A block of mass $$5 \mathrm{~kg}$$ is placed on a rough inclined surface as shown in the figure. If $$\overrightarrow{F_1}$$ is the force required to just move the block up the inclined plane and $$\overrightarrow{F_2}$$ is the force required to just prevent the block from sliding down, then the value of $$\left|\overrightarrow{F_1}\right|-\left|\overrightarrow{F_2}\right|$$ is : [Use $$\left.\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2\right]$$

A light string passing over a smooth light fixed pulley connects two blocks of masses $$m_1$$ and $$m_2$$. If the acceleration of the system is $$g / 8$$, then the ratio of masses is:

In the given arrangement of a doubly inclined plane two blocks of masses $$M$$ and $$m$$ are placed. The blocks are connected by a light string passing over an ideal pulley as shown. The coefficient of friction between the surface of the plane and the blocks is 0.25. The value of $$m$$, for which $$M=10 \mathrm{~kg}$$ will move down with an acceleration of $$2 \mathrm{~m} / \mathrm{s}^2$$, is: (take $$\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2$$ and $$\left.\tan 37^{\circ}=3 / 4\right)$$

A block of mass $$m$$ is placed on a surface having vertical crossection given by $$y=x^2 / 4$$. If coefficient of friction is 0.5, the maximum height above the ground at which block can be placed without slipping is: