AP EAPCET 07-Jul-2022 Shift 1 Paper

Block stops on incline plane when its initial K.E is used in doing work against friction and partly it is converted into potential energy. As angle of incline is given $45^{\circ}$ so, distance covered on incline $(s)$ is related to height of block (h) as, $\Rightarrow \sin 45^{\circ}=\frac{h}{s}$ or $s=\sqrt{2} h$ Let block covers a distance $s$ over block and reaches up-to height $h$. Then, $\text{Initial K.E of block} = \text{Work done against friction} + \text{Potential energy gained by block}$ $\Rightarrow \frac{1}{2} m u^2 = \mu m g \cos \theta \cdot s + m g h$ $\Rightarrow \frac{1}{2} m u^2 = \mu m g \cos \theta \cdot \sqrt{2} \cdot h + m g h$ $\Rightarrow h = \frac{\; \frac{1}{2} m u^2}{\sqrt{2} \mu m g \cos \theta + m g}$ Here, initial $\text{K.E} = \frac{1}{2} m u^2 = 100\text{ J}$ $m=5\text{ kg},\ \theta=45^{\circ},\ g=10\text{ m/s}^2$ So, $h = \frac{100}{\sqrt{2} \times 0.5 \times 5 \times 10 \times \frac{1}{\sqrt{2}} + 5 \times 10}$ $= \frac{100}{50 \times 1.5} = \frac{4}{3}$ Hence, distance covered over block, $s=\sqrt{2} h$ $\Rightarrow s = \frac{\sqrt{2} \times 4}{3}\text{ m} = \frac{4\sqrt{2}}{3}\text{ m}$