A block of mass $10\, kg$ moving at $10\,m/s$ is released to slide on rough surface having coefficient of friction $0.2.$ It will stop by travelling distance ........ $m$
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$S=\frac{V^{2}}{2 \mu g}=\frac{10^{2}}{2 \times 0.2 \times 10}=25 \mathrm{m}$
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