Question

Prove that 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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Answer 1

Explanation:

Let the velocity of the object be vB and the object has fallen through distance x. ... Thus, it can be concluded that the kinetic energy of a freely falling object on reaching the ground is nothing but the transformation of its initial potential energy.

Answer 2

Answer:

ur answer .....

Explanation:

the total energy is

                    $U_{A} +K_{A} =mgh.$    ....(1)

suppose the ball is allowed to fall. the only force on it during the fall is gravity (free fall), and hence, its acceleration is a=g downwards. consider any point B in its path. suppose its distance from A is x. The height of the point B from the ground is (h-x). when the ball B, its potential energy is

                                $U_{B}$ = mg(h-x)

the velocity of the ball at point B is given by

                     $v^{2} = u^{2} + 2gx= 2gx$

the kinetic energy of the ball at point B is

                        $K_{B}=\frac{1}{2} mv^{2} =\frac{1}{2} m(2gx)=mgx$

the total energy at point B is

                                                $U_{B} +K_{B}=mg(h-x)+mgx=mgh$   ...(2)

comparing (1) and (2), we see that

                          $U_{A} +K_{A} =U_{B} +K_{B} .$

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