Practice portal › Laws of Motion › Friction: Blocks, Belts and Inclines
Asked in JEE Main 2005 · Stopping distance and retardation
Idea: friction is the only retarding force, so the deceleration is μ g and v²=2as finishes it.
a=μₖg=0.5×10=5 m/s².
s=(v²)/(2a)=(100²)/(2×5)=(10000)/(10)=1000 m.
(With g=9.8 it is 1020 m, still nearest 1000 m.)
A kilometre to stop is the point of the question: stopping distance grows as the square of the speed, so doubling the speed quadruples it.
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