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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 \(g=10 \mathrm{~ms}^{-2}\) )

  1. A \(7 \sqrt{2} \mathrm{~m}\)
  2. B \(\frac{9}{\sqrt{2}} \mathrm{~m}\)
  3. C \(10 \sqrt{2} \mathrm{~m}\)
  4. D \(5 \sqrt{2} \mathrm{~m}\)
Verified Solution

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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