Question

Show that the sum of kinetic energy and potential energy (i.e., total mechanical energy) is always conserves in the case of a freely falling body under gravity (with air resistance neglected) 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.

Accepted Answer

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

(i) At the position A (at height h above the ground):
Initial velocity of body, $u = 0 m s^{- 1}$ (since body is at rest at A)
Hence, kinetic energy $K = 0 m s^{- 1}$ .
Potential energy, $U = m g h$
Hence, total energy, $E = K + U = m g h . . . . . (1)$
(ii) At the position B (when it has fallen a distance x):
Let $v_{1}$ be the velocity acquired by the body at B after falling through a distance $x$ . Then:

$u = 0, S = x, a = g$ . Using the kinematic equation, we get:

${v_{1}}^{2} = 0 + 2 g x$
Hence, Kinetic energy, $K = \frac{1}{2} m {v_{1}}^{2} = \frac{1}{2} m \times 2 g x = m g x . . . . . (2)$
Now at B, height of body above the ground is $h - x$
Hence, potential energy, $U = m g (h - x)$
Hence, total energy, $T = U + K = m g (h - x) + m g x . . . . . (3)$
(iii) At the position C (on the ground):
Let the velocity acquired by the body on reaching the ground be $v .$ Then

$u = 0, S = h, a = g$
Using the kinematic equation, we get:
$v^{2} = 0 + 2 g h$
Hence, kinetic energy, $K = \frac{1}{2} m v^{2} = \frac{1}{2} m \times 2 g h = m g h$
And potential energy, $U = 0$ (at the ground when $h = 0$ )
Hence, total energy $T = K + U = m g h + 0 = m g h . . . . . (4)$
Thus, from the equations formed above, 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.

R P Goyal and S P Tripathi Solutions for Chapter: Work, Energy and Power, Exercise 7: EXERCISE-2(D)

R P Goyal Physics Solutions for Exercise - R P Goyal and S P Tripathi Solutions for Chapter: Work, Energy and Power, Exercise 7: EXERCISE-2(D)

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