Question

6. Show that the sum of kinetic energy and potential
energy the total mechanical energy) is always
conserved in the case of a freely falling body under
gravity from a height H by finding it when
1) the body is at the top ,
2) the body has fallen a distance x
3) the body has reached the ground ​

Answer 1

Answer:

Answer

Let a body of mass in be falling freely under gravity from a height h above the ground (i.e., from position A). Let at now calculate the sum of kinetic energy K and potential energy U at various positions. say at A (at Malt it above the pound), at B (when it has fallen through a distance x) and at C (on the around).

(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

=u

2

+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.

solution

Answer 2

Explanation:

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