$A$ block of mass $m$ is at rest on an inclined plane with an angle of inclination $\theta$ and coefficient of friction $\mu$,as shown in the figure. The frictional force acting on the block is:

  • A
    $mg \cos \theta$
  • B
    $mg \sin \theta$
  • C
    $\mu mg \sin \theta$
  • D
    $\mu mg \cos \theta$

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$A$ block of mass $m_1=1 \ kg$ and another mass $m_2=2 \ kg$ are placed together (see figure) on an inclined plane with an angle of inclination $\theta$. Various values of $\theta$ are given in List $I$. The coefficient of friction between the block $m_1$ and the plane is always zero. The coefficient of static and dynamic friction between the block $m_2$ and the plane are equal to $\mu=0.3$. In List $II$,expressions for the friction on block $m_2$ are given. Match the correct expression of the friction in List $II$ with the angles given in List $I$,and choose the correct option. The acceleration due to gravity is denoted by $g$. [Useful information: $\tan(5.5^{\circ}) \approx 0.1; \tan(11.5^{\circ}) \approx 0.2; \tan(16.5^{\circ}) \approx 0.3$]
List $I$ List $II$
$P. \theta=5^{\circ}$ $1. m_2 g \sin \theta$
$Q. \theta=10^{\circ}$ $2. (m_1+m_2) g \sin \theta$
$R. \theta=15^{\circ}$ $3. \mu m_2 g \cos \theta$
$S. \theta=20^{\circ}$ $4. \mu(m_1+m_2) g \cos \theta$

$A$ block of mass $5 \ kg$ is kept against an accelerating wedge with a wedge angle of $45^{\circ}$ to the horizontal. The coefficient of friction between the block and the wedge is $\mu = 0.4$. What is the minimum absolute value of the acceleration of the wedge to keep the block steady? Assume $g = 10 \ m \ s^{-2}$.

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