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Question-54841




Question Number 54841 by peter frank last updated on 12/Feb/19
Answered by mr W last updated on 12/Feb/19
50g−T=50a   ...(i)  T−20g=20a   ...(ii)  (i)+(ii):  30g=70a  ⇒a=((3g)/7)≈4.3 m/s^2   from (ii):  ⇒T=20(a+g)=20×((10g)/7)=((200g)/7)≈286 N
$$\mathrm{50}{g}−{T}=\mathrm{50}{a}\:\:\:…\left({i}\right) \\ $$$${T}−\mathrm{20}{g}=\mathrm{20}{a}\:\:\:…\left({ii}\right) \\ $$$$\left({i}\right)+\left({ii}\right): \\ $$$$\mathrm{30}{g}=\mathrm{70}{a} \\ $$$$\Rightarrow{a}=\frac{\mathrm{3}{g}}{\mathrm{7}}\approx\mathrm{4}.\mathrm{3}\:{m}/{s}^{\mathrm{2}} \\ $$$${from}\:\left({ii}\right): \\ $$$$\Rightarrow{T}=\mathrm{20}\left({a}+{g}\right)=\mathrm{20}×\frac{\mathrm{10}{g}}{\mathrm{7}}=\frac{\mathrm{200}{g}}{\mathrm{7}}\approx\mathrm{286}\:{N} \\ $$

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