Answer
The equilibrium point is a distance of $~~0.525~m~~$ from the top of the incline.
Work Step by Step
The equilibrium point is the point where the component of the gravitational force on the block directed down the incline is equal in magnitude to the spring force.
We can find the distance the spring is stretched at the equilibrium point:
$mg~sin~\theta = kx$
$x = \frac{mg~sin~\theta}{k}$
$x = \frac{(14.0~N)~sin~40.0^{\circ}}{120~N/m}$
$x = 0.0750~m$
Since the spring's unstretched length is $0.450~m$, the equilibrium point is a distance of $~~0.525~m~~$ from the top of the incline.
