Q. 1 C3.5( 35 Votes )

# Write detailed an

Let us assume an object of mass m, is kept at rest initially at a height h above the ground, and object is suddenly released , so it will gain speed , and let the Velocity of object just before hitting the ground be v

The situation has been shown in figure

Initially object of mass m was at a height h above the ground so potential energy of the object is

P.E. = mgh

g is acceleration due to gravity

now when object approaches ground, just before hitting the ground let the velocity of object be v, so its kinetic energy will be

Now , applying newton’s third equation of motion for motion of object

v2 – u2 = 2aS

Here v is final velocity of object,

u is the initial velocity of the object,

S is the displacement covered by object and

a is its acceleration during motion

Now initially object was at rest, so its initial velocity

u = 0 m/s

the object is coming downwards due to gravitational force applied by earth, so it’s acceleration is g, i.e. acceleration due to gravity i.e.

a = g

due to acceleration final velocity of object changed to v, in moving from height h to ground, so its displacement is

S = h

And putting the values in the equation we get

V2 – 02 = 2gh

Or we get value of square of final velocity as

v2 = 2gh

Now, putting the value of v2 in equation of final kinetic energy of the object we get

Solving we get final kinetic energy of the object as

K.E. = mgh

Initial potential energy of the object is

P.E. = mgh

Now we can see final kinetic energy of the object is same as the initial potential energy of the object

And since initially object was at rest, so its initial kinetic energy was zero, and finally object was just near the ground i.e. its height from surface of earth is zero, so its final potential energy is also zero

This conveys that all the kinetic energy of a freely falling object on reaching the ground is nothing but the transformation of its initial potential energy

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