What does the law of conservation of energy state?

$A$ ball at rest is dropped from a height of $12 \,m$. It loses $25 \%$ of its kinetic energy on striking the ground and bounces back to a height '$h$'. Then the value of '$h$' is (in $\,m$)

$A$ bead $P$ slides on a frictionless semi-circular wire $(ACB)$. It is at point $S$ at $t = 0$, and at this instant, the horizontal component of its velocity is $v$. Another bead $Q$ of the same mass as $P$ is ejected from point $A$ at $t = 0$ along the horizontal wire $AB$ with speed $v$. Friction between the beads and the respective wires may be neglected in both cases. Let $t_P$ and $t_Q$ be the respective times taken by beads $P$ and $Q$ to reach point $B$. Then the relation between $t_P$ and $t_Q$ is:

Consider an elliptically shaped rail $PQ$ in the vertical plane with $OP = 3 \ m$ and $OQ = 4 \ m$. $A$ block of mass $1 \ kg$ is pulled along the rail from $P$ to $Q$ with a force of $18 \ N$,which is always parallel to the line $PQ$ (see the figure). Assuming no frictional losses,the kinetic energy of the block when it reaches $Q$ is $(n \times 10) \ J$. The value of $n$ is (take acceleration due to gravity $g = 10 \ m/s^2$):

$A$ stone with weight $w$ is thrown vertically upward into the air from ground level with initial speed $v_0$. If a constant force $f$ due to air drag acts on the stone throughout its flight,the maximum height attained by the stone is

$A$ knife of mass $m$ is dropped from a height $h$ onto a wooden floor. If the blade penetrates a distance $d$ into the wood,what is the average resistance offered by the wood to the knife?