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The potential energy for a force filed v...

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)` is

A

(a) `-1/sqrt2(hati+hatj)`

B

(b) `1/sqrt2(hati+hatj)`

C

(c) `(1/2hati+sqrt3/2hatj)`

D

(d) `(1/2hati-sqrt3/2hatj)`

Text Solution

Verified by Experts

The correct Answer is:
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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