Question

Show that the total mechanical energy of a freely falling body under gravity is conserved.

spam mat karna varna jaan se mar dungi.....❌❌❌​

Answer 1

Energy conservation of a free falling body :-

Let a body of mass m is at rest at a height h from the earth's surface, when it start falling, after a distance x ( point B ) it's velocity becomes v and at earth's surface its velocity is v'

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Mechanical energy of the body :

At point A :

➩ $\rm E_A$ = Kinetic energy + Potential energy

➩ $\rm E_A = \frac{1}{2} m 0^2 + mgh$

➩ $\rm E_A = mgh\:\:\:\:\:\: - i$

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At point B :

$\rm E_B = \frac{1}{2} m v^2 + mg ( h - x )$

From third equation of motion at points A and B

➩ $\rm v^2 = u^2 + 2gx$

➩ $\rm v^2 = 2gx$

On putting the value of v² in equation $\rm E_B = \frac{1}{2} m (2gx) + mgh- mgx$

➩ $\rm E_B = mgh \:\:\:\:\: -ii$

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At point C :-

➩ $\rm E_C = \frac{1}{2} m (v')^2 + mg \times 0$

➩ $\rm E_C = \frac{1}{2} m (v')^2$

From third equation at point A and C

➩ $\rm (v')^2 = u^2 + 2gh$

➩ $\rm (v')^2 = 2gh$

Substituting the value of (v')^2 in equation

➩ $\rm E_C = \frac{1}{2} m ( 2gh)$

➩ $\rm E_C = mgh \:\:\:\:\: -iii$

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From equation i , ii and iii

$\boxed{\rm E_A = E_B = E_C }$

Hence, the mechanical energy of a free falling body will be constant.