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A body of mass 10 kg is moving with an initial speed of 20 m/s. The body stops after 5 s due to friction between body and the floor. The value of the coefficient of friction is (Take acceleration due to gravity g=10 m s⁻²)

Asked in JEE Main 31st Jan 2nd Shift 2023 · Stopping distance and retardation

Answer: (2) 0.4

Step-by-step solution

Idea: the retardation comes from the kinematics, and friction is the only thing causing it, so μ g=a.

a=(Δ v)/(Δ t)=(20-0)/5=4 m s⁻².

On level ground the frictional retardation is

(μ mg)/m=μ g.

μ g=4, so μ=4/(10)=0.4.

The 10 kg cancels, as it always does on level ground: a heavier body has more friction to overcome but proportionally more inertia, and stops in the same time.

Why the other options are wrong

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