Question Number 101051 by aurpeyz last updated on 30/Jun/20
Commented by Dwaipayan Shikari last updated on 30/Jun/20
$${Initial}\:{K}.\:{E}=\:{P}.{E}\:+\:{work}\:\:{done}\:{by}\:{frictional}\:{force} \\ $$$$\:\:\frac{\mathrm{1}}{\mathrm{2}}{mv}^{\mathrm{2}} ={mgdsin}\mathrm{30}°+{f}.{d} \\ $$$$\:\:{d}=\frac{\mathrm{90}}{\mathrm{29}.\mathrm{5}}=\mathrm{3}.\mathrm{05}\:{metre} \\ $$
Answered by mr W last updated on 30/Jun/20
$$\frac{\mathrm{1}}{\mathrm{2}}{mv}^{\mathrm{2}} −{fs}={mgs}\:\mathrm{sin}\:\theta \\ $$$${s}=\frac{{mv}^{\mathrm{2}} }{\mathrm{2}\left({mg}\:\mathrm{sin}\:\theta+{f}\right)}=\frac{\mathrm{5}×\mathrm{6}^{\mathrm{2}} }{\mathrm{2}\left(\mathrm{5}×\mathrm{10}×\mathrm{0}.\mathrm{5}+\mathrm{4}.\mathrm{5}\right)} \\ $$$$=\mathrm{3}.\mathrm{05}\:{m} \\ $$
Commented by aurpeyz last updated on 06/Jul/20
$$\mathrm{thanks}\:\mathrm{Sir}> \\ $$