$A$ force acts on a $2\,kg$ object so that its position is given as a function of time as $x = 3t^2 + 5$. What is the work done by this force in the first $5\,seconds$? ................ $J$

Work equal to $25\,J$ is done on a mass of $2\,kg$ to set it in motion. If the whole of it is used to increase the kinetic energy,then the velocity acquired by the mass is ............ $m/s$.

The position $x$ of a particle moving in one dimension under the influence of a constant force is given by $t = \sqrt{x} + 3$, where $x$ is in meters and $t$ is in seconds. Find the work done in the first $6$ seconds. (in $\text{ J}$)

Explain the work-energy theorem.

$A$ ball is moving to and fro about the lowest point $A$ of a smooth hemispherical bowl. If it is able to rise up to a height of $20 \,cm$ on either side of $A$,its speed at $A$ must be .......... $m/s$ (Take $g = 10 \,m/s^2$,mass of the body = $5 \,g$).

The pulley and spring are massless,and friction is absent everywhere. $A$ $5\, kg$ block is released from rest. Find the speed of the $5\, kg$ block when the $2\, kg$ block leaves contact with the ground. (Take the force constant of the spring $K = 40\, N/m$ and $g = 10\, m/s^2$.)