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Q.

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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a

-12(i^+j^)

b

12(i^+j^)

c

−12i^

d

0

answer is D.

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Detailed Solution

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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