An athlete in the Olympic Games covers a distance of $100 \mathrm{m}$ in $10 \mathrm{s}$. His kinetic energy can be estimated to be in the range

A shell explodes in mid-air. It's total:

A body is moving at a speed of $1.0 {\mathrm{ms}}^{-1}$. A force  $F$ is required to stop it within a distance $x$. If the speed of the body increases to $3.0 {\mathrm{ms}}^{-1}$, then the force required to stop it within the same distance $x$ will be:

The relation between the displacement $x$ and the time $t$ for a body of mass $2 \mathrm{kg}$ moving under the action of a force is given by $x=t^3/3$, where $x$ is in metre and $t$ in second. The work done (in joule) by the body in the first two seconds is:

A ball of mass $0.2 \mathrm{kg}$ is thrown vertically upward by applying a force by hand. If the hand moves $0.2 \mathrm{m}$ while applying the force and the ball goes up to $2 \mathrm{m}$ height further, find the magnitude of the force. Consider $g=10 \mathrm{m} {\mathrm{s}}^{-2}$.

A particle is free to move along the $X$-axis has potential energy given by $U\left(x\right)=k\left[1-e^{-x^2}\right]$ for $-\infty \le x\le +\infty$,  where $k$ is a positive constant of appropriate dimensions. Then:

When a spring is stretched through a distance $s$, its potential energy is $10 \mathrm{joule}$. The work (in joule) required to stretch it further through $s$ will be:

Two springs of force-constants  $k_1$ and $k_2$ are stretched by the same force. The ratio of potential energies stored in them is:

A ball of mass$0.2 \mathrm{kg}$ rests on a vertical post of height  $5 \mathrm{m}$. A bullet of mass $0.01 \mathrm{kg}$, travelling with velocity $v \mathrm{m}/\mathrm{s}$ in a horizontal direction, hits the centre of the ball. After the collision, the ball and bullet travel independently. The ball hits the ground at a distance of $20 \mathrm{m}$and the bullet at a distance of $100 \mathrm{m}$ from the foot of the post. The initial velocity of the bullet is: