f )}` which depends on the end points.

✍️ A third definition states that the work done by this force in a closed path is zero. This is once again apparent from Eq. (6.11) since

`x_i = x_f`

`=>` Thus, the principle of conservation of total mechanical energy can be stated as

The total mechanical energy of a system is conserved if the forces, doing work on it, are conservative.


`➢` The above discussion can be made more concrete by considering the example of the gravitational force once again and that of the spring force in the next section. Fig. 6.5 depicts a ball of mass m being dropped from a cliff of height H.

`color{red}☞` The total mechanical energies `E_0, E_h`, and `E_H` of the ball at the indicated heights zero (ground level), `h` and `H`, are

`color {blue}{E_H =mgH}`..........(6.11 a)

`color {blue}{E_h=mgh + 1/2 + 1/2 mv_h^2}`...........(6.11 b)

`color {blue}{E_0 = (1//2) mv^2}`

.........(6.11 c)

`➢` The constant force is a special case of a spatially dependent force `F(x)`. Hence, the mechanical energy is conserved. Thus

`E_H =E_0`

or `mgH= 1/2 mv_f^2`

`v_f = sqrt(2gH)`



a result that was obtained in section 3.7 for a freely falling body.

Further `E_H =E_h`

which implies,

`color {blue}{v_h^2 = 2g (H-h)}`

..........(6.11 d)

and is a familiar result from kinematics. At the height `H`, the energy is purely potential.

`➢` It is partially converted to kinetic at height `h` and is fully kinetic at ground level. This illustrates the conservation of mechanical energy.

Q 3109878718

A bob of mass m is suspended by a light string of length L . It is imparted a horizontal velocity vo at the lowest point A such that it completes a semi-circular trajectory in the vertical plane with the string becoming slack only on reaching the topmost
point, C. This is shown in Fig. 6.6. Obtain an expression for (i) `v_o`; (ii) the speeds at points B and C; (iii) the ratio of the kinetic energies
`(K_B//K_C)` at B and C. Comment on the nature of the trajectory of the bob after it reaches the point C.

Solution:

There are two external forces on the bob : gravity and the tension (T) in the string. The latter does no work since the displacement of the bob is always normal to the string. The potential energy of the bob is thus associated with the gravitational force only. The total mechanical energy `E` of the system is conserved. We take the potential energy of the system to be zero at the lowest point A. Thus,
at A :

`color{pink} {E=1/2 mv_0^2}`..........(6.12)

`color{green} {T_A -mg =(mv_0^2)/L}` [Newton’s Second Law]

where `T_A` is the tension in the string at `A`. At the highest point `C`, the string slackens, as the tension in the string (`T_C`) becomes zero.
Thus, at `C`

`color{pink} {E= 1/2 mv_c^2 + 2mgL}`............(6.13)

`color{green} {mg = (mv_c^2)/L}` [Newton’s Second Law] (6.14)

where `v_C` is the speed at `C`. From Eqs. (6.13) and (6.14)

`E= 5/2 mgL`

Equating this to the energy at A

`5/2 mgL =m/2 v_0^2`

or, `v_0= sqrt(5gL)`

(ii) It is clear from Eq. (6.14)

`v_C = sqrt(gL)`

At `B`, the energy is

`E= 1/2 mv_B^2 +mgL`

Equating this to the energy at `A` and employing the result from (i), namely `v_0^2 = 5gL`

`color{purple} {1/2 mb_B^2 +mgL=1/2 mv_0^2}`

`=5/2 mgL`

`∴v_B = sqrt(3gL)`

(iii) The ratio of the kinetic energies at B and C is :

`(K_B)/(K_C) = (1/2 mv_B^2)/(1/2 mv_C^2) =3/1`



At point C, the string becomes slack and the velocity of the bob is horizontal and to the left. If the connecting string is cut at this instant, the bob will execute a projectile motion with horizontal projection akin to a rock kicked horizontally from the edge of a cliff. Otherwise the bob will continue on its circular path and complete the revolution.
 
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