Determine the maximum acceleration in $m/s^2$ of the train in which a box lying on its floor will remain stationary,given that the coefficient of static friction between the box and the train's floor is $0.15$.

What is the maximum value of the force $F$ such that the block shown in the arrangement does not move? (in $N$)

$A$ block of mass $m=2 \ kg$ is initially at rest on a horizontal surface. $A$ horizontal force $F_1=(6 \ N) \hat{i}$ and a vertical force $F_2=(10 \ N) \hat{j}$ are then applied to the block. The coefficients of static friction and kinetic friction for the block and the surface are $0.4$ and $0.25$,respectively. The magnitude of the frictional force acting on the block is (assume $g=10 \ m/s^2$): (in $N$)

$A$ $300 \ kg$ crate is dropped vertically onto a conveyor belt that is moving at $1.20 \ m/s$. $A$ motor maintains the belt's constant speed. The belt initially slides under the crate,with a coefficient of friction of $0.400$. After a short time,the crate is moving at the speed of the belt. During the period in which the crate is being accelerated,find the work done by the motor which drives the belt.

$A$ block of mass $m$ slides along a floor while a force of magnitude $F$ is applied to it at an angle $\theta$ as shown in the figure. The coefficient of kinetic friction is $\mu_{K}$. Then,the block's acceleration $a$ is given by: ($g$ is acceleration due to gravity)

The coefficient of friction between the block and the surface is $0.03$. Find the acceleration of the system in $m/s^{2}$. $(g = 10\,m/s^{2})$