Question

the potential energy function of a particle in a region of space is given as

Answer 1

\[U=(2x ^{2}+3y ^{3}+2z) J.Here x,y and z are in metres.Find the force acting on the particle at point P (1m,2m,3m)

Answer 2

\[U=(2x ^{2}+3y ^{3}+2z) \rightarrow \bar F(x,y,z)=-\nabla U(x,y,z)=-\frac{ \delta U }{ \delta x }\hat i-\frac{ \delta U }{ \delta y }\hat j-\frac{ \delta U }{ \delta z } \hat k\]

Answer 3

\[F=\frac{ -dU}{ dr }\]

Answer 4

Force is "minus the gradient of the potential" The inverted triangle is called nabla operator and means exactly what you have written

Answer 5

And it is the right way to represent it

Answer 6

so it will come out to be -(4x+9y^2+2)

Answer 7

awesome!

Answer 8

force is -(gradiant potential)