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The potential energy of a force field is given by $U = \sin (x + y)$ .The force acting on the particle at position $(\frac{π}{4}, \frac{π}{4})$ is

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-12(i^+j^)

12(i^+j^)

−12i^

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detailed solution

Correct option is D

F→=−∂U∂xi^−∂U∂yj^−∂U∂zk^ F→=−cos(x+y)i^−cos(x+y)j^F→=−cos(x+y) (i^+j^) At (π4,π4)F→=−cos(π4+π4)(i→+j^) F→=0→ F=0

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The potential energy of a force field is given by U=sin(x+y).The force acting on the particle at position (π4,π4) is

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The potential energy of a force field is given by $U = \sin (x + y)$ .The force acting on the particle at position $(\frac{π}{4}, \frac{π}{4})$ is