Science, asked by sudeep200422, 1 year ago

show that the energy of a freely falling, body is conserved

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Answered by ally7070
3
Object Falling from Rest. As an objectfalls from rest, its gravitational potential energy is converted to kineticenergy. Conservation of energy as a tool permits the calculation of the velocity just before it hits the surface. K.E. = J, which is of course equal to its initial potential energy.
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Answered by subhransumaharaj18
6

Suppose a ball of mass ‘m’ falls under the effect of gravity as shown in figure.

Let us find the kinetic and the potential energy of the ball at various points of its free fall. Let the ball fall from point A at a height h above the surface of the earth.




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.

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