Class 11 > Unit # 05: Work, Energy & Power > Law of Conservation of Energy

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Work and Energy - Law of Conservation of Energy

Q. State and Prove Law of Conservation of Energy.

Statement

"Energy can neither be created nor destroyed; however, energy can be converted from one form to another."

Figure: Freely falling body of mass $m$ at different heights demonstrating mechanical energy conservation.

Explanation & Mathematical Proof

Consider a body of mass $m$ at height $h$ above the ground. Let us calculate its total mechanical energy at three different points: point $A$ (at the top), point $B$ (after falling a distance $x$), and point $C$ (just before striking the ground).

1. At Point A (Initial Position)

At point $A$, the body is at rest ($v_i = 0$) at height $h$:

• Kinetic Energy ($\text{K.E.}_A$):
  $\text{K.E.}_A = \frac{1}{2}mv^2 = \frac{1}{2}m(0)^2 = 0 \quad \text{--- (i)}$
• Potential Energy ($\text{P.E.}_A$):
  $\text{P.E.}_A = mgh \quad \text{--- (ii)}$
• Total Energy ($E_A$):
  $E_A = \text{K.E.}_A + \text{P.E.}_A = 0 + mgh = mgh \quad \text{--- (A)}$

2. At Point B (During Free Fall)

Suppose the body is released from point $A$ and falls through a distance $x$. Its new height above the ground becomes $(h - x)$.

Using the third equation of motion ($2as = v_f^2 - v_i^2$) where $a = g$, $s = x$, and $v_i = 0$:

$2gx = v^2 - 0 \implies v^2 = 2gx$

• Kinetic Energy ($\text{K.E.}_B$):
  $\text{K.E.}_B = \frac{1}{2}mv^2 = \frac{1}{2}m(2gx) = mgx$
• Potential Energy ($\text{P.E.}_B$):
  $\text{P.E.}_B = mg(h - x) = mgh - mgx$
• Total Energy ($E_B$):
  $E_B = \text{K.E.}_B + \text{P.E.}_B = mgx + (mgh - mgx) = mgh \quad \text{--- (B)}$

3. At Point C (Just Before Striking Ground)

When the body reaches point $C$ just before striking the ground, its height becomes $h = 0$.

Using the third equation of motion where $s = h$ and $v_i = 0$:

$2gh = v_f^2 - 0 \implies v^2 = 2gh$

• Potential Energy ($\text{P.E.}_C$):
  $\text{P.E.}_C = mg(0) = 0$
• Kinetic Energy ($\text{K.E.}_C$):
  $\text{K.E.}_C = \frac{1}{2}mv^2 = \frac{1}{2}m(2gh) = mgh$
• Total Energy ($E_C$):
  $E_C = \text{P.E.}_C + \text{K.E.}_C = 0 + mgh = mgh \quad \text{--- (C)}$

From equations (A), (B), and (C), we conclude that:
$E_A = E_B = E_C = mgh$

Hence, the total mechanical energy of the system remains constant at all points during free fall, verifying the Law of Conservation of Energy.

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