Answer:
F = (mg - kx) i ^
Explanation:
Force and potential are related
F = - ([tex]\frac{dU}{dx}[/tex] i ^ + \frac{dU}{dy} j ^ + \frac{dU}{dz} k ^)
in this exercise we are told that the potential is
U = - m g x + ½ kx²
let's make the drive
\frac{dU}{dx} = -mg + kx
\frac{dU}{dy}= 0
\frac{dU}{dz} = 0
we substitute
F = (mg - kx) i ^
