a block of mass m = 12 kg is released from rest on a frictionless incline of angle θ = 30°. Below the block is a spring that can be compressed 2.0 cm by a force of 270 N. The block momentarily stops when it compresses the spring by 5.5 cm. (a) How far does the block move down the incline from its rest position to this stopping point? (b) What is the speed of the block just as it touches the spring?

a) We can use the conservation of energy here. We have gravitational potential energy initially (point A), kinetic energy before the block touch the spring we (we call this point B) and elastic potential energy when the block compresses the spring 5.5 cm (point C).

Let's use the conservation of energy between A and C

[tex]mgh=1/2k\Delta x^{2}[/tex] (1)

m is the mass of the block 12 kg
k is the spring constant
Δx is the compress of the spring, 5.5 cm or 0.055 m
h is the height of the inclined plane.

We need to find k. Let's use Hook's law

[tex]F=k\Delta x[/tex]

[tex]k=\frac{F}{\Delta x}=\frac{270}{0.02}=13500 N/m[/tex]

Now, we can find h using (1)

[tex]h=\frac{k\Delta x^{2}}{2mg}[/tex]

[tex]h=\frac{13500*0.055^{2}}{2*12*9.81}[/tex]

[tex]h=0.17 m[/tex]

The distance from the rest position A and C is the hypotenuse of the inclined plane, we call this distance X. We can use sin() function.

[tex]sin(30)=\frac{h}{X}[/tex]

[tex]X=\frac{0.17}{sin(30)}[/tex]

[tex]X=0.34 m[/tex]

b) Let's use the conservation of energy between A and B

[tex]mgh=1/2mv^{2}[/tex] (2)

[tex]v=\sqrt{2gh}[/tex]

[tex]v=1.83 m/s[/tex]

I hope it helps you!