The total work done on a particle is equal to the change in its kinetic energy. This is applicable:

The kinetic energies of two similar cars $A$ and $B$ are $100 \ J$ and $225 \ J$ respectively. On applying brakes,car $A$ stops after $1000 \ m$ and car $B$ stops after $1500 \ m$. If $F_{A}$ and $F_{B}$ are the forces applied by the brakes on cars $A$ and $B$,respectively,then the ratio $F_{A} / F_{B}$ is:

Consider an elliptically shaped rail $PQ$ in the vertical plane with $OP = 3 \ m$ and $OQ = 4 \ m$. $A$ block of mass $1 \ kg$ is pulled along the rail from $P$ to $Q$ with a force of $18 \ N$,which is always parallel to the line $PQ$ (see the figure). Assuming no frictional losses,the kinetic energy of the block when it reaches $Q$ is $(n \times 10) \ J$. The value of $n$ is (take acceleration due to gravity $g = 10 \ m/s^2$):

$A$ particle of mass $20\,g$ is released with an initial velocity $5\,m/s$ along the curve from the point $A,$ as shown in the figure. The point $A$ is at height $h = 10\,m$ from point $B.$ The particle slides along the frictionless surface. When the particle reaches point $B,$ its angular momentum about $O$ will be ......... $kg \cdot m^2/s$. [Take $g = 10\,m/s^2$]

$A$ bullet when fired at a target with a velocity of $100\,m/s$ penetrates $1\,m$ into it. If the bullet is fired at a similar target with a thickness of $0.5\,m$,then it will emerge from it with a velocity of:

In the presence of which of the following is the work-energy theorem valid?