Question

State work-energy theorem. Prove it for a variable force.

Answer 1

Work energy theorem gives the relationship between change in kinetic energy and the work done by a force. Work is said to be done when the force acting on a particle changes its position. Upon differentiating the formula for kinetic energy, we get the equation in terms of force. And then by integrating it we can prove that work done by a force is equal to the change in kinetic energy.
Formula used:
$K=\dfrac{1}{2}m{{v}^{2}}$
$W=\int\limits_{{{x}_{i}}}^{{{x}_{f}}}{F.dx}$
$F=ma$

Complete step-by-step solution:
The work-energy theorem states that the work done by the net force acting on a body is equal to the change produced in the kinetic energy of the body.
Let $F$ be the variable force.
We have,
$W=\int\limits_{{{x}_{i}}}^{{{x}_{f}}}{F.dx}$ --------- 1
Where,
${{x}_{i}}$ is the initial position of the body
${{x}_{f}}$ is the final position of the body
And we know that,
Kinetic energy, $K=\dfrac{1}{2}m{{v}^{2}}$
Where,
$v$ is the velocity of the object
Then,
$\dfrac{dK}{dt}=\dfrac{d\left( \dfrac{1}{2}m{{v}^{2}} \right)}{dt}=mv\dfrac{dv}{dt}$
Velocity is change in displacement divided by change in time. And change in velocity over time is acceleration. Then,
$\dfrac{dK}{dt}=ma\dfrac{dx}{dt}$
Where,
$m$ is the mass of the object
$a$ is the acceleration
According to Newton's law, $F=ma$.
$\dfrac{dK}{dt}=F\times \dfrac{dx}{dt}$
$dK=Fdx$
Integrating on both sides,
$\int\limits_{{{K}_{i}}}^{{{K}_{f}}}{dK}=\int\limits_{{{x}_{i}}}^{{{x}_{f}}}{Fdx}$
Where,
${{K}_{i}}$is the initial kinetic energy of the body
${{K}_{f}}$ is the final kinetic energy of the body
From equation 1,
$\int\limits_{{{K}_{i}}}^{{{K}_{f}}}{dK}=W$
$\Delta K=W$
Therefore, the work done by the net force acting on a body is equal to the change in kinetic energy.

Note: If the kinetic energy of a particle is increasing, work done is said to be positive. And if the kinetic energy of a particle is decreasing, work done is said to be negative.