Question Number 29216 by NECx last updated on 05/Feb/18
$${A}\:{horizontal}\:{force}\:{of}\:\mathrm{0}.\mathrm{8}{N}\:{is} \\ $$$${required}\:{to}\:{pull}\:{a}\:\mathrm{5}{kg}\:{block}\:{across} \\ $$$${a}\:{table}\:{top}\:{at}\:{a}\:{constant}\:{speed}. \\ $$$${With}\:{the}\:{block}\:{initially}\:{at}\:{rest},{a} \\ $$$$\mathrm{20}{g}\:{bullet}\:{fired}\:{horizontally}\:{into} \\ $$$${the}\:{block}\:{to}\:{slide}\:\mathrm{1}.\mathrm{5}{m}\:{before} \\ $$$${coming}\:{to}\:{rest}\:{again}.{Determine} \\ $$$${the}\:{speed}\:{v}\:{of}\:{the}\:{bullet},{where} \\ $$$${the}\:{bullet}\:{is}\:{assumed}\:{to}\:{be} \\ $$$${embedded}\:{in}\:{the}\:{block}. \\ $$
Answered by ajfour last updated on 05/Feb/18
$${f}=\mathrm{0}.\mathrm{8}\:=\mathrm{50}\mu\:\:\Rightarrow\:\:\:\mu=\mathrm{0}.\mathrm{016} \\ $$$${mv}=\left({M}+{m}\right){V} \\ $$$$\mu\left({M}+{m}\right){gs}=\frac{\mathrm{1}}{\mathrm{2}}\left({M}+{m}\right){V}^{\:\:\mathrm{2}} \\ $$$$\Rightarrow\:\:\:{V}=\sqrt{\mathrm{2}\mu{gs}} \\ $$$${v}=\:\frac{\left({M}+{m}\right){V}}{{m}}\:=\:\left(\frac{{M}}{{m}}+\mathrm{1}\right)\sqrt{\mathrm{2}\mu{gs}} \\ $$$$\:\:=\left(\mathrm{251}\right)\sqrt{\mathrm{2}×\mathrm{0}.\mathrm{016}×\mathrm{10}×\mathrm{1}.\mathrm{5}} \\ $$$$\Rightarrow\:\:\:{v}\:=\mathrm{251}×\sqrt{\mathrm{48}}\:{m}/{s} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:=\mathrm{251}×\mathrm{7}\left(\mathrm{1}−\mathrm{0}.\mathrm{02}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:=\mathrm{1757}−\mathrm{35}\:=\:\mathrm{1722}{m}/{s}\:. \\ $$