A ring of mass m free to slide on a fixed smooth horizontal rod is attached to a particle of mass M kg by a inextensible string of length $l$. Initially, both M and m are at rest and the string is vertical. A horizontal velocity $v_0$ is imparted to the particle. The maximum height up to which block will rise w.r.t its initial position is (M = 2m)

A small block starts slipping down from a point $B$ on an inclined plane $AB$, which is making an angle $\mathrm{\theta }$ with the horizontal section $BC$ is smooth and the remaining section $CA$ is rough with a coefficient of friction $\mu$. It is found that the block comes to rest as it reaches the bottom (point A) of the inclined plane. If $BC\mathit{=}\mathit{2}AC$, the coefficient of friction is given by $\mathrm{\mu }=\mathrm{k} \mathrm{tan\theta }$. The value of $k$ is .......

In the diagram below the spring has a force constant of $K$, the block has a mass of $m$, and the height of hill is $h$. Determine the compression of the spring $x$ such that the block just reaches to the top of the hill.
(Assume that there are no non-conservative forces involved.)