Question

Show how the energy of a freely falling object remains conserved.??​

Answer 1

Answer:

At Point A: At point A, the ball is stationary; therefore, its velocity is zero.

Therefore, kinetic energy, T = 0 and potential energy, U = mgh

Hence, total mechanical energy at point A is

E = T + U = 0 + mgh = mgh … (i)

At Point B : Suppose the ball covers a distance x when it moves from A to B. Let v be the velocity of the ball point B. Then by the equation of motion v^2-u^2 = 2aS, we have

v^2 - 0 = 2gx or v^2 = 2gx Therefore,

Kinetic energy, T = 1/2 mv^2 = 1/2 x m x (2gx)

= mgx

And Potential energy, U = mg (h - x)

Hence, total energy at point B is

E = T + U = mgx + mg(h-x) = mgh …(ii)

At Point C : Suppose the ball covers a distance h when it moves from A to C. Let V be the velocity of the ball at point C just before it touches the ground. Then by the equation of motion v^2 - u^2 = 2aS, we have V^2 - 0 = 2gh or V^2 = 2gh.

Therefore,

Kinetic energy,

T = 1/2 mV^2 = 1/2 x m x (2gh) = mgh

and Potential energy, U = 0

Hence, total energy at point E = T + U

= mgh + 0 = mgh … (iii)

Thus, it is clear from equations (i), (ii) and (iii), that the total mechanical energy of a freely falling ball remains constant.