Given $m_1 = 4m_2$ and the acceleration of $m_1$ is $a$. How much distance in $cm$ will $m_2$ cover in $0.4 \, s$?

An electric lift with a maximum load of $2000\,kg$ (lift + passengers) is moving up with a constant speed of $1.5\,m/s$. The frictional force opposing the motion is $3000\,N$. The minimum power delivered by the motor to the lift in watts is: $(g=10\,m/s^2)$

The masses of blocks $A$ and $B$ are $m$ and $M$ respectively. Between $A$ and $B$,there is a constant frictional force $F$ and $B$ can slide on a smooth horizontal surface. $A$ is set in motion with velocity $v_0$ while $B$ is at rest. What is the distance moved by $A$ relative to $B$ before they move with the same velocity?

Work done by frictional force is:

$A$ body of mass $2 \,kg$ is acted upon by two forces each of magnitude $1 \,N$, making an angle of $60^{\circ}$ with each other. The net acceleration of the body (in $m/s^2$) is

$A$ bead of mass $m$ can slide on a frictionless fixed ring of radius $r$. With the help of two identical springs of force constant $k$,it is connected to two diametrically opposite nails $A$ and $B$,each of which is at a distance $0.5r$ from the centre $O$ of the ring. The relaxed length of each spring is negligible compared to the radius of the ring. The bead is given a small velocity. What can you predict for the further motion of the bead before any of the springs strikes a nail?