TS EAMCET · Physics · Laws of Motion
A block of mass \(10 \mathrm{~kg}\), initially at rest, makes a downward motion on \(45^{\circ}\) inclined plane. Then the distance travelled by the block after \(2 \mathrm{~s}\) is (Assume the coefficient of kinetic friction to be 0.3 and \(g=10 \mathrm{~ms}^{-2}\) )
- A \(7 \sqrt{2} \mathrm{~m}\)
- B \(\frac{9}{\sqrt{2}} \mathrm{~m}\)
- C \(10 \sqrt{2} \mathrm{~m}\)
- D \(5 \sqrt{2} \mathrm{~m}\)
Answer & Solution
Correct Answer
(A) \(7 \sqrt{2} \mathrm{~m}\)
Step-by-step Solution
Detailed explanation
\(a = g (\sin\theta - \mu_k \cos\theta)\) \(a = 10 \left(\sin 45^{\circ} - 0.3 \cos 45^{\circ}\right) = 10 \left(\frac{1}{\sqrt{2}} - 0.3 \frac{1}{\sqrt{2}}\right) = \frac{10}{\sqrt{2}} (1 - 0.3) = \frac{7}{\sqrt{2}} \mathrm{~ms}^{-2}\) \(s = ut + \frac{1}{2}at^2\)…
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