A rough inclined plane $B C E$ of height $\left(\frac{25}{6}\right) \mathrm{m}$ is kept on a rectangular wooden block $A B C D$ of height $10 \mathrm{~m}$, as shown in the figure. A small block is allowed to slide down from the top $E$ of the inclined plane. The coefficient of kinetic…

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A rough inclined plane $B C E$ of height $\left(\frac{25}{6}\right) \mathrm{m}$ is kept on a rectangular wooden block $A B C D$ of height $10 \mathrm{~m}$, as shown in the figure. A small block is allowed to slide down from the top $E$ of the inclined plane. The coefficient of kinetic friction between the block and the inclined plane is $\frac{1}{8}$ and the angle of inclination of the inclined plane is $\sin ^{-1}(0.6)$. If the small block finally reaches the ground at a point $F$, then $D F$ will be (Acceleration due to gravity, $g=10 \mathrm{~ms}^{-2}$)

Aicharya · Accepted Answer

According to question, a small block is slide down from top $E$ of inclined plane as shown in figure,

Force equation of a block, $\Rightarrow \quad m g \sin \theta-f=m a$ $\because$ friction force applied one block, $f=\mu_k R$ or $f=\mu_k R(m g \cos \theta)$ (From figure) where, $\mu_k=$ coefficient of kinetic friction from Eq. (i), we get $\begin{aligned} & \Rightarrow \quad m g \sin \theta-\mu_k m g \cos \theta=\text { ma } \\ & \Rightarrow a=10 \sin \theta-1 / 8 \times 10 \cos \theta \ldots \text { (ii) }\left(\because \mu_k=\frac{1}{8} \text {, given }\right) \end{aligned}$ $\because$ Given, angle of the inclined plane, $\theta=\sin ^{-1}(0.6)$ or $\sin \theta=0.6$ $\begin{array}{rlrl} \because & \cos \theta =\sqrt{1-\sin ^2 \theta}=\sqrt{1-(0.6)^2} \\ \therefore & \cos \theta =0.8 \end{array}$ Form Eq. (ii), $a=10(0.6)-\frac{1}{8} \times 10(0.8) \text { or } a=5 \mathrm{~ms}^{-2}$ when, block reached to point $C$, then from third equation of the motion, $v^2=u^2+2 a s$ where, $v=$ final velocity of the block at point $C$ $u=$ initial velocity of block at point $E$ or $v=\sqrt{2 a s} \quad(\because u=0)$ From $\Delta \mathrm{EBC}, E C=\frac{B E}{\sin \theta} \Rightarrow s=\frac{25 / 6}{0.6}=\frac{25}{6 \times 0.6}$ or $\begin{aligned} & s=\frac{125}{18} \mathrm{~m} \\ & v=\sqrt{2 \times 5 \times \frac{125}{18}}=\frac{25}{3} \mathrm{~ms}^{-1} \end{aligned}$ From second equation of motion of the block at point C,

In $y$-direction, $\begin{array}{ccc} \Rightarrow & h=u t+\frac{1}{2} g t^2 \\ \Rightarrow & 10=u \sin \theta t+\frac{1}{2} g t^2 \\ \Rightarrow & 10=\frac{25}{3} \times 0.6 t+\frac{1}{2} \times 10 \times t^2 \\ \Rightarrow & \frac{25}{3} \times \frac{6}{10} t+\frac{1}{2} \times 10 \times t^2=10 \\ \Rightarrow & 5 t+5 t^2=10 \\ \Rightarrow & t^2+t-2=0 \text { or } t=1 \mathrm{sec} \end{array}$ Now, again from second Eqs. of motion in $x$-direction, $\Rightarrow \quad D F=v \cos \theta . t+0 \text { or } D F=\frac{25}{3} \times 0.8=\frac{20}{3} \mathrm{~m}$

A rough inclined plane \(B C E\) of height \(\left(\frac{25}{6}\right) \mathrm{m}\) is kept on a rectangular…

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