Question

prove that K+U = constant for a freely falling body​

Answer 1

Answer:

Explanation:

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Law of conservation of energy states that the energy can neither be created nor destroyed but can be transformed from one form to another.

Let us now prove that the above law holds good in the case of a freely falling body.

Let a body of mass 'm' placed at a height 'h' above the ground, start falling down from rest.

In this case we have to show that the total energy (potential energy + kinetic energy) of the body at A, B and C remains constant i.e, potential energy is completely transformed into kinetic energy.

At A, 

Potential energy = mgh

Kinetic energy = 1/2 mv² = 1/2 * m* 0Kinetic energy = 0 [the velocity is zero as the object is initially at rest]

\ Total energy at A = Potential energy + Kinetic energy

= mgh + 0 

Total energy at A = mgh       …(1)

At B,

Potential energy = mgh

= mg(h - x)            [Height from the ground is (h - x)]Potential energy = mgh - mgx

Kinetic energy = 1/2 mv²

The body covers the distance x with a velocity v. We make use of the third equation of motion to obtain velocity of the body.

v2 - u2 = 2aSHere, u = 0, a = g and S = x

v² - 0 = 2gx 

Here, u = 0, a = g and S = x

v² -  0 = 2gx 

v² = 2gx 

kinetic energy = 1/2 mv²

= 1/2 m2gx   

Kinetic energy = mgx

 Total energy at B = Potential energy + Kinetic energy

= mgh - mgx + mgx

Total energy at B = mgh …(2)

At C,

Potential energy = m x g x 0 (h = 0)

Potential energy = 0Kinetic energy = 1/2 mv²

The distance covered by the body is hv2 - u2 = 2aS

Here, u = 0, a = g and S = h

v² - 0 = 2gh 

v² = 2gh 

kinetic energy = 1/2 mv² 

= 1/2 m * 2gh 

Kinetic energy = mgh

 Total energy at C = Potential energy + Kinetic energy= 0 + mgh

Total energy at C = mgh …(3)It is clear from equations 1, 2 and 3 that the total energy of the body remains constant at every point. Thus, we conclude that law of conservation of energy holds good in the case of a freely falling body.

Attachment is given in reverse so start from 4 page

Answer 2

Explanation:

At A (when object in rest)velocity $= 0 = u$

KE$= \frac{1}{2} m {v}^{2}$

PE$= mgh$

TE=PE+KE$= mgh$

At B,(when object in motion)

Velocity v:

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

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

KE$= \frac{1}{2} m {v}^{2} = mgx$

PE$= mg(h - x)$

Total energy TE=PE+KE=$mg(h - x) + mgx = mgh$

At C,velocity v;

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

${v}^{2} = 2gh$

KE=$\frac{1}{2} m {v}^{2} = mgh$

PE=$0$

Through out the motion Total energy be conserved.

Therefore Sum of kinetic energy and potential energy always constant.

So,K+U=Constant[ Proved]