Wednesday, January 11, 2012

Derivation of Kinetic energy formula and worked examples





      Where there are no opposing forces, a moving body needs noforce to keep it moving with a steady velocity ( Newton's first law of motion ). 

If, however, a resultant force does act on a moving body in the direction of its motion, then it will accelerate ( Newton's second law of motion ) and the work done by the force will become converted into increased kinetic energy in the body.

     In order to calculate the kinetic energy of a body of mass m moving with velocity v, we begin by supposing that the body starts from rest and is acted upon by a force F ( no friction or other forces acting ).

  This force will give the body a uniform acceleration a, and will acquire a final velocity v, after travelling a distance x. These quantities, a,v and x will be related by the equation
                          v² = u² + 2 ax

 
   In accordance with the law of conservation of energy, the work done by the force F in pushing the body through distance x will become converted into kinetic energy of motion in the body.
Thus,


                          work done = force × distance
                                           = F × x
but                             
                                            F = ma

therefore, substituting for F,  
                                       
                          work done = ma × x .......................( 1 )
Applying the equation   v² = u² + 2 ax and remembering that u = 0
                                    v² = 0 + 2 ax
whence                          a =   v² / 2x 
 

                               
substituting this value of a in equation ( 1 ), we obtain,

                          work done = m × v² / 2x × x = kinetic energy
or kinetic energy ( k.e.) = 1/2  m  v² 
                        
  This expression, 1/2  m  v², gives the energy in joules or ergs according to the system of units used.

  Worked examples :


  1. A stone of mass 500 g is thrown vertically upwards with a velocity of 15 m /s.
  Find : ( a) the potential energy at greatest height; ( b ) the kinetic energy on reaching the ground ( Assume g = 10 m / s ² and neglect air resistance. )

  Solution
    To solve this problem we use the equation of motion, v² = u² + 2 ax, replacing a by g since we are dealing with gravitational acceleration. Thus,

    v² = u² + 2 gx
in which u = 15 m / s
             v = 0   m / s
             g = - 10 m / s ²

hence, by substution,

             v² = 15² + 2 ( -10) × x
whence   x  = - 15² /  2 ( -10) = 11.25 m

              Potential energy = weight × height
                                        = mg × x ( m in kg, of course)
                                        = 0.5 × 10 × 11.25 = 56.25 J

    In accordance with the principle of conservation of energy, the whole of this potential energy becomes converted to kinetic energy when the stone reaches the ground again.

Hence
kinetic energy on reaching the gorund
=  56.25 J.
   
    2. During a shunting operation, a truck of total mass 15 metric tons ( t ) moving at 1 m / s, collides with a stationary truck of mass 10 t. If the two trucks are automatically connected so that they move off together, find their velocity. Also calculate the kinetic energy of the trucks: (a) before; ( b ) after collision. Explain why these are not equal.
Solution
By the principle of conservation of momentum,
momentum before collision = momentum after collision  
                                                                           
    Let v = common velocity after collision, then using t m / s units of momentum,

    ( 15 × 1 ) + ( 10 × 0 ) = ( 15 + 10 ) × v
    or                          = 15 / 25 = 0.6 m

    Using the formula K.E = 1/2  m  v² ( m in kg; v in m / s )
    K.E. before collision     = 1/2  m  v² = 1/2 × 15000 × 1² = 7500 J
    K.E. after collision        = 1/2  m  v² = 1/2 × 25000 × 0.6 = 4500 J
    In accordance with the principle of conservation of energy, the total energy after collision is the same as that before.

    Before collision the whole of the energy is kinetic in the moving truck, but when collision occurs part of this becomes converted into internal energy in both trucks ( k.e. and p.e. of molecules) and part into sound energy ( k.e. and p.e. of air molecules). The remainder is left as mechanical kinetic energy in both trucks. Consequently, mechanical kinetic energy after collision is less than mechanical kinetic energy before collision.

 3. Water is pumped through a hose-pipe at the rate of 75 litres/min and issues from the nozzle with a velocity of

20 m / s.

Find: ( a ) the force of reaction on the nozzle in newtons; ( b) the useful power of the pump in watts. ( Assume 1 litre of water has a mass of 1 kg ).
   
Solution
    The reaction on the nozzle is equal to the force required to set the water in motion. ( Newton's third law of motion )

    We know that - I hope - F = ma ( = the rate of change in momentum)
    this may be written       F = mv / t
    in which m = 75 kg
                  t = 60 s

Substituting these values we obtain,
reaction on the nozzle = F = 75 × 20 / 60 = 25 N

  The useful power of the pump may be found from the kinetic energy of the issuing water.

  kinetic energy supplied per second = 1/ 2 × ( mass of water / s ) velocity ²
                                                         = 1/ 2 × 75/60 × 20²
                                                         =250 J /s
  i.e., useful power = 250 W.