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Question Number 215975 by Tawa11 last updated on 23/Jan/25

Answered by Tawa11 last updated on 23/Jan/25

v = 1.98 m/s at the back of the book.

$$\mathrm{v}\:\:=\:\:\mathrm{1}.\mathrm{98}\:\mathrm{m}/\mathrm{s}\:\:\:\:\:\mathrm{at}\:\mathrm{the}\:\mathrm{back}\:\mathrm{of}\:\mathrm{the}\:\mathrm{book}. \\ $$

Answered by mr W last updated on 26/Jan/25

Commented by mr W last updated on 26/Jan/25

l_0 =1.42 m l_1 =1.68 m Δl_1 =1.68−1.42=0.26 m (l_2 /2)=(√(0.35^2 +(0.5×1.68)^2 ))=0.91 m l_2 =2×0.91=1.82m Δl_2 =1.82−1.42=0.4 m sin θ=((0.35)/(0.91))=(5/( 13)) 2T sin θ=mg ⇒T=((mg)/(2 sin θ))=((2×9.81×13)/(10))=25.506 N Δl_2 =(T/k) k=(T/(Δl_2 ))=((25.506)/(0.4))=63.765 N/m energy conservation: ((k×Δl_1 ^2 )/2)+mg×0.35=((k×Δl_2 ^2 )/2)+((mv^2 )/2) ((k(Δl_1 ^2 −Δl_2 ^2 ))/m)+2g×0.35=v^2 ⇒v=(√(((63.765(0.26^2 −0.4^2 ))/2)+2×9.81×0.35)) =1.98 m/s

$${l}_{\mathrm{0}} =\mathrm{1}.\mathrm{42}\:{m} \\ $$$${l}_{\mathrm{1}} =\mathrm{1}.\mathrm{68}\:{m} \\ $$$$\Delta{l}_{\mathrm{1}} =\mathrm{1}.\mathrm{68}−\mathrm{1}.\mathrm{42}=\mathrm{0}.\mathrm{26}\:{m} \\ $$$$\frac{{l}_{\mathrm{2}} }{\mathrm{2}}=\sqrt{\mathrm{0}.\mathrm{35}^{\mathrm{2}} +\left(\mathrm{0}.\mathrm{5}×\mathrm{1}.\mathrm{68}\right)^{\mathrm{2}} }=\mathrm{0}.\mathrm{91}\:{m} \\ $$$${l}_{\mathrm{2}} =\mathrm{2}×\mathrm{0}.\mathrm{91}=\mathrm{1}.\mathrm{82}{m} \\ $$$$\Delta{l}_{\mathrm{2}} =\mathrm{1}.\mathrm{82}−\mathrm{1}.\mathrm{42}=\mathrm{0}.\mathrm{4}\:{m} \\ $$$$\mathrm{sin}\:\theta=\frac{\mathrm{0}.\mathrm{35}}{\mathrm{0}.\mathrm{91}}=\frac{\mathrm{5}}{\:\mathrm{13}} \\ $$$$\mathrm{2}{T}\:\mathrm{sin}\:\theta={mg} \\ $$$$\Rightarrow{T}=\frac{{mg}}{\mathrm{2}\:\mathrm{sin}\:\theta}=\frac{\mathrm{2}×\mathrm{9}.\mathrm{81}×\mathrm{13}}{\mathrm{10}}=\mathrm{25}.\mathrm{506}\:{N} \\ $$$$\Delta{l}_{\mathrm{2}} =\frac{{T}}{{k}}\: \\ $$$${k}=\frac{{T}}{\Delta{l}_{\mathrm{2}} }=\frac{\mathrm{25}.\mathrm{506}}{\mathrm{0}.\mathrm{4}}=\mathrm{63}.\mathrm{765}\:{N}/{m} \\ $$$${energy}\:{conservation}: \\ $$$$\frac{{k}×\Delta{l}_{\mathrm{1}} ^{\mathrm{2}} }{\mathrm{2}}+{mg}×\mathrm{0}.\mathrm{35}=\frac{{k}×\Delta{l}_{\mathrm{2}} ^{\mathrm{2}} }{\mathrm{2}}+\frac{{mv}^{\mathrm{2}} }{\mathrm{2}} \\ $$$$\frac{{k}\left(\Delta{l}_{\mathrm{1}} ^{\mathrm{2}} −\Delta{l}_{\mathrm{2}} ^{\mathrm{2}} \right)}{{m}}+\mathrm{2}{g}×\mathrm{0}.\mathrm{35}={v}^{\mathrm{2}} \\ $$$$\Rightarrow{v}=\sqrt{\frac{\mathrm{63}.\mathrm{765}\left(\mathrm{0}.\mathrm{26}^{\mathrm{2}} −\mathrm{0}.\mathrm{4}^{\mathrm{2}} \right)}{\mathrm{2}}+\mathrm{2}×\mathrm{9}.\mathrm{81}×\mathrm{0}.\mathrm{35}} \\ $$$$\:\:\:\:\:\:\:=\mathrm{1}.\mathrm{98}\:{m}/{s} \\ $$

Commented by Tawa11 last updated on 25/Jan/25

Wow, thanks sir. I really appreciate.

$$\mathrm{Wow},\:\mathrm{thanks}\:\mathrm{sir}. \\ $$$$\mathrm{I}\:\mathrm{really}\:\mathrm{appreciate}. \\ $$

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