Question-218013

Question Number 218013 by Tawa11 last updated on 25/Mar/25

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

a) AP=BP=l=(√(1^2 +2.4^2 ))=2.6 m l_0 =2 m Δl=l−l_0 =2.6−2=0.6 m sin θ=(1/(2.6)) 2T sin θ=mg 2T×(1/(2.6))=mg ⇒T=1.3mg T=(λ/l_0 )Δl=((637)/2)×0.6=1.3mg ⇒m=((637×0.6)/(2×1.3×9.81))≈15 kg b) l=(√(0.7^2 +2.4^2 ))=2.5 m Δl_1 =2.5−2=0.5 m sin θ_1 =((0.7)/( 2.5))=0.28 T_1 =(λ/l_0 )Δl_1 ma=mg−2T_1 sin θ_1 a=g−((2T_1 sin θ_1 )/m) =(1−((Δl_1 sin θ_1 )/(Δl sin θ)))g =(1−((0.5×0.28×2.6)/(0.6×1)))×9.81≈3.86 m/s^2

$$\left.{a}\right) \\ $$$${AP}={BP}={l}=\sqrt{\mathrm{1}^{\mathrm{2}} +\mathrm{2}.\mathrm{4}^{\mathrm{2}} }=\mathrm{2}.\mathrm{6}\:{m} \\ $$$${l}_{\mathrm{0}} =\mathrm{2}\:{m} \\ $$$$\Delta{l}={l}−{l}_{\mathrm{0}} =\mathrm{2}.\mathrm{6}−\mathrm{2}=\mathrm{0}.\mathrm{6}\:{m} \\ $$$$\mathrm{sin}\:\theta=\frac{\mathrm{1}}{\mathrm{2}.\mathrm{6}} \\ $$$$\mathrm{2}{T}\:\mathrm{sin}\:\theta={mg} \\ $$$$\mathrm{2}{T}×\frac{\mathrm{1}}{\mathrm{2}.\mathrm{6}}={mg} \\ $$$$\Rightarrow{T}=\mathrm{1}.\mathrm{3}{mg} \\ $$$${T}=\frac{\lambda}{{l}_{\mathrm{0}} }\Delta{l}=\frac{\mathrm{637}}{\mathrm{2}}×\mathrm{0}.\mathrm{6}=\mathrm{1}.\mathrm{3}{mg} \\ $$$$\Rightarrow{m}=\frac{\mathrm{637}×\mathrm{0}.\mathrm{6}}{\mathrm{2}×\mathrm{1}.\mathrm{3}×\mathrm{9}.\mathrm{81}}\approx\mathrm{15}\:{kg} \\ $$$$\left.{b}\right) \\ $$$${l}=\sqrt{\mathrm{0}.\mathrm{7}^{\mathrm{2}} +\mathrm{2}.\mathrm{4}^{\mathrm{2}} }=\mathrm{2}.\mathrm{5}\:{m} \\ $$$$\Delta{l}_{\mathrm{1}} =\mathrm{2}.\mathrm{5}−\mathrm{2}=\mathrm{0}.\mathrm{5}\:{m} \\ $$$$\mathrm{sin}\:\theta_{\mathrm{1}} =\frac{\mathrm{0}.\mathrm{7}}{\:\mathrm{2}.\mathrm{5}}=\mathrm{0}.\mathrm{28} \\ $$$${T}_{\mathrm{1}} =\frac{\lambda}{{l}_{\mathrm{0}} }\Delta{l}_{\mathrm{1}} \\ $$$${ma}={mg}−\mathrm{2}{T}_{\mathrm{1}} \:\mathrm{sin}\:\theta_{\mathrm{1}} \\ $$$${a}={g}−\frac{\mathrm{2}{T}_{\mathrm{1}} \:\mathrm{sin}\:\theta_{\mathrm{1}} }{{m}} \\ $$$$\:\:=\left(\mathrm{1}−\frac{\Delta{l}_{\mathrm{1}} \:\mathrm{sin}\:\theta_{\mathrm{1}} }{\Delta{l}\:\mathrm{sin}\:\theta}\right){g} \\ $$$$\:\:=\left(\mathrm{1}−\frac{\mathrm{0}.\mathrm{5}×\mathrm{0}.\mathrm{28}×\mathrm{2}.\mathrm{6}}{\mathrm{0}.\mathrm{6}×\mathrm{1}}\right)×\mathrm{9}.\mathrm{81}\approx\mathrm{3}.\mathrm{86}\:{m}/{s}^{\mathrm{2}} \\ $$

Commented by Tawa11 last updated on 26/Mar/25

$$\mathrm{Thanks}\:\mathrm{sir}. \\ $$$$\mathrm{I}\:\mathrm{appreciate}. \\ $$

Commented by Tawa11 last updated on 26/Mar/25

I have seen a mistake in their work sir. They used m = 12.5kg. They also proved m = 15kg. so, I don′t know why they now used m = 12.5kg to find the acceleration.? silly error

$$\mathrm{I}\:\mathrm{have}\:\mathrm{seen}\:\mathrm{a}\:\mathrm{mistake}\:\mathrm{in}\:\mathrm{their}\:\mathrm{work}\:\mathrm{sir}. \\ $$$$\mathrm{They}\:\mathrm{used}\:\:\mathrm{m}\:\:=\:\:\mathrm{12}.\mathrm{5kg}. \\ $$$$\mathrm{They}\:\mathrm{also}\:\mathrm{proved}\:\:\mathrm{m}\:=\:\mathrm{15kg}. \\ $$$$\mathrm{so},\:\mathrm{I}\:\mathrm{don}'\mathrm{t}\:\mathrm{know}\:\mathrm{why}\:\mathrm{they}\:\mathrm{now}\:\mathrm{used}\:\mathrm{m}\:=\:\mathrm{12}.\mathrm{5kg} \\ $$$$\mathrm{to}\:\mathrm{find}\:\mathrm{the}\:\mathrm{acceleration}.? \\ $$$$\mathrm{silly}\:\mathrm{error} \\ $$

Commented by Tawa11 last updated on 26/Mar/25

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

we even didn′t need m to find a: a=(1−((Δl_1 sin θ_1 )/(Δl sin θ)))g

$${we}\:{even}\:{didn}'{t}\:{need}\:{m}\:{to}\:{find}\:{a}: \\ $$$${a}=\left(\mathrm{1}−\frac{\Delta{l}_{\mathrm{1}} \:\mathrm{sin}\:\theta_{\mathrm{1}} }{\Delta{l}\:\mathrm{sin}\:\theta}\right){g} \\ $$

Commented by Tawa11 last updated on 26/Mar/25

Yes, I observed it in your workings sir. God bless you always sir.

$$\mathrm{Yes},\:\mathrm{I}\:\mathrm{observed}\:\mathrm{it}\:\mathrm{in}\:\mathrm{your}\:\mathrm{workings}\:\mathrm{sir}. \\ $$$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{always}\:\mathrm{sir}. \\ $$

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