Q.
A particular of mass ‘m’ moves along the quarter section of the circular path whose centre is at the origin. The radius of the circular path is ‘a’. A force F→=yi^−xj^ newton acts on the particle, where x,y denote the coordinates of position of the particle. Calculate the work done by this force in taking the particle from Point A (a,0) to Point B(0,a) along the circular path.
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a
−2πa2J
b
−πa22J
c
-πa2 J
d
-3πa22J
answer is B.
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Detailed Solution
Work done by force F is W=∫F→.dr=∫(yi^−xj^).(dxi^+dyi^)=∫(ydx−xdy)Equation of circular pathX2+y2=a2differentiate∴xdx+ydy=0⇒dx=-ydyx⇒W=∫(y(−ydyx)−xdy)=−∫(x2+y2)xdysubstitute, X2+y2=a2 in above integral W=-∫0aa2a2−y2dy=−πa22 joule
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