$A$ particle $(m = 1 \; kg)$ slides down a frictionless track $(AOC)$ starting from rest at a point $A$ (height $2 \; m$). After reaching $C$,the particle continues to move freely in air as a projectile. When it reaches its highest point $P$ (height $1 \; m$),the kinetic energy of the particle (in $J$) is: (Figure drawn is schematic and not to scale; take $g = 10 \; ms^{-2}$)

$A$ 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 $\,m$)

$A$ particle of mass $100 \ g$ is thrown vertically upwards with a speed of $5 \ m/s$. The work done by the force of gravity during the time the particle goes up is.....$J$

Explain the conservation of mechanical energy using the example of a falling ball.

$A$ body of mass $m$ slides down a smooth hemispherical surface of radius $r$. What will be its velocity at the bottom of the surface?

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 $g = 10 \ m/s^2$):