Show that 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:
(i) The body is at the top.
(ii) The body has fallen a distance x.
(iii) The body has reached the ground. (air resistance neglected)
Answers
Given:
A freely falling body.
(i) The body is at the top.
(ii) The body has fallen a distance x.
(iii) The body has reached the ground. (air resistance neglected)
To find:
Show that the total mechanical energy is always conserved in the case of a freely falling body under gravity from a height 'h'
Solution:
Let us consider a body of mass m placed at A
We should neglect the effect of air resistance on the motion of body.
Velocity at point A = 0
At point A
Mechanical energy at Point A
Potential energy at A = E (pA) = mgh
Kinetic energy at A = E (kA) = 0
Total mechanical energy at A (Ea) = E (pa) + E (kA) = mgh + 0 = mgh ------ (A)
At point B
v2 = 2gh
Kinetic energy at B = E (kB) = 1/2 mv^2 = 1/2 m (2gh) = mgh
Potential energy at B = E (pB) = 0
Total mechanical energy at B= E (B) = E (kB) + E (pB) = mgh + 0 = mgh ----- (B)
At point C
When body moves from A to C, it covers distance 's'. and vc is velocity at C then from,
v^2 - u^2 = 2as
vc^2 - 0 = 2 gs
Kinetic energy at C = E (kC) = 1/2 mv^2 = 1/2 m (2gs) = mgs
Potential energy at C = E (pC) = mg (h - s)
Total mechanical energy at C = E (C) = E (pC) + E (kC) = mg (h-s) + mgs = mgh ------ (C)
From (A), (B) and (C) we have obtained,
E (A) = E (B) = E (C)
As the total mechanical energy was the same at all three points, this proves that the total mechanical energy of the body throughout the free fall is conserved.