Question

Show that the sum of kinetic energy and potential energy is always conserved

in case of a freely falling body under gravity from a height ‘h’ by finding it

when(i) the body is at the top (ii) the body has fallen a distance x (iii) the body

has reached the ground​

Answer 1

Answer:

Explanation: Let a body of mass in be falling freely under gravity from a height h above the ground (i.e., from position A). ... the sum of kinetic energy and potential energy always remain constant at each point of motion and is equal to initial potential energy at height h.

Answer 2

Answer: Mark it as the Brainliest, hope it helps you !!!!

The sum of kinetic energy and potential energy always remain constant at each point of motion and is equal to initial potential energy at height h.

Explanation:

(i) At the position A (at height h above the wound):

Initial velocity of body K=0 (since body is at rat at A)

Hence, kinetic energy =0

Potential energy U=mgh

Hence total energy =K+U=0+mgh=mgh ...(i)

(ii) At the position B (when it has fallen a distance x):

Let vt be the velocity acquired by the body at B after falling through a distance x. Then u=0,S=x,a=g

From equation v

2  = u

2 +2As

v = 1

2 =0+2gx=2gx

Hence, Kinetic energy K= 2 1 mv

1 2

Now at B, height of body above the ground =h−x

Hence, potential energy U=mg(h−x)

Hence total energy =K+U

mgx+mg(h−x)−mgh ...(ii)

(iii) At the position C (on the ground):

Let the velocity acquired by the body on reaching the round be v.

Then u=0,S=ha=g

From equation: v 2 =u2 +2aS

v 2 =0

2+2gh

v 2 =2gh

Hence, kindle energy K= 2 1 mv

2 = 2 1

m=(2gh)=mgh

And potential energy U=0 (a the wound when h=0)

Hence total energy =K+U=mgh−0=mgh ...(iii)

Thus from equation (i), (ii) and (iii), we note that the total mechanical energy i.e.. the sum of kinetic energy and potential energy always remain constant at each point of motion and is equal to initial potential energy at height h.