WORK - POWER - ENERGY

   
                                       = area under the curve BCDA

Thus, work done by a variable, force is numerically equal to the area under the force and displacement curve.  

ENERGY 

        Energy of a body is its capacity to do work. Units and dimensional formula are same as that of work. It is a scalar quantity. Different forms of energy are mechanical, light, heat, electrical, nuclear, ........., etc. In this chapter we discuss only mechanical energy.
Example:
i) A bullet fired into a wall is able to penetrate into the wall by doing work against friction due to its energy.
(ii) Fast blowing wind is able to turn a wheel. Dimensional formula for energy is [ML²T⁻²], SI unit is joule (J), CGS unit is erg. 1 J = 107 erg. Other units are l ev = 1.6 × 10⁻¹⁹ J.
1 KWH = 36 × 10¹² erg = 36 × 10⁵ J.
In mechanics, mechanical energy is divided into two forms.
(i) Kinetic energy (ii) Potential energy.
Kinetic energy: It is the energy possessed by a body by virtue of its motion.
Examples: (i) A vehicle is in motion (ii) Water flowing along a river (iii) A bullet fired from a gun.
        Consider a body of mass 'm' is moving with a velocity 'v'. A uniform force opposes its motion to bring the body to rest in a displacement 's'. The uniform retardation of the body due to force is 'a', which is obtained by the kinematic equation v² − u² = 2as.

Here, initial velocity u = v (given), final velocity (v) = 0
   -v² = + 2 as ⇒ a = − 
From Newton's third law, force applied by the body

                                  
As the body moves against the applied force, its displacement is along a force applied by the body. Hence Work done

    
This work done resides in the form of kinetic energy in the body.

Work - Energy theorem: This theorem gives the interrelation between work and KE of the body. 'The change in KE of the particle is equal to the work done on it by the net force'.
          Consider a particle of mass 'm' is moving with an initial velocity 'u'. When it is under the action of a constant force 'F', it gets an acceleration 'a', velocity 'v' after a displacement 's'. Then v² − u² = 2as
On multiplying both sides with 

  KE_f  − KE_i = W

Work -Energy theorem is proved.

Work-Energy theorem for a variable force: 
The time rate of change of kinetic energy is

 
Integrate on both sides from initial position xi to final position xf.

Thus, the 'Work Energy' theorem is proved for a variable force.
* This theorem is not only applicable for a single particle but also for a system.
* It is also applicable for a system under the action of variable force, conservative and non-conservative forces.
Potential Energy (PE):
The word PE suggests possibility (or) capacity for action. The word potential means 'stored'. Thus, the energy possessed by a body by virtue of its position is called 'PE'. It is defined for conservative forces. It does not exists for non-conservative forces.
Expression for Gravitational Potential Energy:
Consider a body of mass 'm' is on the ground. It is lifted vertically upward to a height of 'h'. If  h < < < R (radius of earth), we can ignore the variation of 'g'. The gravitational force 'mg' is taken to be constant. Work done against gravitational force is 'mgh'. This work gets stored in the form of P.E. of the body. Now PE of the body is a function of 'h' denoted by U(h) = mgh   -------- (1)

The gravitational force 'F' equal to -ve derivative of U(h) with respect to h.

       
Where −ve sign indicates that the gravitational force is attractive and acts towards downwards.

                  
 F (dh) = −dv
Integrate on both sides

    
The work done by a conservative force such as gravity depends on initial and final positions only. Dimensions of potential energy are [ML²T⁻²]. Unit in SI system is joule (J), same as kinetic energy and work.

Examples:
(i) Stone kept at certain height.
(ii) Energy possessed by a bended bow, wounded spring of a watch.
(iii) Energy possessed by water stored in a tank.
PE of a Spring:
 A spring force is an example of a variable force which is conservative. The figure shows one end of massless spring attached to a rigid vertical support and other end to a block of mass 'm' resting on smooth horizontal surface. Let x = 0 denote the position of block, when spring is at its natural length. In an ideal spring, the spring force Fs is directly proportional to 'x'. When 'x' is the displacement of block from equilibrium position. 'x' and Fs may be +ve or − ve as shown in figure.      

This force law for spring is called ''Hook's Law" which states that Fs = −kx, k - spring constant, unit is N − m⁻¹. 'k' measures the stiffness of spring. When the block attached to spring is pulled out, the extension is xm, then work done by the spring force

         
Work done by spring force is always -ve. If block moves from initial position x_i to final position x_f, work done by spring force

           
If the block is pulled from initial position xi and returned to same position, then

Thus, work done by spring force depends on initial and final positions of the spring and independent on path followed. So, the spring force is a conservative force.
Conservative Force: Work done by the force around a closed path is zero and it is independent of the path, such a force is called conservative force.
Work done by the conservative force is stored in the form of potential energy.
e.g.: Gravitational force, electrostatic force and spring force.

Example: 1) In the absence of air resistance, a body is projected vertically up then work done by the gravitational force in moving the body through a height 'h' is W= −mgh and in return journey W = +mgh. On reaching the ground, the network done by the gravitational force in a round trip is zero.
W_Total = −mgh + mgh = 0
Example: 2) A body of mass 'm' lifted to a height 'h' from the ground level in different paths in between two points A & B. The work done by the gravitational force in all paths is same i.e., W₁ = W₂ = W₃ = mgh. So, work done by the conservative force depends on the path followed by the body.

Non-Conservative force: If work done by the force around a closed path is not equal to zero and it is dependent on the path, such a force is called non-conservative force.
e.g.: Frictional force, viscous force.
      Work done by a non-conservative force will not be stored in the form potential energy.
Example - 1: In the presence of air resistance, when a body is projected up, then it reaches a maximum height 'h'. Work done by the air friction in upward journey is '−fh' and in return journey 'fh'. On reaching the ground net workdone by the air friction is −ve, non - zero.      
Example - 2: A block of mass 'm' is dragged on a rough horizontal surface through distance 's' from the the point 'P' to point 'Q' and then back to the point 'P'.

Work done by the frictional force from P to Q is −ve, Q to P is −ve. So the work done by the frictional force around a closed path is -ve and not equal to zero.
W₁ ≠ W₂ ≠ W₃
Law of conservation of energy:
''Total mechanical energy of the system is conserved if the internal forces doing work on it are conservative and external forces do no work". If some of the forces are non-conservative, part of mechanical energy may get transferred into other forms such as heat, light and sound. The total energy of an isolated system remains constant. ''Energy neither be created nor be destroyed, but one form of energy transferred into another form of energy".
The principle of conservation of energy cannot be proved. However, no violation has been observed.
 Total mechanical energy (E) = PE + KE = u + k = constant.

Under only conservative forces mechanical energy of a system is constant.
         E₁ = E₂ ⇒ u₁ + k₁ = u₂ + k₂
      ⇒  k + u = Constant ⇒ ∆ k + ∆ u = Constant 
                                       ⇒ ∆ (k + u) = ∆ (Constant)
                                       ⇒ ∆ (k + u) = 0
                                     k + u = E = Constant
 Mechanical energy (E) = Constant
  i.e., if K.E. of the body increases, its potential energy will decrease by an equal amount and vice versa.
                 k + u = C
                 k = −u + C
Compare with y = mx + c.
Hence slope m = −1 = tan θ
               ⇒ θ = 135°

 Gain in K.E. = loss in P.E. 

Law of conservation of energy in case of a freely falling body:
 A body of mass 'm' is at a height 'h' from the ground as the body falls freely under gravity, the potential energy decreases and kinetic energy increases.
Total energy at Point A:
PE of the body = mgh
velocity of the body, v = 0
KE of the body =  mv² = 0
Total Energy (TE)A = PE + KE
                                = mgh + 0 
                                = mgh  ............... (1)
At point B:
As the body falls freely, consider a point 'B' in its path where AB = x
PE of the body = mg(h − x) = mgh − mgx
Let V₁ be the velocity of the body at 'B'
Initial velocity u = 0 (at  A),
Final velocity v = v₁, acceleration a = g,

Distance travelled s = x
Using v₂ − u₂ = 2as

Total energy at B = PE + KE
                            = mg(h − x) + mgx
                            = mgh    .............   (2)
At point C: The body touches the ground at C with a velocity 'V₂'
PE at C :    = mg(0) = 0      
Again u = 0, v = v₂, a = g & s = h

T.E.C = PE + KE = 0 + mgh = mgh  ------ (3)

.^.. From (1), (2) & (3),  TEA = TEB = TEC
Thus, law of conservation of energy is verified.
Various forms of energy: 
Energy comes in many forms which transform into one form to another form. Few forms of energy are discussed below.
1. Heat: A block of mass 'm' sliding on a rough horizontal surface and comes to rest over a distance by work-energy theorem W= change of K.E., here in this case loss of KE by the block due to friction is transferred as heat energy. This raises the internal energy of the block. In winter, in order to feel warm, we generate heat by vigorously rubbing our palms together. A quantitative idea of transfer of heat energy is obtained by noting that 1 kg of water releases 42,000 J of energy when it cools by 10°C.
2. Chemical energy: 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 separated parts. If the total energy of the reactants is more than the product of the reaction, heat is released and reaction is said to be an exothermic reaction. If the reverse is true, heat is absorbed, and reaction is endothermic coal consists of carbon and a kilogram of it, when burnt releases about 3 × 10⁷ J of energy. Chemical energy is associated with the forces that give rise to the stability of substances.

3. Electrical energy: The energy associated with an electric current is called electrical energy. The flow of electric current causes bulbs to glow, fans to rotate and bells to ring. An urban Indian household consumes about 200 J of energy per second on an average.
4. Nuclear energy: The most destructive weapons (fission and fusion bombs) made by man. Energy output of the sun and energy of stars is due to fusion of four light hydrogen nuclei form a helium nucleus whose mass is less than mass of reactants. This mass difference is called mass defect (∆m) which is the source of energy (E = ∆mc²). In fission, a heavy nucleus (₉₂U²³⁵) splits into two smaller nuclies. The energy released in this reaction can also be related to the mass defect.
The Equivalence of Mass Energy:
        Albert Einstein showed that mass and energy are equivalent and are related by the relation E = mc², c is speed of light in vacuum ( = 3 × 10⁸ m/s). The energy associated with a kg of matter.
        E = (1) (3 × 10⁸)² = 9 × 10¹⁶ J
This is equivalent to the annual electrical output of a large (3000 MW) power generating station.

Power: Workdone per second (or) the time rate of doing work is called power. If 'w' is the total workdone by a foce in a time interval 't', then the average power is

      
Instantaneous power is the dot product of force and velocity of body, provided the force does not change with time.
Relation between instantaneous power and average power:
A particle starts from rest and moving with uniform  acceleration and gains a velocity 'v' in time 't', then

Units of Power: Watt (or) Joule per second (SI),
                             erg per second (CGS).
                             1 horse power = 1 hp = 746 watt.       
Watt: When work is done at the rate of one joule per second, the power is said to be one watt. Dimensional formula is [ML² T⁻³]

* Slope of w-t curve gives instantaneous power.
* Efficiency of a crane or motor is the ratio of output power to the input power.

Collisions: Generally collision is refered to the case when two bodies come in physical contact with each other. Collision between two billiard balls or between two automobiles on a road, cricket bat hitting a ball are few examples of collision from every day life. In certain situations no physical contact but the path of one body is changed by the influence of another body, collision is said to have taken place.

                              
Example - 1: In Rutherford's scattering, the α − particles are scattered due to electrostatic interaction between the α − particle and the nucleus from a distance, (no physical contact between α − particle & nucleus).
Example - 2: Two similarly charged particles separated by a finite distance may collide by interaction through their electric fields.
In a collision, before and after the impact, the interaction forces between the colliding particles becomes effectively zero.
In a collision the effect of external forces such as gravity (or) friction are not taken into account as due to small duration of collision. The average impulsive force is responsible for collision is much larger than external force acting on the system.

"The strong pyhysical interaction among bodies involing exchange of momenta in small interval of time is called a collision"

         
Classification of collisions on the basis of direction of motion of colliding bodies: On the basis of direction of motion of colliding bodies, collisions, are classified into two types.
1. Head on (or) one dimensional (or) direct collision
2. Oblique collision.
Head-on-collision: In a collision if the motion of colliding particles before and after collision are along same straight line. Then collision is said to be head on collision.
Oblique collision: If the motion of colliding objects before and after collision are not along the initial line of motion, such collisions are known as oblique collision. In this collision, if the objects travel in a plane before and after collsion, the collision is called two dimensional collision. If the objects travel in space and collide, it is called three dimensional collision.

            
Types of collisions on the basis of Law of conservation of kinetic Energy: On the basis of conservation of total kinetic energy of colliding objects, collisions are classified into two types. They are (a) Elastic collision (b) Inelastic collision.
Elasitc Collision: The collision, in which both momentum and kinetic energy remains constant are known as elastic collision.
e.g.: (i) Collsions between nuclei
        (ii) Collisions between fundamental particles like electrons, protons, α − particles etc.

(iii) According to kinetic theory of gases, the collisions between gas molecules.
* In elastic collisions, objects regain their shape and size completely. Forces involved during interaction are conservation.
Inelastic Collions: The collisions in which kinetic energy is not conserved but law of conservation of momentum hold good are known as inelastic collisions.
Example: (i) Collision between cricket bal and bat.
                  (ii) Collision between automobiles on a road.
* In this collision the colliding bodies does not regain their shape and size completely after collision.
* Some fraction of mechanical energy is retained by colliding objects in the form of potential energy.
* Some (or) all the forces involved are non conservative in nature.
Perfectly inelastic collisions: If in a collsion, the colliding objects, stick together, and move with common velocity, then collision is called perfectly inelastic collision.
Example: (i) Collision between bullet and block of wood, when bullet is embedded in the block.
                (ii) Collision between clay balls
                (iii) Collision between positively and negatively charged particles.

* In this collision law of conservation of linear momentum is conserved, kinetic energy is not conserved. But total energy is conserved.
Elastic collision in one dimension: Consider two masses m₁, m₂ suppose u₁, u₂, and v₁, v₂ be their velocities before collision and after collision respectively along the same straight line. Apply law of conservation of linear momentum, then momentum of the system before collision is equal to momentum of the system after collision. When two objects collide mutual impulsive forces acting over the collision time ∆t causes a change in their momenta.
     i.e., ∆P₁ = F₁₂ ∆t
     ∆P₂ = F₂₁ ∆t 

F₁₂ is force exerted on the first particle by the second particle.
F₂₁ is force exerted on the second particle by the first particle.

From Newton's third law, F₁₂ = −F₂₁
 ∆P₁ + ∆P₂ = 0 ⇒  P = constant
 m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂
m(u₁ − v₁) = m₂(v₂ − u₂) -------- (1)
According to law of conservation of kinetic energy

u₁ + v₁ = v₂ + u₂
(u₁ − u₂) = (v₂ − v₁) ............ (3)
Thus in head on collision (elastic), "The relative velocity of approach before collision is equal to the relative velocity of separation after collision".
From equation (3), v₁ = v₂ + u₂ − u₁
Substitute this value in equation (1), we get
m₁[u₁ − (v₂ + u₂ − u₁)] = m₂ (v₂ − u₂)
2m₁u₁ − m₁v₂ − m₁u₂ = m₂v₂ − m₂u₂
2m₁u₁ + (m₂ − m₁) u₂ = (m₁ + m₂)v₂

 
From equation (3), v₂ = u₁ − u₂ + v₁                                                                          
Substitute this value in equation (1)
m₁(u₁ − v₁) = m₂ [(u1 − u₂ + v₁) − u₂]
m₁u₁ − m₁v₁ = m₂u₁ − m₂ u₂ + m₂v₁ − m₂u₂
m₁u₁ − m₂u₁ + 2m₂u₂ = (m₁ + m₂)v₁

(4), (5) are final velocities.
Special cases:
Case (i): When two spheres have equal masses, m₁ = m₂ = m then from equations (4) & (5)
                 v₂ = u₁ and v₁ = u₂

           
i.e., if elastic collision takes place between two spheres of equal masses, they simply exchange their velocities after collision.
Case (ii): (a) First sphere is moving with a velocity (u₁) and second body (sphere) is at rest (u₂ = 0).

Then from equations (4) & (5)

(b) If  m₁ = m₂, v₂ = u₁, v₁ = 0
          i.e., the first mass comes to rest and pushes off the second mass with its initial speed on collision.
(c) When a smaller sphere collides with a heavier sphere (m₁< < < m₂) at rest (u₂ = 0), neglect mass 'm' as compared with 'm₂'.
         v₁ = -u₁   and   v₂ = 0
 i.e., when a lighter sphere collides with a much heavier sphere at rest, the heavier sphere almost remains at rest and lighter sphere rebounds with same velocity.
* When a heavier sphere collides with a lighter sphere (m₁ > > > m₂) at rest (u₂ = 0), 
neglect 'm₁' compared with 'm₂' then v₁ = u₁ and v₂ = 2u₁
i.e., The velocity of heavier sphere remains unchanged but the velocity of lighter sphere moves with double the velocity of the heavier sphere.
Case-3: Perfectly inelastic collision:
In perfectly inelastic collision, two spheres stick together after collision and move with common velocity 'v' according to law of conservation of linear momentum.

 m₁u₁ + m₂ u₂ = m₁ v + m₂ v
                         = (m₁ + m₂)v

   
Application
          A body of mass m₁ moves with kinetic energy k undergoes head on collision with another body of mass m₂ at rest. k₁, k₂ and p₁, p₂ are kinetic energies and linear momentums after collision.

     Transfer of KE are momentum in a head - on collision is maximum (100%) when
      m₁ = m₂, m₂ is initially at rest.
COEFFICIENT OF RESTITUTION (e):
          Let u₁ and u₂ be the velocities of the bodies before collision and v₁ & v₂ are their final valocities after colision all along the same line in same direction.
          Then coefficient of restitution is defined as the ratio of relative velocity of separation (v₂ − v₁) after collision to the relative velocity of approach (u₁ − u₂) before collision.

                
'e' depends on nature of colliding bodies, independent of masses and velocities. 'e' has no units and dimensions. When e = 1 means a perfectly elastic collision, e = 0 implies perfectly inelastic collision. Practical values of 'e' lies between '0' and '1'.
Determination of Coefficient of restitution: To determine the coefficient of restitution between two bodies of different materials, one of them is taken in the form of very heavy plate and other body in the form of a small sphere. The small sphere is dropped on to the plate from a height ''h₁''. It hits the plate with a velocity 'u₁'. Let the sphere rebounds to a height h₂ after collision. Let the velocity of plate before and after collision to be zero (u₂ = v₂ = 0).

      t₁, t₂ are time in seconds.
* For perfectly elastic collision, h₁ = -h₂
* For a perfectly inelastic collision, h₂ = 0 
* For other collisions, h₂< h₁
* For any collision h₂ can not be greater than h₁.     

Equation for height attained for freely falling body after number of rebounds on floor:
Let a small sphere be allowed to fall freely from a height 'h' onto the floor. It strikes the floor with a velocity u₁ so that

v₂ − u₂ = 2as 

     −ve sign indicates that after collision the motion of sphere is opposite to its initial direction of motion. If the sphere rebounds to a height 'h₁', then u = v₁
   v₂ − u₂ = 2as
(0)² − e² (2gh) = −2gh1 
Similarly after 2nd rebound, the ball rises to a height h₂ which is given by
h₂ = e²h₁ = e²(e²h)

 Similarly  h₃ = e⁶h

        
The height to which body rebounds after nth rebound is given by 
* Total distance travelled by the sphere before it stops bouncing is

* Total time taken by the ball to stop bouncing is

           
* Average speed of the body during its entire journey is

           
* Average velocity of the body during its entire journey

            
* Total displacement of the body during its entire journey = h
Oblique collision - Application:
 "In oblique collision (elastic) of two particles of equal masses if one is at rest, the recoiling particles move off at right angles to each other".
  In elastic collislon, momentum is conserved. So, conservation of momentum along X-axis yields.

mu = mv₁ cos θ₁ + mv₂ cos θ₂
u = v₁ cos θ₁ + v₂ cos θ₂ ................... (1)
 along Y-axis yields
0 = v₁ sin θ₁ − v₂ sin θ₂ .................. (2)
Squaring and adding (1) and (2), we get

0 = 2v₁v₂ cos (θ₁ + θ₂)
v₁ ≠ 0, v₂ ≠ 0
so, cos (θ₁ + θ₂) = 0

⇒ θ₁ + θ₂ = cos⁻¹ (0) = 90°
 θ₁ + θ₂ = 90°