Question

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

Answer 1

Answer:

Law of conservation of energy states that:−

Energy can neither be created nor be destroyed.

Energy transforms from one form to another form and, potential energy+ kinetic energy= constant

Proof:−

At point A

K.E=0m/s

P.E=mgh

P.E=mgh

At point B,

K.E.,

2gh=V

2

−O

2

V

2

=2gx

K.E=

2

1

mV

2

K.E=

2

1

m.2gx

K.E=mgx-----------------------------(1)

P.E=mgh

P.E=m.g.(h−x)

P.E=mgh−mgx---------------------(2)

from(1) and (2)

K.E+P.E constant

mgx+mgh−mgx=mgh

mgh=mgh

At point C,

P.E=0

2gh=v

2

−u

2

( initial velocity =0 m/s)

2gh=v

2

K.E=

2

1

.m.v

K.E=

2

1

.m.2gh

K.E=mgh

Thus, in all the points the energy is same.

Explanation:

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.

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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