# Work energy and power

In general work is said to done when a body moves in a certain distance when force is applied to that body.

# Work:

The work done by a force acting on the body is defined as the product of the force and the displacement of the body in the direction of the force.

## Constant Force :

If the force is constant and it displaces the body through a distance (s) in the same direction, then work done.

W = F a

But if the displacement $\overrightarrow{s}$ makes an angle $\theta$ with
the direction of force; then Constant force

component of displacement along the direction of force = $s cos \theta$

so,

work done by the force ,

W = F s $\cos \theta = \overrightarrow{F} . \overrightarrow{s}$

i.e. the work done by a force is the scalar product of the force vector and displacement vector.

## 2. Variable Force :

If the force is variable, then for infinitely small displacement, the force is assumed constant, so the work done during small displacement
d W = F dx (one dimensional form)
Then total work is joined by adding such small amounts of work $\therefore W = \sum F dx$

This is found in two ways :

(i) By integration :

Let body be displaced from $x_1 \text{to} x_2$

And $W = \int_{x_1}^{x_2}$ Graph of integration

(ii) By graphical method :

If a graph is plotted between force F and displacement x, then work done by the force from $x_1 to x_2$
= Area enclosed by F – x curve and x-axis from $x_1 \text{to} x_2$  = area of shaded portion.

Unit of work  in M.K.S. and S.I. system is joule.

Dimensions of work

= [ F ] [s] = [ $MLT^{-2}$ ] [L] = [ $M L^2 T^{-2}$ ]

In C.G.S system the unit of work is erg $1 \text{Joule} = 10^7 \text{erg}$

Sometimes the unit of work is written as kilogram-metre

1 kilogram-metre

= g joule = 9.8 joule

Remark : Work done by a force acting on a body may be zero in the following cases

The force is constant in direction and the displacement is zero.

Potential energy by the relation $F = - \dfrac{\triangle U}{\triangle r}$

## Non-conservative forces.

If the work done by a force acting on a particle depends on path followed between two given points or is not equal to zero in a complete round trip is called non-conservative force. The frictional force, the viscous force are examples of non-conservative forces.

## Work-Energy Theorem :

This theorem states that the work done by the net force acting on a body is equal to change is kinetic energy of the body.

i.e. Work = gain in Kinetic Energy $\dfrac{1}{2} m {v_2}^2 - \dfrac{1}{2} m {v_1}^2$

Minimum distance in stopping a car for a given velocity.

If a car of of mass M stops due to retardation of frictional force $\mu R = \mu Mg$, after
traversing a minimum distance $s_{min}$ then work energy theorem gives $\dfrac{1}{2} m v^2 = \mu Mg s_{min} \rightarrow s_{min} = \dfrac{v^2}{2 \mu g}$

### Conservation of energy :

If a system is acted on by conservative forces, the total mechanical
theory of the system is conserved i.e. Mechanical energy E = KE + PE constant

i.e E = T + U = constant
(under conservative forces).
Other forms of energy :

In addition to mechanical energy, there are other forms of energy e.g. thermal energy, light energy, sound energy, electrical energy, chemical energy, nuclear energy etc.
If the system is under the action of non-conservative forces, the conservation law for mechanical energy does not hold and a more general law stated as “The total energy of the universe remains constant” holds. This simply means that energy may be transformed from one form to another. For example in loudspeakers, electric bells the electrical energy is converted into sound energy while in electromagnet electrical energy is converted into magnetic energy and for a ball falling on earth, the mechanical energy is converted into heat energy etc.

## Einstein’s Mass Energy Equivalence :

According to Einstein neither mass nor energy of the universe is conserved separately: but mass and energy are interconvertible according to relation $E = mc^2$

This relation is called Einstein’s mass energy equivalence. Accordingly,
“Total (mass + energy) of universe is conserved”. This is the most general law of conservation of energy.
According to mass energy equivalence the body of mass $m_0$ at rest has rest energy $m_0 c^2$ therefore the kinetic energy of the body $T = mc^2 - {m_o}^2 = { ( m - m_o ) }^2$

Where m is the mass of the body moving with speed v given by $m = \dfrac{m_0}{\sqrt{1 - \dfrac{v^2}{c^2}}}$

Collisions

Coefficient of Restitution :

When the two bodies collide directly, then the ratio of relative velocity after collision to the relative velocity before collision is a fixed quantity. The quantity $e = - \dfrac{v_1 - v_2}{u_1 - u_2}$

is called the coefficient of restitution. Here u1, U2 are initial velocities and $v_1 \, , v_2$ are final velocities of two bodies. The value of e lies between O and l. For a perfectly elastic collision e = 1 and for a perfectly inelastic collision e = 0.
If a body falls freely from a height H on a floor and bounces back to a height h. Then initial velocity u is given by, $\dfrac{1}{2} m u^2 = mg H \, or \, u = \sqrt{2gH}$

and the final velocity v is given by $\dfrac{1}{2} m v^2 = mgh$

or $v = \sqrt{2gh}$

Elastic and Inelastic Collisions :

## Elastic collisions :

A collision is said to be elastic if the total kinetic energy before and collision remains the same. Thus if two bodies of masses $m_1 \text{and } m_2$ moving with velocities $\overrightarrow{u_1} \text{and} \overrightarrow{u_2}$ collide in the absence of external forces and the collision is elastic and $\overrightarrow{v_1} \text{and} \overrightarrow{v_2}$ are their velocities after collision, then the principle of conservation of linear momentum gives $m_1 \overrightarrow{u_1} + m_2 \overrightarrow{u_2} = m_1 \overrightarrow{v_1} + m_2 \overrightarrow{v_2}$     ………(1)
and the principle of conservation of kinetic energy gives $\dfrac{1}{2} m_1 {u_1}^2 + \dfrac{1}{2} m_1 {u_2}^2 = \dfrac{1}{2} m_1 {v_1}^2 + \dfrac{1}{2} m_1 {v_2}$   ………………………(2)

Solving (1) and (2), we get the following important results $v_1 - v_2 = - ( u_1 - u_2 )$  …………………………..(3)

This equation tells that in an one-dimensional elastic collision, the relative velocity of approach before collision is equal to the relative velocity of separation after collision.
Also the velocities after collision are $v_1 = ( \dfrac{m_1 - m_2}{m_1 + m_2} ) u_1 + ( \dfrac{2 m_2}{m_1 + m_2} ) u_2$………(4) $v_2 = ( \dfrac{2 m_1}{m_1 + m_2} ) u_1 + ( \dfrac{m_2 - m_1}{m_1 + m_2} ) u_2$………(5)

Special cases: :

(i)When the colliding particles have equal masses.
i.e. $m_1 = m_2$ , then (4) and (5) give $v_1 = u_2 \, \text{and} \, v_2 = u_1$ ………………………(6)

Thus in one-dimensional elastic collision, the particles simply exchange velocities during collision.
(ii)If the particle $m_2$ is initially at rest, i.e $u_2 = 0$, then equations (4) and (5) give $v_1 = \dfrac{m_1 - m_2}{m_1 + m_2} u_1$ and $v_2 = \dfrac{2m_1}{m_1 + m_2} u_1$……………………………(7)

(a) Again if $m_1 = m_2$, then $v_1 = 0$ and $v_2 = u_1$ i.e. in one dimensional elastic collision between two particles of equal masses, if second particle is at rest; the first particle is stopped and the second particle takes of the original velocity of the first particle.

(b) If $m_2 >> m_1$ (i.e. If second particle is too heavy). then equations (7) give $v_1 = \dfrac{(m_1 / m_2) - 1}{(m_1 / m_2 ) + 1} u_1 = - u_1$

And $v_1 = \dfrac{2 (m_1 / m_2)}{(m_1 / m_2 ) + 1} u_1 = 0$

That is when a light particle collides with a much heavier particle at rest, the velocity of light particle is nearly reversed and the heavier particle remains nearly at rest.
For example if a ball strikes with a wall elastically, the ball rebounds with a reversed velocity, the wall still remaining at rest.

(c) If $m_2 >> m_1$ (i.e. incident particle is too heavy), then we get
That is in the collision of a heavy particle with a light particle at rest, the velocity of heavy particle remains unchanged; while light particle runs with nearly twice the velocity of the incident particle.

## 2. Inelastic collisions :

In an inelastic collision the total kinetic energy does not remain constant, but may be less than or greater than as the case may be. For example, in the presence frictional forces, the final kinetic energy is less than initial kinetic energy; the difference being convened to heat; while in the explosion of a bomb the final kinetic energy is greater than initial kinetic energy, the difference of kinetic energy being released from internal (atomic) energy.
In an inelastic collision, the momentum remains conserved and we have $m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2$ …………………(10)

And $e = - ( \dfrac{v_1 - v_2}{u_1 - u_2}$
or $v_1 - v_2 = - ( u_1 - u_2 )$ ……………………..(11)

where e is called the coefficient of restitution.
(For an elastic collision e = 1)

Solving (10) and (11), we get $v_1 = \dfrac{1}{m_1 + m_2} [ ( m_1 u_1 - m_2 u_2 ) - em_2 ( u_1 - u_2 ) ]$…….(12) $v_1 = \dfrac{1}{m_1 + m_2} [ ( m_1 u_1 - m_2 u_2 ) - em_1 ( u_1 - u_2 ) ]$………(13)

In a completely inelastic collision, the two colliding particles stick together after collision, so that they will have a final common velocity v. Then the conservation of linear momentum only, gives $m_1 u_1 + m_2 u_2 = ( m_1 + m_2 ) v$…..(14)

Remarks :

(1)    If the collision is two dimensional, then we apply the principle of conservation of linear momentum along the two dimensions separately and get the equations

For X-axis $m_1 u_{1x} + m_2 u_{2x} = m_1 v_{1x} + m_2 v_{2x}$

For Y-axis $m_1 u_{1y} + m_2 u_{2y} = m_1 v_{1y} + m_2 v_{2y}$…………..15

(2)    In two-body collision problems it is convenient to use two equations.

(i) Conservation of momentum $m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2$

(iii)               Newton’s experimental formula $v_1 - v_2 = - ( u_1 - u_2 )$

Equilibrium and Potential Energy:

In translational equilibrium $\sum \overrightarrow{F} = 0 \, i.e \, \sum F_x = 0 , \sum F_y = 0 , \sum F_2 = 0$
There are three types of equilibrium :

### 1.Stable Equilibrium :

If on displacing a body a little from its equilibrium position, the restoring forces are developed in such a way so as to bring the body back to its equilibrium position, then the equilibrium of the body is said to be the stable equilibrium.
In stable equilibrium position the potential energy of the body is minimum . A particle at the lowest position of a well is in stable equilibrium.

### 2.Unstable equilibrium :

If on displacing a body from its equilibrium position, the restoring forces are developed in such a way so as to take the body away from equilibrium position, then the equilibrium of the body is said to be unstable equilibrium.
In unstable equilibrium position, the potential energy of the body is maximum.
A particle at the highest position of a mountain is-in unstable equilibrium.

### 3. Neutral Equilibrium :

If on displacing a body from its equilibrium position, no restoring forces are developed, then the equilibrium is said to be neutral equilibrium.
In neutral equilibrium position, the potential energy is stationary.
A particle (e.g. pen, book) placed on a table is in neutral equilibrium.

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