Question

show that when a body is dropped from a certain height the mehanical energy at any instant during its fall is constant

Answer 1

We know that Mechanical energy is the sum of kinetic and potential energy

case 1

Let the body is at height h before being dropped

potential energy = mgh

kinetic energy = 0

Hence

Mechanical energy = kinetic + potential = 0+ mgh = mgh

case 2

let the body has fallen a distance x and is at a height (h-x)

we know that,

v² = u² + 2as = 0 + 2gx = 2gx

kinetic energy = mv²/2 = 2mgx/2= mgx

potential energy = mg(h-x) = mgh - mgx

Hence

Mechanical energy = kinetic + potential = mgx + mgh - mgx = mgh

case 3

The object has just fallen to the ground covering h distance

v² = u² + 2as = 0 + 2gh = 2gh

kinetic energy = mv²/2 = m2gh/2 = mgh

potential energy = 0

Hence

Mechanical energy = kinetic + potential = mgh + 0 = mgh

Hence from all the 3 cases we may see that the mechanical energy of a falling particle is constant.