Q. 1 C3.5( 35 Votes )

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Answer :

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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