A block of mass $5 \mathrm{~kg}$ starts up a $45^{\circ}$ incline plane with initial kinetic energy of $100 \mathrm{~J}$. If the coefficient of friction between block and plane is 0.5 then the distance covered by the block before it stops is (Acceleration due to gravity $=10 \mathrm{~ms}^{-2}$ )

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Aicharya · Accepted Answer

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 retaled to height of block (h) as, $\Rightarrow \quad \sin 45^{\circ}=\frac{h}{s}$ or $s=\sqrt{2} . h$ Let block covers a distance $s$ over block and reaches upto height $h$. Then Initial K.E of block = Work done against friction + Potential energy gained by block $\begin{aligned} & \Rightarrow \frac{1}{2} m u^2=\mu m g \cos \theta \cdot s+m g h \\ & \Rightarrow \frac{1}{2} m u^2=\mu m g \cos \theta \cdot \sqrt{2} \cdot h+m g h\end{aligned}$ $\Rightarrow \quad h=\frac{\frac{1}{2} m u^2}{\sqrt{2} \mu m g \cos \theta+m g}$ Here, initial K. $\mathrm{E}=\frac{1}{2} m u^2=100 \mathrm{~J}$ $m=5 \mathrm{~kg}, \theta=45^{\circ}, g=10 \mathrm{~m} / \mathrm{s}^2$ 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, $s=\sqrt{2} h$ $\Rightarrow \quad s=\frac{\sqrt{2} \times 4}{3} \mathrm{~m}=\frac{4 \sqrt{2}}{3} \mathrm{~m}$

A block of mass $5 \mathrm{~kg}$ starts up a $45^{\circ}$ incline plane with initial kinetic energy of $100…

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