Q.A curved (wavy) rigid track lies in a vertical plane. Reading from left to right it starts at a high point A, descends to a low valley point B, rises again to a crest C, and then descends to a lower end point D; point A is at a slightly greater height than the crest C. The portion of the track from B through C to D is perfectly smooth (frictionless), while the portion from A down to B is rough. Three solid spherical balls of identical radii and masses are released from rest, one at a time, from A. On the rough part AB, ball 1 experiences friction large enough to roll down without slipping, ball 2 experiences only a small friction (so it slips as it moves), and ball 3 experiences negligible friction.

Conservation of Mechanical Energy

The Intuition First

Imagine you're holding a heavy stone at shoulder height. Your arm is tired. That stone has potential energy — energy stored because of its position. Now let it go. As it falls, it speeds up. The potential energy is turning into kinetic energy — the energy of motion. Just before it hits the ground, all the original potential energy has become kinetic energy.

Now imagine the reverse: you throw a ball straight up. It leaves your hand fast (lots of kinetic energy), rises, slows down, stops for an instant at the top (zero kinetic energy), then falls back. At the top, all the kinetic energy you gave it has turned back into potential energy.

This back-and-forth transformation — potential ↔ kinetic — is the heart of the idea. Energy doesn't disappear; it just changes form. That's conservation.

The Precise Statement

Total mechanical energy is the sum of kinetic energy ($K$) and potential energy ($U$):

The law says:

Or, in symbols:

What This Means in Practice

Let's go back to the falling stone. Suppose you hold it 5 metres above the ground. Its mass is 2 kg. Take $g = 10\ \text{m/s}^2$.

• At the top (initial):

  $K_i = 0$ (not moving)

  $U_i = mgh = 2 \times 10 \times 5 = 100\ \text{J}$

  $E_{\text{mech}} = 0 + 100 = 100\ \text{J}$

• Just before hitting ground (final):

  $U_f = 0$ (height = 0)

  $K_f = \frac{1}{2}mv^2$

  Conservation says $K_f = 100\ \text{J}$, so $\frac{1}{2} \times 2 \times v^2 = 100$, giving $v = 10\ \text{m/s}$.

You never needed to know the time of fall or acceleration. Energy conservation gave you the speed directly.

The Two Critical Conditions

If friction is present, some mechanical energy turns into heat (thermal energy). The total energy of the universe is still conserved, but mechanical energy alone is not.

A Simple Example to Cement It …