A tool is moving in the $ x - y $ plane under a quality control process under the action of various forces. One force is $ \overrightarrow F = - \alpha x{y^2}\,\widehat j $ a force in the negative $ y $ -direction whose magnitude depends on the position of the tool. The constant $ \alpha $ value is $ 2.50\,N/{m^3} $ . Consider the displacement of the tool from the origin to point $ x = 3.00m $ and $ y = 3.00m $ . Calculate the work done on the tool by the force $ \overrightarrow F $ if this displacement is along the straight line $ x = y $ that connects the two points.
(A) $ - 50.6J $
(B) $ - 67.5J $
(C) $ - 77.5J $
(D) $ - 60.5J $

Hint: To answer this question we have to understand the directions in which force and displacement are aligned. Then we have to use the formula of work done and put the given values in it. Also, it is important to find the angle between the force and displacement by analyzing the question carefully.

Complete Step By Step Answer:
Here, force is in negative $ y $ -direction but while solving we would take the magnitude of the force as the sign only shows the direction. Also, the displacement is on the line $ x = y $ , which implies that displacement can be taken in $ x $ as well as in $ y $ - direction if we resolve our components. For simplicity, we take the direction of the displacement in the $ y $ - direction as the force is variable and the displacement taken is infinitesimally small.
Now, if the displacement is in the $ y $ -direction and the question says that force is in the negative $ y $ -direction then we can figure out that, the angle between them will be $ {180^\circ } $ .
According to the question, the values we have are:
 $ \overrightarrow F = - \alpha x{y^2}\,\widehat j $
 $ \left| {\overrightarrow F } \right| = \;\;\alpha x{y^2} $
 $ \alpha = 2.50N/{m^3} $
 $ (x,y) = (0,0)\, to \,(3m,3m) $
The formula for work done for a variable force applied on a body and it shows an infinitesimally small displacement is given by:
 $ dW = \overrightarrow {F.} \,\overrightarrow {dr} $
By integrating the above equation we can get the total work done.
 $ W = \int {\overrightarrow F \,.\overrightarrow {dr} } $
 $ \Rightarrow W = \int {Fdr\cos \theta } .......(1) $
Where,
 $ W $ is the total work done.
 $ F $ is the force applied
 $ dr $ is an infinitesimally small displacement
 $ \theta $ is the angle that the force and the displacement have in between them.
As the displacement is in the $ y $ - direction, there is no motion along x and hence its value can be taken as $ 3 $ as per the question and we replace $ dr $ by $ dy $ . While integrating, the limits will be from $ 0 $ to $ 3 $ as in $ y $ - direction the body moves from $ (0,0) $ to $ (3m,3m) $
So, we will now put all the known values in equation (1), we get,
 $ W = \int_0^3 {\alpha x{y^2}\cos 180^\circ dy......(2)} $
We know that, $ \cos 180^\circ = - 1 $ and $ \alpha = 2.50N/{m^3} $ .
Putting the values in equation (2) we get,
 $ W = \int_0^3 {2.50 \times 3 \times \left( { - 1} \right) \times {y^2}dy} $
 $ \Rightarrow W = - 7.5\int_0^3 {{y^2}dy} .......(3) $
Integrating equation (3) we get,
 $ W = - 7.5{\left[ {\dfrac{{{y^3}}}{3}} \right]^3}_0 $
We now put the limits in the equation,
 $ W = - 7.5\left[ {\dfrac{{{3^3}}}{3} - 0} \right] $
Therefore the value to total work done is:
 $ W = - 67.5J $
So, option B is the correct answer.

Note:
The work done by a force is independent of the type of motion for a particular displacement i.e. whether it moves with constant velocity, constant acceleration or retardation work done will be the same.
Also, work done is independent of time. Work will be the same for the same displacement whether the time taken is large or small. The change in kinetic energy is the effect of the work of the particle or system.