Question

A force $F=-k(x\hat i+y\hat j)$[ where $k$ is positive constant] acts on a particle moving is the x-y plane. Starting from origin, the particle is taken to $(a,a)$ and then to $\left(\dfrac{a}{\sqrt2},0\right)$. The total work done by the force $F$ on the particle is

& A.-2k{{a}^{2}} \\\ & B.2k{{a}^{2}} \\\ & C.\dfrac{3k{{a}^{2}}}{4} \\\ & D.\dfrac{-k{{a}^{2}}}{4} \\\ \end{aligned}$$

Answer 1

We know that the work is defined as the product of force and displacement. Here we have a force which displaces the particle on the x-y plane. Hence to calculate the work, we must calculate the displacement of the particle and then use scalar multiplication to find the work done.
Formula: $W=F\cdot d$

Complete answer:
We know that work done is the scalar product of force $F$ and the displacement $d$. And is given as $W=F\cdot d=Fdcos\theta$ where $\theta$ is the angle between the force $F$ and the displacement $d$.
Here, it is given that the $F=-k(x\hat i+y\hat j)$
Here we have two points, say $A$ is $(a,a)$ and $B$ is $\left(\dfrac{a}{\sqrt2},0\right)$.
Let us assume that the $dx\hat i+dy\hat j$ is a small displacement, then then small work $d\;W$ is given as
$dW=F\cdot d=-k(x\hat i+y \hat j)\cdot dx\hat i+dy\hat j$
$\implies dW=-k(xdx+ydy)$
To get the total work done, we integrate the small work done, using limits, it is given as
$W=-k\int\limits_{a}^{\dfrac{a}{\sqrt{2}}}{xdx}-k\int\limits_{a}^{0}{ydy}$
Integrating on both sides, we get
$\implies W=-k\left( \left. \dfrac{{{x}^{2}}}{2} \right|_{a}^{\dfrac{a}{\sqrt{2}}}+\left. \dfrac{{{y}^{2}}}{2} \right|_{a}^{0} \right)$
Applying appropriate limits we get,
$\implies W=-\dfrac{k}{2}\left( \dfrac{{{a}^{2}}}{2}-{{a}^{2}}+0-{{a}^{2}} \right)$
On simplification, we get
$\implies W=-\dfrac{k{{a}^{2}}}{4}$
Thus, the total work done by the particle in moving from A to B when $F$ force acts on it, is found to be $\dfrac{-ka^{2}}{4}\;units$

Hence clearly the answer is option $D.\dfrac{-k{{a}^{2}}}{4}$.

Note:
The scalar multiplication can be used to multiply a scalar and a vector or two vectors. If there are two vectors $\vec A$ and $\vec B$, then the scalar product is defined as $\vec A \cdot \vec B=|\vec A ||\vec B|cos\theta$, where $\theta$ is the angle between the$\vec A$ and $\vec B$. Here we are integrating to get the work done, the student must know some basic integration to solve this sum. The scalar multiplication is also known as the dot product. Also, the resultant of scalar multiplication is always a scalar.