Class 11 > Unit # 05: Work, Energy & Power > Work Energy Theorem

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Work and Energy - Work-Energy Theorem

Q. State and prove Work-Energy Theorem.

Statement

"The total work done on a body is equal to the change in its kinetic energy, provided the body is confined to move horizontally and no dissipative forces are operating."

Figure: A body of mass $m$ accelerating from initial velocity $v_1$ to final velocity $v_2$ under external force $F$.

Proof

Consider a body of mass $m$ moving with an initial velocity $v_1$. After travelling through a displacement $s$ under the action of a constant external force $F$, its final velocity becomes $v_2$.

According to Newton's second law of motion, the external force acting on the body is:

$F = ma$

The work done ($W$) by the force in moving the body through a displacement $s$ is:

$W = F \cdot s = (ma)s$

To eliminate acceleration $s$, we use the third equation of motion:

$2as = v_2^2 - v_1^2 \implies as = \frac{v_2^2 - v_1^2}{2}$

Substituting $as$ into the work equation:

$W = m(as) = m \left( \frac{v_2^2 - v_1^2}{2} \right)$

Expanding and rearranging the terms:

$W = \frac{1}{2} m v_2^2 - \frac{1}{2} m v_1^2$

$W = \text{Final K.E.} - \text{Initial K.E.} = \Delta \text{K.E.}$

This expression proves that the total work done on a body is equal to the change in its kinetic energy.

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