The potential energy for a force field →F is given by U(x, y) = sin(x + y). Magnitude of the force acting on the particle of mass m at (0,π4) is
A
1
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B
√2
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C
1√2
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D
0
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Solution
The correct option is A 1 →F=(x,y)=−∇U(x,y)∇U(x,y)=∂U∂xˆx+∂U∂yˆy=cos(x+y)(ˆx+ˆy)F(x,y)=|∇U(x,y)|F(0,π4)=|∇U(x,y)|=∣∣cos(π4)∣∣|ˆx+ˆy|F(0,π4)=1√2√12+11=1unit