An ideal massless spring can be compressed $2.33\ \mathrm{cm}$ by a force of $268\ \mathrm{N}$. A block of mass $m=3.18\ \mathrm{kg}$ is released from rest at the top of a frictionless incline of angle $32.0^\circ$. The block comes to rest momentarily after it has compressed this spring by $5.48\ \mathrm{cm}$.
- How far has the block moved down the incline at this moment?
- What is the speed of the block just as it touches the spring?
I really don't know where to start.
You can draw a schematic diagram to analyze it and suppose that elastic force is $F=-kx$ by Hooke's law and the gravity is $mg$.
For part (1.) think about this: just as the spring comes to rest momentarily all of the energy in this system is stored in potential energy of the spring itself. First find the spring constant $k$ using Hooke's law. From there you can find the potential energy stored in the spring. You then know that just as the mass comes in contact with the spring the kinetic of the mass is equal to the potential energy of the spring when it is fully compressed. from there you can use the kinetic energy formula:
$$K=\frac{1}{2}mv^2$$
To get the velocity of the block just as it hits the mass. So... in starting part (1) we actually need to answer part (2) first.
To finish off part (1) think about potential energy in the form of:
$$U = Mgh$$
you know the height of the vertical component, all you need to do now is use some trigonometry to get the vector sum of the vertical and horizontal component.
