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Question Number 101051 by aurpeyz last updated on 30/Jun/20

Commented by Dwaipayan Shikari last updated on 30/Jun/20

$$InitialK.E=P.E+workdonebyfrictionalforce$$$$\frac{1}{2}{mv}^2=mgdsin30°+f.d$$$$d=\frac{90}{29.5}=3.05metre$$

Answered by mr W last updated on 30/Jun/20

$$\frac{1}{2}{mv}^2-fs=mgs\sin \theta$$$$s=\frac{{mv}^2}{2(mg\sin \theta +f)}=\frac{5\times 6^2}{2(5\times 10\times 0.5+4.5)}$$$$=3.05m$$

Commented by aurpeyz last updated on 06/Jul/20

$$\mathrm{thanks}\mathrm{Sir}>$$

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