Question

The potential energy for a force field `vecF` is given by `U(x,y) = sin(x+y)`. The magnitude of force acting on the particle of mass m at `(0,pi/4)` is?

A) 1
B) `sqrt2`
C) `1/sqrt2`
D) `0`

Answer 1

`vec F(x,y) = - \grad U(x,y)`
`F(x,y) = |\gradU(x,y)|; \qquad F(0,pi/4) = 1`

Explanation:

`vec F(x,y) = -\grad U(x,y)`

`\grad U(x,y)= \frac{\delU}{\del x} hat x + \frac{\delU}{\dely} hat y = cos(x+y) (hatx+haty)`

`F(x,y) = |\gradU(x,y)|`
`F(0,pi/4) = |\gradU(0,pi/4)| = |cos(pi/4)||hatx+haty|`
`F(0,pi/4)= 1/sqrt{2}\sqrt{1^2+1^2} = 1 "unit"`