1. State and prove law of conservation of energy in case of freely falling body?
Answers
Answer:
Law of conservation of energy states that the energy of a system is always constant. In other words, we can say that energy can neither be created nor destroyed. In the case of a freely falling body, it is the mechanical energy of the system that is conserved. ... Therefore, E=K+U=constant.
Statement :-
Energy can neither be created nor be destroyed but it can transform from one form of energy another at constant Temperature.
Proof :-
Let us consider a body is freely falling from the height 'h' above the ground at three different points A,B,C.
At point A :-
P.E = mgh
K.E = 1/2 mv²
K.E = 1/2 m (0)
K.E = 0
Total energy at A = P.E + K.E
= mgh + 0
= mgh - - - - - - >(1)
At point B :-
P.E = mg(h-x)
P.E = mgh - mgx
Since,
v² - u² = 2as
v² - (0)² = 2as
v² = 2as
Acceleration, a = g
Displacement, s = (x)
There Fore,
K.E = 1/2 mv²
K.E = 1/2 m(2gx)
K.E = mgx
Total Energy at point B = P.E + K.E
= mgh - mgx + mgx
= mgh - - - - - - >(2)
At point C :-
P.E = mgh
P.E = mg(0)
P.E = 0
Since,
v² - u² = 2as
v² - (0)² = 2as
v² = 2as
Acceleration, a = g
Displacement, s = h
There Fore,
K.E = 1/2 mv²
K.E = 1/2 m (2gh)
K.E = mgh
Total Energy at point C :- P.E + K.E
= 0 + mgh
= mgh - - - - - - > (3)
From, (1),(2) and (3)
T.E A = T.E B = T.E C = mgh
Hence Proved Law Conservation Of Energy in the case of Freely Falling Body.
