$A$ bead of mass $\frac{1}{2} \, kg$ starts from rest from point $A$ to move in a vertical plane along a smooth fixed quarter ring of radius $5 \, m$,under the action of a constant horizontal force $F = 5 \, N$ as shown. The speed of the bead as it reaches point $B$ is ................ $m/s$. [Take $g = 10 \, m/s^2$]

$A$ block weighing $20 \ N$ travels down a smooth curved track $AB$ joined to a rough horizontal surface (figure). The rough surface has a friction coefficient of $0.40$ with the block. If the block starts slipping on the track from a point $1.0 \ m$ above the horizontal surface,the distance it will move on the rough surface is: (in $m$)

$A$ particle of mass $m$ is moving along a circle of radius $R$ such that its tangential acceleration $a_t$ varies with distance covered $x$ as $a_t = \alpha x^2$,where $\alpha$ is a constant. The kinetic energy $K$ of the particle varies with the distance as $K = \beta x^c$,where $\beta$ and $c$ are constants. The values of $\beta$ and $c$ are:

$A$ particle is released from a height $H$. At a certain height $h$,its kinetic energy is two times its potential energy. The height $h$ and the speed $v$ of the particle at that instant are:

What does the law of conservation of energy state?

$A$ 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 $g = 10 \; ms^{-2}$)