MHT CET · Physics · Laws of Motion
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
- A \(\frac{g}{2 \sqrt{2}}\)
- B \(\frac{g}{2}\)
- C \(\frac{3 g}{2 \sqrt{2}}\)
- D \(\frac{g}{\sqrt{2}}\)
Answer & Solution
Correct Answer
(C) \(\frac{3 g}{2 \sqrt{2}}\)
Step-by-step Solution
Detailed explanation
Retardation
\(\begin{aligned} &=g(\sin \theta+\mu \cos \theta) \\ &=g\left(\frac{1}{\sqrt{2}}+0.5 \frac{1}{\sqrt{2}}\right) \\ &=\frac{g}{\sqrt{2}}(1+0.5) \\ &=\frac{1.5 g}{\sqrt{2}}=\frac{3 g}{2 \sqrt{2}} \end{aligned}\)
\(\begin{aligned} &=g(\sin \theta+\mu \cos \theta) \\ &=g\left(\frac{1}{\sqrt{2}}+0.5 \frac{1}{\sqrt{2}}\right) \\ &=\frac{g}{\sqrt{2}}(1+0.5) \\ &=\frac{1.5 g}{\sqrt{2}}=\frac{3 g}{2 \sqrt{2}} \end{aligned}\)
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