Answer :
Answer:
a) [tex]X=0.34 m[/tex]
b) [tex]v=1.83 m/s[/tex]
Explanation:
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!