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CBSE Class 11-science Answered

explain all the derivations of chapter-work power energy
Asked by ABHILASHA | 16 Sep, 2019, 06:55: PM
answered-by-expert Expert Answer
(1) Work-Energy Theorem  :-
The change in kinetic energy of a particle is equal to the work done on it by the net force.  
 
Kf - Ki = W
 
where Ki amd Kf are initial and final kinetic energy of the body. W is the workdone on the body.
 
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(2) conservation of mechanical energy
 
if ΔK is change in kinetic energy and ΔV is change in potential energy, then ΔK + ΔV = Δ(K+V) = 0
 
i.e., sum of kinetic energy and potential energy,  (K+V) is constant
 
that leads the conservation of energy relation,   Kf + Vf = Ki + Vi  
 
In the above relation K stands for kinetic energy and V is for potential energy.
Subscript i refers initial state and f refers final state.
 
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Potential energy V stored in spring ,    V = (1/2) k x2
 
where k is spring constant defined as  F = kx ,
 
i.e.  spring is elongated or compressed by a length x due to the application of force F.
 
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Instantaneous power = Force × instantaneous velocity  = F × v
 
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Inelastic collision :-  Momentum is conserved but energy is not conserved
 
Loss of kinetic energy in Inelastic collision ,  begin mathsize 12px style increment K space equals space 1 half fraction numerator m subscript 1 space m subscript 2 over denominator open parentheses space m subscript 1 space plus space m subscript 2 space close parentheses end fraction v subscript i superscript 2 end style
m1 and m2 are respective mass of  colliding particles. vi is initial velocity of one particle before collision
while another particle is at rest before collision.
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elastic collision :- Both momentum and energy are conserved
 
Let m1 is mass of a particle moving with initial speed v1i is colliding with another particle of mass m2
which is at rest, then final speeds of both the particles are given by,
 
begin mathsize 12px style v subscript 1 f end subscript space equals space fraction numerator open parentheses m subscript 1 space minus space m subscript 2 close parentheses over denominator open parentheses m subscript 1 plus m subscript 2 close parentheses end fraction v subscript 1 i end subscript space space space space space space a n d space space space space space v subscript 2 f end subscript space equals space fraction numerator 2 space m subscript 1 over denominator open parentheses m subscript 1 plus m subscript 2 close parentheses end fraction space v subscript 1 i end subscript end style
Special case (1) :-  if m1 = m2 , then v1f = 0  and v2f = v1i
 
if both masses are equal, particle of mass m1 will come rest after collision and other particle of mass m2
will move with initial velocity of particle of mass m1
 
Special case (2) :-  if  m2 > m1 ,  particle of mass m1 will move in opposite direction after collision.
 
 
 
 
Answered by Thiyagarajan K | 19 Sep, 2019, 08:52: AM
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