AP EAMCET · PHYSICS · Laws of Motion
A block of mass \(5 \mathrm{~kg}\) starts up a \(45^{\circ}\) incline plane with initial kinetic energy of \(100 \mathrm{~J}\). If the coefficient of friction between block and plane is 0.5 then the distance covered by the block before it stops is
(Acceleration due to gravity \(=10 \mathrm{~ms}^{-2}\) )
- A \(\frac{4 \sqrt{2}}{3} m\)
- B \(\frac{3}{\sqrt{2}} m\)
- C \(2 \sqrt{2} \mathrm{~m}\)
- D \(\frac{6}{5} \sqrt{2} \mathrm{~m}\)
Answer & Solution
Correct Answer
(A) \(\frac{4 \sqrt{2}}{3} m\)
Step-by-step Solution
Detailed explanation
Block stops on incline plane when its initial K.E is used in doing work against friction and partly it is converted into potential energy. As angle of incline is given \(45^{\circ}\) so, distance covered on incline \((s)\) is retaled to height of block (h) as,…
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