Question

Show that for a freely falling body total energy is conserved

Answer 1

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Answer 2

$\huge\bold {\fbox{{Correct \: Question:-}}}$

Show that the energy of a freely falling body is conserved.

$\huge \bold{\fbox{Answer:- }}$

Energy of a freely falling body is conserved:

Let a ball of mass "m" at height from the ground level starts falling down from rest.

At point A:

We know that,

$\fbox{Potential \: energy = mgh}$

Since the velocity here is zero.

Therefore kinetic energy=0

Therefore, total energy=mgh+0

=mgh ...(i)

At point B:

It falls through a distance "s" from A to B.

Hence potential energy=mg(h-s)

Now, v²=u²+2gs

or, v²=2gs (u=0)

$\sf \underline{Kinetic \: energy= \frac{1}{2}m {v}^{2} = \frac{1}{2}m \times 2gs = mgh}$

Total energy=mg(h-s)+mgs

=mgh. ..(ii)

At point C:

v²-u²=2gh

or. v²=2gh. ..(u=0)

$\sf \underline{Kinetic \: energy= \frac{1}{2}m {v}^{2} = \frac{1}{2}m \times 2gs = mgh}$

and, potential energy=0

Total energy=mgh+0=mgh. ..(iii)

It is clear from expression(i),(ii) and (iii) that the total energy of ball is constant at every point.

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