Two masses $m_1 = 10 \text{ kg}$ and $m_2 = 20 \text{ kg}$ are connected by an inextensible string over a frictionless pulley,as shown in the figure. The coefficient of friction of the horizontal surface is $\mu = 0.15$. The minimum mass $m$ that should be placed on top of $m_2$ to stop the motion is:

$A$ block of mass $10\, kg$ starts sliding on a surface with an initial velocity of $9.8\, m/s$. The coefficient of friction between the surface and the block is $0.5$. The distance covered by the block before coming to rest is: [use $g = 9.8\, m/s^2$].........$m$

Explain kinetic friction. Write the laws of kinetic friction. Define the coefficient of kinetic friction.

Find the work done against friction in $J$ when a block of mass $50\, kg$ moves a distance of $1\, m$ on a rough horizontal surface with a coefficient of friction of $0.2$.

$A$ body of weight $64 \,N$ is pushed with just enough force to start it moving across a horizontal floor and the same force continues to act afterwards. If the coefficients of static and dynamic friction are $0.6$ and $0.4$ respectively, the acceleration of the body is (Acceleration due to gravity $= g$)

$A$ particle of mass $M$ moves along a horizontal $x$-axis from $x=0$ to $x=L$. The coefficient of kinetic friction varies as a function of $x$ as $\mu(x) = \mu_0 - \alpha x$, where $\mu_0$ and $\alpha$ are constants of appropriate dimensions, such that $\mu(L)=0$. The total work done by the frictional force during the motion is $n\mu_0 MgL$, where $g$ is the acceleration due to gravity. The value of $n$ is: