Question

A force $\overrightarrow F = 2xy\hat i + {x^2}\hat j$ acts on a particle moving in the $x - y$ plane. Starting from a point$\left( {0,2} \right)$ , the particle is taken along a straight line to $\left( {2,4} \right)$ . The work done by the force is
A. $16{\text{units}}$
B. $Zero$
C. $8{\text{units}}$
D. $8{\text{units}}$

Answer 1

The force is given as a vector term. So use the work-done formula by integrating the dot product of the force vector and the displacement vector.
Find the displacement from the given initial points from where the particle starts its journey to the final points of the plane.
Use the integration limit by taking the points on the x-axis and the y-axis.

Formula used:
Work done, $W = \int {\overrightarrow F .\overrightarrow {dr} }$
$\overrightarrow F = ({F_x}\hat i + {F_y}\hat j)$
$\overrightarrow {dr} = (dx\hat i + dy\hat j)$
Straight line equation,
$\dfrac{{y - {y_1}}}{{x - {x_1}}} = \dfrac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}}$

Complete step by step answer:
The work done is the scalar product of the force vector and the displacement vector or the position vector. This can be represented as, $W = \int {\overrightarrow F .\overrightarrow {dr} }$
Where, $\overrightarrow F = ({F_x}\hat i + {F_y}\hat j)$ and
So, we can write, $W = \int {({F_x}\hat i + {F_y}\hat j)} .(dx\hat i + dy\hat j)$
$\Rightarrow W = \int {{F_x}} dx + \int {{F_y}} dy$

Given that, $\overrightarrow F = 2xy\hat i + {x^2}\hat j$
So, ${F_x} = 2xy$ and ${F_y} = {x^2}$
$\Rightarrow W = \int {2xy} dx + \int {{x^2}} dy$
Now we have to find the relation between $x$ and $y$ from the given points using the straight-line equation, $\dfrac{{y - {y_1}}}{{x - {x_1}}} = \dfrac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}}$

Given, ${x_1} = 0,{x_2} = 2$ and ${y_1} = 2,{y_2} = 4$
Putting these values we get,
$\dfrac{{y - 2}}{{x - 0}} = \dfrac{{4 - 2}}{{2 - 0}}$
$\Rightarrow y = x + 2$
Use this relation in the integration part we get,
$\Rightarrow W = \int {2x(x + 2)} dx + \int {{{(y - 2)}^2}} dy$
Use the initial points and final points in limits,
$\Rightarrow W = 2\int\limits_0^2 {({x^2} + 2x)dx} + \int\limits_2^4 {({y^2} - 4y + 4)dy}$
$\Rightarrow W = 2\left[ {\dfrac{{{x^3}}}{8} + {x^2}} \right]_0^2 + \left[ {\dfrac{{{y^3}}}{3} - 2{y^2} + 4y} \right]_2^4$
Putting the limit values we get, $W = 16$
Thus, the force from which the work is done is $16{\text{ unit}}$.

Hence, the correct answer is option (A).

Additional information:
If in $A$ is the point of a 3d plane whose coordinate points $(x.y,z)$ and the main point is $O$
Then, the position vector of $A$ is, $\overrightarrow r = \overrightarrow {OA} = x\hat i + y\hat j + z\hat k$
$\therefore r = \sqrt {{x^2} + {y^2} + {z^2}}$
If $\overrightarrow r$ makes the angles $\alpha ,\beta ,\gamma$ with the axis $x.y,z$respectively, the direction cosine of $\overrightarrow r$will be
$\cos \alpha = \dfrac{x}{r}$ , $\cos \beta = \dfrac{y}{r}$ and $\cos \gamma = \dfrac{z}{r}$

Note: The force vector is given by, $\overrightarrow F = 2xy\hat i + {x^2}\hat j$
We write the force vector as the sum of two components, such as
$\overrightarrow F = ({F_x}\hat i + {F_y}\hat j)$
Where ${F_x}$ is the component of the force vector along the x-axis and ${F_y}$ is the component along the y-axis.
Here, ${F_x} = 2xy$ and ${F_y} = {x^2}$.
The position vector is represented as, $\overrightarrow {dr} = (dx\hat i + dy\hat j)$