When a body is placed on a rough plane inclined at an angle $\theta$ to the horizontal,its acceleration is

$A$ 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 $g = 10 \ m \ s^{-2}$.

$A$ block is stationary on a rough inclined plane. How many forces are acting on the block?

$A$ block of wood resting on an inclined plane of angle $30^o$ just starts moving down. If the coefficient of friction is $0.2$,its velocity (in $ms^{-1}$) after $5\, s$ is: $(g = 10\, ms^{-2})$

Consider three masses $m_1, m_2$ and $m_3$ $(m_1 > m_2 > m_3)$ that are at rest on an inclined plane as shown in the figure. The angle of inclination $(\theta)$ of the plane is gradually increased until the masses just begin to slide. (Assume the coefficient of static friction between the masses and the surface is constant). Then,which of the following statements is correct?

If $300\,J$ of work is done to move a $2\,kg$ block up a rough inclined plane to a height of $10\,m$,find the work done against friction in $J$. (Take $g = 10\,m/s^2$)