$A$ 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 $5\,seconds$? ................ $J$

$A$ bullet is fired on a target with velocity $V$. Its velocity decreases from $V$ to $V/2$ when it penetrates $30 \text{ cm}$ in a target. Through what thickness will it penetrate further in the target before coming to rest (in $\text{ cm}$)?

If mechanical work is done on an object,does its kinetic energy increase or decrease?

$A$ body of mass $m$ starts moving from rest along the $x-$axis such that its velocity varies as $v = a\sqrt{s}$,where $a$ is a constant and $s$ is the distance covered by the body. The total work done by all the forces acting on the body in the first $t$ seconds after the start of the motion is:

$A$ body of mass $1 \ kg$ is under a force, which causes a displacement given by $x = \frac{t^3}{3}$ (in $m$). Find the work done by the force in the first second (in $J$).

$A$ particle is placed at the point $A$ of a frictionless track $ABC$ as shown in the figure. It is gently pushed toward the right. The speed of the particle when it reaches the point $B$ is: (Take $g = 10 \ m/s^2$).