As shown in the figure,there is a spring-block system. $A$ block of mass $500\,g$ is pressed against a horizontal spring fixed at one end to compress the spring by $5.0\,cm$. The spring constant is $500\,N/m$. When released,calculate the horizontal distance from the edge of the table where it will hit the ground $4\,m$ below the spring. $(g = 10\,m/s^2)$

$A$ particle of mass $m$ is moving in a circular path of constant radius $r$ such that its centripetal acceleration $a$ varies with time $t$ as $a = k^{2} r t^{2}$,where $k$ is a constant. The power delivered to the particle by the force acting on it is given as:

$A$ particle of mass $m$ is moving in the $xy$-plane such that its velocity at a point $(x, y)$ is given as $\vec{v} = \alpha(y \hat{i} + 2x \hat{j})$,where $\alpha$ is a non-zero constant. What is the force $\vec{F}$ acting on the particle?

$A$ stone thrown with velocity $u$ at angles $\theta$ and $(90^{\circ}-\theta)$ with the horizontal reaches maximum heights $H_1$ and $H_2$ respectively. Its horizontal range is

$A$ player kicks a football at an angle of $30^{\circ}$ with the horizontal with an initial speed of $30 \,ms^{-1}$. $A$ second player, standing at a distance of $21 \sqrt{3} \,m$ from the first player in the direction of the kick, starts running to catch the ball at the same instant it is kicked. What is the minimum speed of the second player to catch the ball before it hits the ground? (Take acceleration due to gravity $g = 10 \,ms^{-2}$)

If a particle of mass $m$ covers half of the horizontal circle with constant speed $v$,then the change in its kinetic energy is