Now that an extra model spring is attached, as shown, what will the vector force due to the additional spring be? Can someone explain why the equation of motion is
mx+5kx=-mgsin(alpha) +11kl0
There is a weight on the block directed downward, there is a normal force on the block directed out of the plane of the slope, there is a spring force acting down the incline and a spring force acting up the incline.
The spring extension of the lower spring is presumably $\displaystyle x - l_0$? Since the top spring has an equilibrium length of $\displaystyle 2l_0$, the spring extension of the upper spring is $\displaystyle (5l_0 - 2l_0) - x$. So Newton's 2nd tells us
$\displaystyle \sum F_x = -mg~sin(\alpha) + 3k(3l_0 - x) - 2k(x - l_0) = ma$
This gives the required equation of motion.
-Dan