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

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

$$a)$$$$AP=BP=l=\sqrt{1^2+2.4^2}=2.6m$$$$l_0=2m$$$$\mathrm{\Delta }l=l-l_0=2.6-2=0.6m$$$$\sin \theta =\frac{1}{2.6}$$$$2T\sin \theta =mg$$$$2T\times \frac{1}{2.6}=mg$$$$\Rightarrow T=1.3mg$$$$T=\frac{\lambda }{l_0}\mathrm{\Delta }l=\frac{637}{2}\times 0.6=1.3mg$$$$\Rightarrow m=\frac{637\times 0.6}{2\times 1.3\times 9.81}\approx 15kg$$$$b)$$$$l=\sqrt{0.7^2+2.4^2}=2.5m$$$$\mathrm{\Delta }{l}_1=2.5-2=0.5m$$$$\sin {\theta }_1=\frac{0.7}{2.5}=0.28$$$$T_1=\frac{\lambda }{l_0}\mathrm{\Delta }{l}_1$$$$ma=mg-2T_1\sin {\theta }_1$$$$a=g-\frac{2T_1\sin {\theta }_1}{m}$$$$=(1-\frac{\mathrm{\Delta }{l}_1\sin {\theta }_1}{\mathrm{\Delta }l\sin \theta })g$$$$=(1-\frac{0.5\times 0.28\times 2.6}{0.6\times 1})\times 9.81\approx 3.86m/s^2$$

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

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

Commented by Tawa11 last updated on 26/Mar/25

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

$$weeven{didn}^{\prime }tneedmtofinda:$$$$a=(1-\frac{\mathrm{\Delta }{l}_1\sin {\theta }_1}{\mathrm{\Delta }l\sin \theta })g$$

Commented by Tawa11 last updated on 26/Mar/25

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

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