Search any question & find its solution
Question:
Answered & Verified by Expert
A body is projected up along a rough inclined plane of inclination $45^{\circ}$. The coefficient of friction is $0.5$. Then the retardation of the block is
Options:
Solution:
2288 Upvotes
Verified Answer
The correct answer is:
$\frac{3 g}{2 \sqrt{2}}$
Retardation
\begin{aligned}
&=g(\sin \theta+\mu \cos \theta) \\
&=g\left(\frac{1}{\sqrt{2}}+0.5 \frac{1}{\sqrt{2}}\right) \\
&=\frac{g}{\sqrt{2}}(1+0.5) \\
&=\frac{1.5 g}{\sqrt{2}}=\frac{3 g}{2 \sqrt{2}}
\end{aligned}
\begin{aligned}
&=g(\sin \theta+\mu \cos \theta) \\
&=g\left(\frac{1}{\sqrt{2}}+0.5 \frac{1}{\sqrt{2}}\right) \\
&=\frac{g}{\sqrt{2}}(1+0.5) \\
&=\frac{1.5 g}{\sqrt{2}}=\frac{3 g}{2 \sqrt{2}}
\end{aligned}
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.