For a body of mass $1 \ kg$,the $U-x$ graph is shown in the figure. If the body is released from rest at $x = 2 \ m$,then its speed when it crosses $x = 5 \ m$ is:

$A$ bead $P$ slides on a frictionless semi-circular wire $(ACB)$. It is at point $S$ at $t = 0$, and at this instant, the horizontal component of its velocity is $v$. Another bead $Q$ of the same mass as $P$ is ejected from point $A$ at $t = 0$ along the horizontal wire $AB$ with speed $v$. Friction between the beads and the respective wires may be neglected in both cases. Let $t_P$ and $t_Q$ be the respective times taken by beads $P$ and $Q$ to reach point $B$. Then the relation between $t_P$ and $t_Q$ is:

$A$ particle which is experiencing a force,given by $\vec F = 3\hat i - 12\hat j$,undergoes a displacement of $\vec d = 4\hat i$. If the particle had a kinetic energy of $3\, J$ at the beginning of the displacement,what is its kinetic energy (in $J$) at the end of the displacement?

$A$ body falls freely under gravity. Its speed is $v$ when it has lost an amount $U$ of the gravitational potential energy. Then its mass is

Explain the conservation of mechanical energy using the example of a falling ball.

$A$ force of $5\,N$,making an angle $\theta$ with the horizontal,acts on an object and displaces it by $0.4\,m$ along the horizontal direction. If the object gains a kinetic energy of $1\,J$,the horizontal component of the force is......$N$. (in $.5$)