8) State and prove work-energy theorem for variable force.

Work-energy theorem for variable force states that " The work done by a variable force on a body is equal to the kinetic energy gained by the body."

Suppose:

m = mass of the body
u = initial velocity of the body
v = final velocity of the body
a = acceleration 
K_i = Initial kinetic energy
K_f = final kinetic energy

the rate of change of kinetic energy is given by:

$$\begin{gathered} \frac{dK}{dt} = \frac{d}{dt}\left(\frac{1}{2}m{v}^2\right) = m \frac{dv}{dt}\times v \\ \frac{dK}{dt} = Fv \left(\because F = m \frac{dv}{dt} (newton\text{'}w second law)\right) \\ \therefore \frac{dK}{dt} = F\frac{dx}{dt} \\ dK = F dx \\ on integrating from the initial position \left(X_A\right) to the final position \left(X_B\right), we have \\ {\int }_{k_i}^{k_f}dk ={\int }_{X_A}^{X_B} F\hspace{0.17em}dx \\ {K}_f - K_i = {\int }_{X_A}^{X_B} F\hspace{0.17em}dx = Work done \\ \therefore Work done on the body = Increase in the KE of the body \end{gathered}$$