$A$ body of mass $5 \,kg$ is thrown vertically up with a kinetic energy of $490 \,J$. The height at which the kinetic energy of the body becomes half of the original value is (acceleration due to gravity $=9.8 \,ms^{-2}$) (in $\,m$)

Under the action of a force, a $2 \,kg$ body moves such that its position $x$ as a function of time $t$ is given by $x = \alpha t^2 / 2$, where $x$ is in meters, $t$ is in seconds, and $\alpha = 1 \,m/s^2$. The work done by the force in the first two seconds is: (in $\,J$)

$A$ body thrown vertically upwards from the ground reaches a maximum height $h$. The ratio of the kinetic and potential energies of the body at a height $40 \%$ of $h$ from the ground is

$A$ body of mass $2 \, kg$ slides down a curved track which is a quadrant of a circle of radius $1 \, m$ as shown in the figure. All the surfaces are frictionless. If the body starts from rest,its speed at the bottom of the track is ............. $m/s$.

$A$ particle of mass $m$ moves on a straight line with its velocity increasing with distance according to the equation $v = \alpha \sqrt{x}$,where $\alpha$ is a constant. The total work done by all the forces applied on the particle during its displacement from $x = 0$ to $x = d$ will be:

$A$ uniform chain of mass $M$ and length $L$ is released from rest from a smooth horizontal surface as shown in the figure. The velocity of the chain at the instant it completely comes off the table will be: