Question

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

The law of conservation of energy:

Energy can neither be created nor destroyed. Verification of the energy conservation law in the event of a body falling.

Consider a body of mass (m) raised above the ground to position A at a height (h).

(i) At A

P.E. = mgh

K.E. = 0      ...[∵ body is at rest]

∴ P.E. + K.E. = mgh + 0 = mgh   ...(i)

Let it fall from A to B through a distance (x)

In doing so v² - u² = 2gx

v² - 0 = 2gx     ...[∴ u at A = 0]

Body is still at height (h - x) above ground.

(ii) At B, P.E. = mg (h - x)

K.E. = `1/2 mv^2`

= `1/2 m (2gx)`

= mgx

∴ P.E. + K.E. = mg (h - x) + mgx

= mgh - mgx + mgx

P.E. + K.E. = mgh

When a body falls on ground, the velocity with which it falls v² - u² = 2gh

v² = 2gh

(iii) At C the ground

P.E. = 0      ...[∵ h = 0]

K.E. = `1/2 xx mv^2`

= `1/2 m (2gh)`

= mgh

P.E. + K.E. = 0 + mgh = mgh      ...(iii)

This we find in all three cases (i), (ii) and (iii)

The sum total of P.E. + K.E. = mgh which remains constant and hence verifies the law.