Variation of Kinetic Energy 
Question: Which of the following graphs correctly describes the variation of kinetic energy with time of a block when it slides down a smooth inclined plane from rest? 
The answer is C but I do not understand why. How do we know if the velocity increases as it slides down the inclined plane?
 A: The conserved mechanical energy is:
$E=KE+V=KE+mgh$
where $h$ is the height of the body. As the body slides down due to the acceleration of gravity, the distance travelled would be $d=\tfrac{1}{2}a{{t}^{2}}$, where $a$ is the acceleration caused by the addition of the weight of the body and the vertical reaction force. If $L$ is the length of the inclined plane and $\theta $ the inclination angle, then:
$h=\left( L-d \right)\sin \theta =L\sin \theta -\tfrac{1}{2}a{{t}^{2}}\sin \theta $
Plugging this to the mechanical energy formula and rearranging the terms we get:
$KE=A+B{{t}^{2}}$
where $A=E-mgL\sin \theta $ and $B=\tfrac{1}{2}mga\sin \theta $. Since it is only the energy difference that is of physical importance, we can set the mechanical energy level such that $A=0$. So:
$KE=B{{t}^{2}}$
which is the equation for a convex parabola that crosses the origin.
A: Consider the following free body diagram:

The net force acting on the body along the surface = $mg \sin x$
Therefore, the net acceleration in that direction is $g \sin x$.
So the velocity of the body at a time t after starting to slide down the inclined plane is $ v = gt \sin x$
So, K.E. (kinetic energy) of the body is $$\frac{1}{2}mv^2 = \frac{1}{2}mg^2t^2 \sin^2 x$$
Hence we conclude that $K.E. \propto t^2$.
The graph of K.E. and t is a parabola. So option C is correct.
A: Its a graph of $y = {v(t)}^2= kt^2$. 
Graph of $x^2$ is:  
graph http://www4b.wolframalpha.com/Calculate/MSP/MSP29252043e2d43bebg8bb000049f405gb0d606d9b?MSPStoreType=image/gif&s=62&w=300.&h=190.&cdf=RangeControl
When the object moves down from the top, it gets accelerated due to gravititional force or its component so its velocity changes(increases) instant after instant; this means $v^2$ in $\frac{m}{2} v^2$ is a function of time squared. Thus, the variation of KE is a square. 

A: the value of ke increases exponentially as ke = 1/2 x m x v^2, so the v^2 means that the ke value keeps on increasing at a faster rate
