The potential energy for a force field \(\vec{F}\) is given by :
\(U(x, y) = \cos ({x+y})\)
The force acting on a particle at position given by coordinates (0, π/4) is :
Answer & explanation
Correct answer: option 3
\(F_x = -\frac{\delta U}{\delta x} = \sin ({x+y}) = \frac{1}{\sqrt{2}}\)
\(F_y = -\frac{\delta U}{\delta y} = \sin ({x+y}) = \frac{1}{\sqrt{2}}\)
\(\vec{F} = \frac{1}{\sqrt{2}} (\hat{i} + \hat{j})\)