Question

State the work-energy theorem and prove it for a variable force.

Answer 1

Step 1. Statement of the work-energy theorem,

The work-energy theorem states that the work done by the net force acting on a body is equal to the change in the kinetic energy of the body.

Step 2: Proof

Let $F$ be a variable force and $t=$time.

$\therefore$Work done by the force, $W={\int }_{x_0}^{x_1}F·dx$ …..$\left(i\right)$

Where $x_0$ is the initial position and $x_1$ is the final position.

We know that kinetic energy $K=\frac{1}{2}m{v}^2$.

where $m=$the mass of the object, $v=$velocity of the object.

Initially when the position is $x_0$ then the kinetic energy is $K_0$ and when the position is $x_1$ the kinetic energy is $K_1$.

Now differentiating $K$ with respect to $t$:

$$\begin{gathered} \Rightarrow \frac{dK}{dt}=mv\frac{dv}{dt} \\ \Rightarrow \frac{dK}{dt}=ma\frac{dx}{dt}\left[\because v=\frac{dx}{dt},\mathrm{acceleration}\left(a\right)=\frac{dv}{dt}\right] \\ \Rightarrow \frac{dK}{dt}=F\frac{dx}{dt}\left[\because F=ma\right] \\ \Rightarrow \mathrm{d}K=F\mathrm{d}x \\ \Rightarrow {\int }_{K_0}^{K_1}\mathrm{d}K={\int }_{x_0}^{x_1}Fdx \\ \Rightarrow △K=W\left[\mathrm{From}\left(\mathrm{i}\right)\right] \end{gathered}$$

Where $△K$ is the change in kinetic energy.

Hence proved.