Prove using Newton's second law and kinematics.

Work-Energy Theorem — Net Work = Change in KE

Question

Prove the Work-Energy Theorem: Show that the net work done on an object equals the change in its kinetic energy.

W_{net} = \Delta KE = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2

Solution — Step by Step

We know that the net force on an object is $F_{net} = ma$ . This is our starting point — we want to connect force (and therefore work) to motion.

The strategy is to eliminate time from the equations, since work depends on displacement, not time.

Here’s the key move. From kinematics, we have $v^2 = v_0^2 + 2as$ , which gives us:

a = \frac{v_f^2 - v_i^2}{2s}

We chose this equation specifically because it contains displacement $s$ , not time $t$ .

Plug our expression for $a$ back into $F = ma$ :

F_{net} = m \cdot \frac{v_f^2 - v_i^2}{2s}

Work is defined as $W = F \cdot s$ (for constant force along displacement). So we multiply both sides by $s$ :

F_{net} \cdot s = m \cdot \frac{v_f^2 - v_i^2}{2s} \cdot s

W_{net} = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2

The term $\frac{1}{2}mv^2$ is kinetic energy by definition. So:

\boxed{W_{net} = KE_f - KE_i = \Delta KE}

Proved. Net work done on a body equals the change in its kinetic energy.

Why This Works

The theorem works because Newton’s second law is essentially a statement about how energy transfers into motion. When a net force acts on a body over a displacement, it’s doing net work — and that work has nowhere to go except into changing the body’s speed.

The trick in this proof is the kinematic substitution in Step 2. By using $v^2 = v_0^2 + 2as$ instead of $v = v_0 + at$ , we trade time for displacement — exactly what we need to bring work ( $W = Fs$ ) into the picture. This is a technique worth memorising: whenever you want to connect force to energy, eliminate time using the $v^2$ kinematic equation.

One important caveat: the proof above assumes constant force. For variable forces, you’d integrate — $W = \int F \, dx$ — and the result is the same, but that version is tested in JEE Advanced, not JEE Main.

Alternative Method

Using the Work-Energy Theorem directly from integration (for variable force):

If force varies with position, we cannot use $W = Fs$ directly. Instead:

W_{net} = \int_{x_i}^{x_f} F \, dx = \int_{x_i}^{x_f} ma \, dx

Now use the chain rule: $a = \frac{dv}{dt} = \frac{dv}{dx} \cdot \frac{dx}{dt} = v\frac{dv}{dx}$

W_{net} = \int_{x_i}^{x_f} mv \frac{dv}{dx} \, dx = \int_{v_i}^{v_f} mv \, dv = \left[\frac{1}{2}mv^2\right]_{v_i}^{v_f}

W_{net} = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2

The chain-rule substitution $a = v\frac{dv}{dx}$ is a scoring trick that appears in both JEE Main and NEET derivation questions. Write it down once on your formula sheet — it converts acceleration into a form that integrates cleanly with respect to position.

Common Mistake

Using total work instead of net work. Students often apply the theorem as $W = \Delta KE$ but plug in only one force’s work (say, only the applied force), forgetting friction or normal components. The theorem holds only for net work — the sum of work done by ALL forces acting on the body.

For example: a block pushed across a rough floor with applied force $F$ over distance $d$ . Work by $F$ is $Fd$ , but work by friction is $-\mu mg d$ . Net work is $(F - \mu mg)d$ , and that equals $\Delta KE$ . Using just $Fd = \Delta KE$ gives the wrong answer and is one of the most common errors in JEE Main MCQs on this topic.

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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