The centre of mass of a system cannot change its state of motion, unless there is an external force acting on it. Yet the internal force of the brakes can bring a car to rest. Then

A block of mass $5 \mathrm{k}\mathrm{g}$ is (i) pushed in case $\left(\mathrm{A}\right)$ and (ii) pulled in case $\left(\mathrm{B}\right),$ by a force $\mathrm{F}=20\mathrm{ }\mathrm{N},$ making an angle of ${30}^o$ with the horizontal, as shown in the figures. The coefficient of friction between the block and floor is $\mu =0.2.$ The difference between the accelerations of the block, in case $\left(\mathrm{B}\right)$ and case $\left(\mathrm{A}\right)$ will be:
$\left(\mathrm{g}=10\mathrm{m}\mathrm{ }{\mathrm{s}}^{-2}\right)$

A uniform rope of total length $l$ is at rest on a table with fraction $f$ of its length hanging (see figure). If the coefficient of friction between the table and the chain is $\mu$, then

Consider the system shown below.

A horizontal force $F$ is applied to a block $X$ of mass $8 \mathrm{kg}$ such that the block $Y$ of mass $2 \mathrm{kg}$ adjacent to it does not slip downwards under gravity. There is no friction between the horizontal plane and the base of the block $X$. The coefficient of friction between the surfaces of the blocks $X$ and $Y$ is $0.5$. Take acceleration due to gravity to be $10 \mathrm{m} {\mathrm{s}}^{-2}$. The minimum value of $F$ is