A force acts on a $30 \text{ gm}$ particle in such a way that the position of the particle as a function of time is given by $x = 3t - 4t^2 + t^3$, where $x$ is in metres and $t$ is in seconds. The work done during the first $4 \text{ seconds}$ is:

A $100 \text{ g}$ iron ball having velocity $10 \text{ m/s}$ collides with a wall at an angle $30^\circ$ and rebounds with the same angle. If the period of contact between the ball and wall is $0.1 \text{ second}$, then the force experienced by the wall is:

An object of mass $500 \text{ g}$ initially at rest is acted upon by a variable force whose $x$-component varies with $x$ in the manner shown. The velocities of the object at the points $x=8 \text{ m}$ and $x=12 \text{ m}$ would have the respective values of nearly:

When a body moves with a constant speed along a circle

Force $F$ on a particle moving in a straight line varies with distance $d$ as shown in the figure. The work done on the particle during its displacement of $12$ m is:

A particle moves with a velocity $(5\hat i-3\hat j+6\hat k)\text{ ms}^{-1}$ horizontally under the action of a constant force $(10\hat i+10\hat j+20\hat k)\text{ N}$. The instantaneous power supplied to the particle is:

In a circus stuntman rides a motorbike in a circular track of radius $R$ in the vertical plane. The minimum speed at highest point of track will be:

The minimum work done in pulling up a block of wood weighing $2 \text{ kN}$ for a length of $10 \text{ m}$ on a smooth plane inclined at an angle of $15^\circ$ with the horizontal is (given $\sin 15^\circ = 0.2588$):

A man fires a bullet of mass $200 \text{ g}$ at a speed of $5 \text{ m/s}$. The gun is of one kg mass. By what velocity the gun rebounds backwards?