AP EAMCET Engineering 20 Apr 2019 Shift 1 Paper

The figure below shows the free body diagram of the block.The friction force acting on the block is given as, $f = \mu_{k} (m g \cos \theta)$ The force equation for the block along the plane of inclination, $m g \sin \theta - f \;\; = m a$ $m g \sin \theta - \mu_{k} (m g \cos \theta) \;\; = m a$ $a \;\; = 10 \times 0.6 - \frac{1}{8} \times 10 \times 0.8$ $a \;\; = 5 \text{ m/s}^2$ The distance $s$ is given from $\Delta EBC$ $E C = s = \frac{BE}{\sin \theta} = \frac{\frac{25}{6}}{0.6} = \frac{125}{18} \text{ m}$ The velocity at final position is give as, $v^{2} = 2 a s$ Substituting the values in above expression, we get $v^{2} = 2 \times 5 \text{ m/s}^2 \times \frac{125}{18} \text{ m}$ $v = \frac{25}{3} \text{ m/s}$ The component of velocity at point $C$ is given as,Using second equation of motion, we get $h \;\; = u t + \frac{1}{2} g t^{2}$ $10 \;\; = u t \sin \theta + \frac{1}{2} g t^{2}$ $5 t^{2} + 5 t - 10 \;\; = 0$ $t^{2} + t - 2 \;\; = 0$ On solving the time comes out to be $t = 1$ s The distance DF is, $DF = v \cos \theta = \frac{25}{3} \times 0.8 = \frac{20}{3} \text{ m}$