Validity of work-energy theorem in presence of non-conservative forces? How can the work-energy theorem be valid in presence of non-conservative forces since conservation of energy is not there?
 A: 
How can the work energy theorem be valid in presence of
  non-conservative forces since conservation of energy is not there?

The work-energy principle simply states that work is the net increase of KE

The principle of work and kinetic energy (also known as the
  work-energy principle) states that the work done by all forces acting
  on a particle (the work of the resultant force) equals the change in
  the kinetic energy of the particle


Imagine that the black force on the right (14 N) is the force of gravity acting on a body B, and that the red opposing force on the left (-10 N) is air drag, which is not conservative. The net resulting force (obtained by vector addition is +4N and this will determine an increase of KE. The principle just says that that increase is the work done on B.
The energy lost has been transformed into other forms of energy (heat, sound etc), and the general principle of conservation of energy is valid, but has nothing to to with the work-energy principle
A: The work-energy principle is valid regardless of the presence of any non conservative forces. As long as you are using the work done by the resultant force (and resultant moment when involving rigid bodies) in the equation (or equivalently adding the work done by each force/moment), the work energy principle is valid. This can be shown by considering a simple 1-dimensional example.
Consider a particle of mass $m$ that travels between points 1 and 2. It travels in a straight line, and has displacement for an origin $x$, a velocity $v$. It is acted upon by some forces, including conservation forces, $F_c$ and non-conservative forces, $F_{nc}$. Therefore, a net force $F_{net} = F_c + F_{nc}$ acts on the particle. By Newton's 2nd Law, note $F_{net} = ma$, where $a$ is acceleration.
Work is defined as the product of the force multiplied by the component of displacement in the direction of force, which is $F_{net} \cdot \Delta x$ for a constant force. If force is to vary over the path, then you must instead use $\int^{x_2}_{x_1}{F_{net} \cdot dx}$.
$$Work, W = \int^{x_2}_{x_1}{F_{net} \cdot dx} = m \int^{x_2}_{x_1}{a \cdot dx}$$
$$a = \frac{dv}{dt} = \frac{dx}{dt}\frac{dv}{dx} = v\frac{dv}{dx}$$
$$\therefore a \cdot dx = v \cdot dv$$
$$W = m \int^{v_2}_{v_1}{v \cdot dv} = m \left[ \frac{1}{2}v^2 \right]^{v_2}_{v_1} $$
$$\therefore W = \frac{1}{2}mv^2_2 - \frac{1}{2}mv^2_1 = \Delta KE$$
While this example is set for a 1-D problem, it can be expanded to 2 or 3-dimensions by setting the work as $\int^{x_2}_{x_1} \vec F_{net} \cdot \vec{dx}$ instead, using vector calculus, to obtain the same result.
