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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, \(g=10 \mathrm{~ms}^{-2}\))
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Verified Answer
The correct answer is:
\(\frac{20}{3} \mathrm{~m}\)
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,
\(\Rightarrow \quad D F=v \cos \theta . t+0 \text { or } D F=\frac{25}{3} \times 0.8=\frac{20}{3} \mathrm{~m}\)
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,
\(\Rightarrow \quad D F=v \cos \theta . t+0 \text { or } D F=\frac{25}{3} \times 0.8=\frac{20}{3} \mathrm{~m}\)
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