Question

A cutting force under microprocessor control has several forces acting on it. One force is $ {{\vec F = - \alpha x}}{{{y}}^{{2}}}{{\hat j}} $ , a force in the negative y-direction whose magnitude depend on the position of the tool. The constant is $ \alpha = 2.50N/{m^3} $ . Consider the displacement of the tool from the origin to the point x=3.00m, y=3.00m. Calculate the work done on the tool by $ {{\vec F}} $ if the tool is first moved out along the x- axis to the point x=3.00m, y=0m and then moved parallel to the y -axis to x=3.00m, y=3.00m
(A) 67.5 $ {{J}} $
(B) 85 $ {{J}} $
(C) 102 $ {{J}} $
(D) 7.5 $ {{J}} $

Answer 1

Hint: We are looking for the work done. This could be calculated with the basic formula $ {{dW = f}}{{.ds}} $ and integrating it up to the limit as force is not constant throughout.

Formula used : To calculate the amount of force, we use the equation:
 $ {{dW = f}}{{.ds}} $
Here, $ dw $ is the small amount of work being done,
 $ {{f}} $ is the force applied.
 $ {{ds}} $ is the small displacement.

Complete step by step answer
It is already known that the cutting tool is being moved from one position to another.
And, $ {{dW = f}}{{.ds}} $
On integrating,
 $ {{W = }}\int {{{\vec F}}} {{.d\vec s}} $
And the force is only on y-axis,
Work done along x-axis is 0.
Work done in moving from (3,0) to (3,3) is
 $ {{W = }}\int\limits_{{0}}^{{3}} {{{ - \alpha x}}{{{y}}^{{2}}}{{.dycos18}}{{{0}}^{{0}}}} $
On solving further,
 $ {{W = }}\int\limits_{{0}}^{{3}} {{{\alpha \times 3 \times }}{{{y}}^{{2}}}{{dy}}} $
 $ \Rightarrow {{W = 2}}{{.5 \times 3 \times }}\int\limits_{{0}}^{{3}} {{{{y}}^{{2}}}{{dy}}} $
On integrating,
 $ {{W = 7}}{{.5 \times }}\left( {\dfrac{{{{{3}}^{{2}}}}}{{{3}}}{{ - 0}}} \right) $
 $ \Rightarrow {{W = 67}}{{.5J}} $
So, we need to see from the above options, and select the correct value.
Thus, the correct answer is option A.

Additional Information
A force is said to do work when it acts on a body so that there is a displacement of the point of application in the direction of the force. Thus, a force does work when it results in movement. The same integration approach can be also applied to the work done by a constant force. This suggests that integrating the product of force and distance is the general way of determining the work done by a force on a moving body.

Note
The examples of variable force can be seen everywhere as all forces in the world are not constant. Constant force is just an ideal situation used in solving questions. Work may still be calculated as the area under the force- displacement curve.