State and prove the law of conservation of energy in the case of a freely falling object - Physics - Work Energy And Power

Question

State and prove the law of conservation of energy in the case of a
freely falling object

Aditi · Accepted Answer

m = mass of body

g = Acceleration due to gravity

Suppose a body is at point A initially. Then it falls freely, during the motion mechanical energy of the body is conserve.

At point A,

AC = height of object from ground

= h

Initial speed u = 0

So, P.E. = mgh

$K.E. = \frac{1}{2} m v^{2} = \frac{1}{2} m (0)^{2} = 0$

TE = P.E + K.E = mgh + 0

(TE)_A= mgh ... (i)

At point B,

PE = mg (BC)

= mg (h – x) = mgh – mgx

Use kinematics equation

$v^{2} = u^{2} + 2 g x \Rightarrow v'^{2} = 0 + 2 g x K.E. = \frac{1}{2} m v'^{2} = \frac{1}{2} m (2 g x) = m g x$

(TE)_B = P.E. + K.E = mgh – mgx + mgx

= mgh ...(ii)

At point C,

P.E. = 0

$K.E. = \frac{1}{2} m v^{2}$

Use kinematics equation

$v^{2} = u^{2} + 2 g h$

⇒ v ² = 2gh

So, $K.E. = \frac{1}{2} m (2 g h) = m g h$

(T.E.)_C= K.E + P.E. = mgh +0 = mgh ...(iii)

(TE)_A =(TE)_B=(TE)_C

Hence, the total energy of the body is conserve during free fall.

State and prove the law of conservation of energy in the case of a freely falling object.