TS EAMCET · Physics · Laws of Motion
A small block starts sliding down an inclined plane forming an angle \(45^{\circ}\) horizontal. The coefficient of friction \(\mu\), varies with distance \(s\) as \(\mu=c s^2\) where, \(c\) is a constant of appropriate dimensions, then distance covered by the block before it stops is
- A \(\sqrt{\frac{3}{C}}\)
- B \(\sqrt{3 C}\)
- C \(\sqrt{C}\)
- D \(\sqrt{\frac{1}{C}}\)
Answer & Solution
Correct Answer
(A) \(\sqrt{\frac{3}{C}}\)
Step-by-step Solution
Detailed explanation
\[ \text { Given that, } \] \(\mu=C s^2 \quad\{\because s=\) distance and \(C=\) constant \(\}\) Net force on the block is…
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