Three blocks $A, B$ and $C,$ of masses $4 \, kg, 2 \, kg$ and $1 \, kg$ respectively,are in contact on a frictionless surface,as shown. If a force of $14 \, N$ is applied on the $4 \, kg$ block,then the contact force between $A$ and $B$ is .......... $N$.

Two fixed frictionless inclined planes making an angle $30^\circ$ and $60^\circ$ with the vertical are shown in the figure. Two blocks $A$ and $B$ are placed on the two planes. What is the relative vertical acceleration of $A$ with respect to $B$?

$A$ particle of mass $m$ is acted upon by a force $F$ given by the empirical law $F = \frac{R}{t^2} v(t)$. If this law is to be tested experimentally by observing the motion starting from rest,the best way is to plot:

One end of a spring of force constant $150 \text{ dyne cm}^{-1}$ is connected to a block of mass $0.2 \text{ kg}$ kept on a rough horizontal surface of coefficient of friction $0.3$. The other end of the spring is connected to a rigid support as shown in the figure and the spring is initially undeformed. The maximum velocity $v$ that can be given to the block so that it travels only in one direction is . . . . . . $\text{ms}^{-1}$. (Acceleration due to gravity $= 10 \text{ ms}^{-2}$)

The figure shows a man standing stationary with respect to a horizontal conveyor belt that is accelerating with $1 \; m/s^2$. What is the net force on the man? If the coefficient of static friction between the man's shoes and the belt is $0.2$,up to what acceleration of the belt can the man continue to be stationary relative to the belt? (Mass of the man $= 65 \; kg.)$

$A$ block of mass $m$ is on an inclined plane of angle $\theta$. The coefficient of friction between the block and the plane is $\mu$ and $\tan \theta > \mu$. The block is held stationary by applying a force $P$ parallel to the plane. The direction of force pointing up the plane is taken to be positive. As $P$ is varied from $P_1 = mg(\sin \theta - \mu \cos \theta)$ to $P_2 = mg(\sin \theta + \mu \cos \theta)$,the frictional force $f$ versus $P$ graph will look like: