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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. `-(1)/(sqrt2)(hat(i)+hat(j))`B. `(1)/(sqrt2)(hat(i)+hat(j))`C. `((1)/(2)hat(i)+(sqrt3)/(2)hat(j))`D. `((1)/(2)hat(i)-(sqrt3)/(2)hat(j))` |
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Answer» Correct Answer - B `F_(x)=-(delU)/(delx)=sin(x+y)` and `F_(y)=-(delU)/(dely)=sin(x+y)` so `F_(x)=[sin(x+y)]_(((0,pi/4)))=(1)/(sqrt2)` `F_(y)=[sin(x+y)]_(((0,pi/4)))=(1)/(sqrt2)` `vec(F)=(1)/(sqrt2)[hat(i)+hat(j)]` |
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