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The potential energy for a force filed `vecF` is given by `U(x,y)=cos(x+y)`. The force acting on a particle at position given by coordinates `(0, pi//4)` isA. (a) `-1/sqrt2(hati+hatj)`B. (b) `1/sqrt2(hati+hatj)`C. (c) `(1/2hati+sqrt3/2hatj)`D. (d) `(1/2hati-sqrt3/2hatj)` |
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Answer» Correct Answer - B `F_x=-(delU)/(delX)=sin(x+y)=1/sqrt2` `F_y=-(delU)/(delY)=sin(x+y)=1/sqrt2` `vecF=1/sqrt2[hati+hatj]` |
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