AP EAMCET · PHYSICS · Laws of Motion
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, \(g=10 \mathrm{~ms}^{-2}\))
- A \(\frac{5}{3} m\)
- B \(\frac{10}{3} \mathrm{~m}\)
- C \(\frac{13}{3} m\)
- D \(\frac{20}{3} \mathrm{~m}\)
Answer & Solution
Correct Answer
(D) \(\frac{20}{3} \mathrm{~m}\)
Step-by-step Solution
Detailed explanation
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…
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