Question Number 103792 by aurpeyz last updated on 17/Jul/20
Answered by Dwaipayan Shikari last updated on 17/Jul/20
$${v}_{{f}} ^{\mathrm{2}} ={v}_{\mathrm{0}} ^{\mathrm{2}} +\mathrm{2}{as} \\ $$$$\mathrm{400}=−\mathrm{2}{a}.\mathrm{120} \\ $$$${a}=−\frac{\mathrm{400}}{\mathrm{240}}\:\frac{{m}}{{s}^{\mathrm{2}} }=−\frac{\mathrm{5}}{\mathrm{3}}\:\frac{{m}}{{s}^{\mathrm{2}} }\left({negative}\:{means}\:{deaccleration}\right) \\ $$$${Force}\:{on}\:{puck}=−{m}.\frac{\mathrm{5}}{\mathrm{3}}\:{N} \\ $$$${Is}\:{equal}\:{tofrictional}\:{force}\:{f}=−\mu{mg}\:\left({negative}\:{depends}\:{on}\:{sign}\:{convention}\right. \\ $$$$−{m}\frac{\mathrm{5}}{\mathrm{3}}=−\mu{mg} \\ $$$$\mu=\frac{\mathrm{5}}{\mathrm{30}}=\frac{\mathrm{1}}{\mathrm{6}}=\mathrm{0}.\mathrm{1666666} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$
Answered by Dwaipayan Shikari last updated on 17/Jul/20
$$\frac{\mathrm{1}}{\mathrm{2}}{mv}^{\mathrm{2}} ={f}.{d}\:\:\:\:\:\left\{{Change}\:{in}\:{kinetic}\:{energy}=\frac{\mathrm{1}}{\mathrm{2}}{m}\left({v}^{\mathrm{2}} −\mathrm{0}\right)=\frac{\mathrm{1}}{\mathrm{2}}{mv}^{\mathrm{2}} \right. \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}{mv}^{\mathrm{2}} =\mu{mgd}\:\:\:\left\{{Work}\:{done}\:{by}\:{the}\:{frictional}\:{force}={f}.{d}\right. \\ $$$$\mu=\frac{{v}^{\mathrm{2}} }{\mathrm{2}{gd}}=\frac{\mathrm{400}}{\mathrm{2400}}=\frac{\mathrm{1}}{\mathrm{6}}=\mathrm{0}.\mathrm{16666666}.. \\ $$
Commented by aurpeyz last updated on 17/Jul/20
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{Sir}.\:\mathrm{is}\:\mathrm{there}\:\mathrm{any}\:\mathrm{method}\:\mathrm{apart}\:\mathrm{from}\:\mathrm{conservstion}\:\mathrm{of}\:\mathrm{energy}\:\mathrm{that}\:\mathrm{can}\:\mathrm{be}\:\mathrm{used}? \\ $$