A variable force, given by the 2 dimensional vector $\overrightarrow{F}=\left( 3{{x}^{2}}\widehat{i}+4\widehat{j} \right)$, acts on a particle. The force is in Newton and x is in metre. What is the change in the kinetic energy of the particle as it moves from the point with coordinates (2,3) to (3,0)?
(coordinates are in metres).
A. -7J
B. zero
C. 7J
D. 19J

Hint: Resolve the force into its horizontal and vertical components. Find the work done by each of the forces in the horizontal and vertical direction. Then use the work energy theorem that says that the total work done on a particle is equal to the change in its kinetic energy.
Formula used:
$W=\int{F.dx}$
W=Fd

Complete answer:
To find the change in the kinetic energy of the particle we will use the work energy theorem.
According to the work energy theorem, the total work done by all the forces on a particle is equal to the change in kinetic energy.
Therefore, we will find the total work done on the particle and that will be the change in kinetic energy of the particle.
Work is defined as the product of the force and the displacement of the particle in the direction of this force.
i.e. W=Fd.
In this case, the force acting on the particle is given as $\overrightarrow{F}=\left( 3{{x}^{2}}\widehat{i}+4\widehat{j} \right)$. We can divide this force into two forces. One force is acting in the horizontal direction (along x-axis) with magnitude of ${{F}_{x}}=3{{x}^{2}}$N.
 The other force is acting in the vertical direction (along y-axis) with magnitude of ${{F}_{y}}$= 4N.
According to the given data, the displacement of the particle in the horizontal direction is x=1m. The displacement in vertical direction is y= -3m because it is along the negative y-axis.
Since the force ${{F}_{x}}$ depends on the x coordinate, it is a variable force. Work done by a variable force is given as $W=\int{F.dx}$.
Here, the limits of x are from x=2 to x=3.
$\Rightarrow {{W}_{x}}=\int\limits_{2}^{3}{F.dx}=\int\limits_{2}^{3}{3{{x}^{2}}.dx}$
$\Rightarrow {{W}_{x}}=3\left. \left( \dfrac{{{x}^{3}}}{3} \right) \right|_{2}^{3}$
$\Rightarrow {{W}_{x}}=3\left( \dfrac{\left( {{3}^{3}}-{{2}^{3}} \right)}{3} \right)=(27-8)=19J$
The work done by the vertical force is ${{W}_{y}}={{F}_{y}}y=4(-3)=-12J$
Hence, the total work done on the particle is $W={{W}_{x}}+{{W}_{y}}=19-12=7J$.
This means that the change in kinetic energy of the particle is 7J.

Hence, the correct option is C.

Note:
The exact definition of work done on a particle is that it is the dot product of the force and the displacement of the particle.
i.e. $W=\overrightarrow{F}.\overrightarrow{d}$.
If the horizontal force was a constant that we could have found the dot product of force and the displacement vector.
Here the displacement of the particle is $\overrightarrow{d}=\widehat{i}-3\widehat{j}$.
And the constant force be $\overrightarrow{F}={{F}_{x}}\widehat{i}+{{F}_{y}}\widehat{j}$.
Then $W=\overrightarrow{F}.\overrightarrow{d}=\left( {{F}_{x}}\widehat{i}+{{F}_{y}}\widehat{j} \right).\left( \widehat{i}-3\widehat{j} \right)=\left( {{F}_{x}}-3{{F}_{y}} \right)J$.