Question

principle
State and prove work-energy
(fore 3 marks)​

Answer 1

$\mathcal{WORK\:\:ENERGY\:THEOREM}$

According to this theorem, work done by net force in displacing a body is equal to change in kinetic energy of the body.

Proof:

As we know, work done on a body by a variable force F in displacing the body from initial position $\sf{x_i}$ to the final position $\sf{x_f}$ is given by

$\sf{W=\displaystyle\int\limits_{x_i}^{x_f}F(x)\,dx.....(1)}$

If m is the mass of the body and $\sf{a_x}$ is acceleration produced in the body along x-axis, then $\sf{F(x)=ma_x}$ Put in (1),

$\sf{W=\displaystyle\int\limits_{x_i}^{x_f}ma_x\,dx.....(2)}$

Now, $\sf{a_x=\dfrac{dv_x}{dt}=\dfrac{dv_x}{dx}.\dfrac{dx}{dt}}$

But $\sf{\dfrac{dx}{dt}=v_x}$

$\sf{\therefore\:a_x=v_x\dfrac{dv_x}{dx}.....(3)}$

Putting in (2) and changing the limits of integration in terms of v,

$\sf{W=\displaystyle\int\limits_{x_i}^{x_f}mv_x\dfrac{dv_x}{dx}\,dx}\\\sf{W=\displaystyle\int\limits_{v_i}^{v_f}mv_x.dv_x}\\\sf{=\left[m\dfrac{v_x^2}{2}\right]_{v_i}^{v_f}}\\\sf{W=\dfrac{1}{2}m\left(v_f^2-v_i^2\right)=\dfrac{1}{2}mv_f^2-\dfrac{1}{2}mv_i^2}\\\sf{work\:done=increase\:in\:K.E.\:of\:the\:body.}$

$\mathbb{HENCE\:PROVED}$