A block of mass 10 kg starts sliding on a surface with an initial velocity of 9.8 m s⁻¹. The coefficient of friction between the surface and block is 0.5. The distance covered by the block before coming to rest is [Use g=9.8 m s⁻²]

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(2) 9.8 m. Idea: friction is the only horizontal force, so the retardation is μ g; then use v²=u²-2as. a=μ g=0.5×9.8=4.9 m s⁻². 0=u²-2as, so s=(u²)/(2a)=((9.8)²)/(2(4.9))=(96.04)/(9.8)=9.8 m. The numbers are chosen so that s=u numerically — a coincidence of u=g and μ=0.5, not a general rule. The 10 kg is not needed: on level ground the stopping distance is independent of mass.

A block of mass 10 kg starts sliding on a surface with an initial velocity of 9.8 m s⁻¹. The coefficient of friction between the surface and block is 0.5. The distance covered by the block before coming to rest is [Use g=9.8 m s⁻²]

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