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A block of mass 0.5kg has an initial velocity of 10 m/s down an inclined plane 30°, the coefficient of friction between the block and the inclined surface is 0.2. |
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Answer» Doward acceleration a = gsin30°-mu×gxcis30° =10×1/2-0.2×10×√3/2 =5-√3 m/s^2 So velocity after moving 10 m downward v^2=u^2+2×a×10 =>v^2= 10^2+2×(5-√3)×10 =v=13m/s (approx) |
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