Question
Question: The work done by the force \(\vec F = A({y^2}\hat i + 2{x^2}\hat j)\), where A is a constant and x a...
The work done by the force F=A(y2i^+2x2j^), where A is a constant and x and y are in meters around the path shown is
A) Zero.
B) Ad.
C) Ad2.
D) Ad3.
Solution
Work is done when a force that is applied to an object moves that object. The work is calculated by multiplying the force by the amount of movement of an object W=F∗D, where W= work done, F= force applied and D= displacement.
Complete step by step answer:
To calculate the work done let us consider some points along the path. Let, these be O, A, B, and C where, O is the initial point and C is the final point. So, the body moves from O to A, then from A to B, from B to C, and finally from C to O.
Let us calculate a small work done at any random point between O and A. This can be written as,
dW=F.dr, where dr is a small displacement along OA.
Again, dr can be written as dr=dxi^+dyj^.
So, we can write the small work done for this displacement as dW=A(y2i^+2x2j^).(dxi^+dyj^)=A(y2dx+2x2dy)
Now, to find total work done along OA can we given by,
WOA=∫dW=A∫y2dx+2x2dy.
But, y=0 along OA. So, we have,
WOA=A0∫02x2dy=0.
Similarly, we can write,
WAB=A∫y2dx+2x2dy.
Now x =d is constant so dx=0.
Thus, we can write,
WAB=A∫y20+2d2dy=2Ad20∫ddy=2Ad3.
Along BC we get,
WBC=A∫y2dx+2x2dy.
Now y =d is constant so dy=0.
So, WBC=Ad2d∫0dx=−Ad3.
For the path C0 we can write,
WCO=A∫y2dx+2x2dy. Now x =0 is constant so dx=0.
Thus, WCO=A.0=0.
Finally, the total work done is
WTotal=WOA+WAB+WBC+WCO.
Putting the values previously calculated we get,
WTotal=0+2Ad3−Ad3+0=Ad3.
So, the total work done is Ad3.
Thus, the right answer is option (D).
Note: The most common mistake done in this type of problem is done while taking the integration. Please take great care while putting in the limit. Also, special care has to be taken about the values of x and y at different points. If we make even a small mistake for any of these values than the solution will give us a different output altogether.