Question

prove that the mechanical energy (KE+PE) of a freely falling body always remain constant​

Answer 1

It may be shown in the absence of external frictional force of total mechanical energy of a body remains constant.

Let a body of mass ‘m' falls from a point A, whish is at a height ‘h' from the ground as shown in fig.

If the body reaches the point C ,

At A ,

Kinetic energy ( KE ) = 0

Potential energy (PE ) = mgh

Total energy E = PE + KE = mgh + 0 = mgh .

During the fall , the body is at position B .The body has removed a distance ‘x' from A .

At B ,

Velocity v² = u² + 2as

So, v² = 0 + 2ax = 2ax

Kinetic energy (KE) = 1/2 mv²

= 1/2 × m × 2gx

Potential energy (PE) = mg (h-x)

Total energy (E) = PE + KE

= mg(h-x) + mgh

= mgh - mgx + mgx

Total energy = mgh

At C,

Potential energy Ep = 0

Velocity of body C is ,

v² = u² + 2as

In this formula, u = 0 , a = g , s = h.

Now , applying v² = 0 + 2gh = 2gh......(1)

Kinetic energy (KE) = 1/2 mv²

= 1/2 × m × 2gh....(from 1 )

Now, Total energy at C ,

E = PE + KE

= mgh + 0

E = mgh

Thus, we have seen that sum of potential energy and kinetic energy of freely falling object at all points remains same. under the force of gravity,the mechanical energy of a body remains constant.

Answer 2

$\dag\:\:\underline{\sf Question:- :}$

Prove that the mechanical energy(KE + PE) of a freely falling object always remains constant.

$\dag\:\:\underline{\sf Answer:- :}$

We know that,

                        Energy can neither be created nor be destroyed, it can be transformed from one form to another. The total energy in the Universe remains constant.

Let a body of mass 'm' be at rest.

Let 'h' be the height of the body from earth's surface.

Let 'v' be the velocity of the body after it has covered distance 'x'.

Let 'v₁' be the velocity of the body at earth's surface.

_________________________________________________

Mechanical energy of the body at point A:-

Energy = Kinetic Energy + Potential Energy₁

             $= \frac{1}{2} mv^{2} + mgh$

             $= \frac{1}{2} m(0)^{2} + mgh$

             $= 0 + mgh$

             $= mgh$

Energy at point A = mgh

Mechanical energy of the body at point B:-

Energy = Kinetic Energy + Potential Energy

             $= \frac{1}{2}mv^{2} + mgh\\= \frac{1}{2} mv^{2} + mg(h -x)$ -----eqⁿ.1

From third equation of motion at point A and B,

v² = u² + 2as

As the body is falling towards the earth, 'a' would become 'g'

v² = u² + 2gs

v² = (0)² + 2gs

v² = 2gs

∵ s = distance covered

s = x

v² = 2gx

Putting the value of v² in eqⁿ.1,

        $= \frac{1}{2}m(2gx) + mg(h - x)\\= \frac{1}{2}m(2gx) + mgh - mgx\\\\= mgx + mgh -mgx\\= mgh$

Energy at point B = mgh

Mechanical energy of the body at point C:-

Energy = Kinetic Energy + Potential Energy

            $= \frac{1}{2}mv^{2} + mgh\\= \frac{1}{2} m(v₁)^{2} + mg(0)\\= \frac{1}{2} m(v₁)^{2} + 0\\= \frac{1}{2} m(v₁)^{2}$----eqⁿ.2

From third equation of motion at point A and C,

v² = u² + 2as

As the body is falling towards the earth, 'a' would become 'g'

v² = u² + 2gs

(v₁)² = (0)² + 2gs

(v₁)² = 2gs

∵ s = distance covered

s = h

(v₁)² = 2gh

Putting the value of (v₁)² in eqⁿ.2,

    $= \frac{1}{2} m(2gh)\\= mgh$

Energy at point C = mgh

As the mechanical energy at all the points is 'mgh', that is same. We infer that the mechanical energy of a freely body will be constant.

Hope this helps! ♡