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Question-168616




Question Number 168616 by Tawa11 last updated on 14/Apr/22
Answered by mr W last updated on 14/Apr/22
M=M_1 +M_2   M_1 v=(M_1 +M_2 )V  V=(5/(5+3))×15=((75)/8) m/s  (a)  ω=(√(k/M))=(√((3000)/8))=5(√(15)) 1/s  (b)  d_(max) =(V_(max) /ω)=((75)/(8×5(√(15))))=0.484m  (c)  a_(max) =ωV_(max) =5(√(15))×((75)/8)=181.5 m/s^2
$${M}={M}_{\mathrm{1}} +{M}_{\mathrm{2}} \\ $$$${M}_{\mathrm{1}} {v}=\left({M}_{\mathrm{1}} +{M}_{\mathrm{2}} \right){V} \\ $$$${V}=\frac{\mathrm{5}}{\mathrm{5}+\mathrm{3}}×\mathrm{15}=\frac{\mathrm{75}}{\mathrm{8}}\:{m}/{s} \\ $$$$\left({a}\right) \\ $$$$\omega=\sqrt{\frac{{k}}{{M}}}=\sqrt{\frac{\mathrm{3000}}{\mathrm{8}}}=\mathrm{5}\sqrt{\mathrm{15}}\:\mathrm{1}/{s} \\ $$$$\left({b}\right) \\ $$$${d}_{{max}} =\frac{{V}_{{max}} }{\omega}=\frac{\mathrm{75}}{\mathrm{8}×\mathrm{5}\sqrt{\mathrm{15}}}=\mathrm{0}.\mathrm{484}{m} \\ $$$$\left({c}\right) \\ $$$${a}_{{max}} =\omega{V}_{{max}} =\mathrm{5}\sqrt{\mathrm{15}}×\frac{\mathrm{75}}{\mathrm{8}}=\mathrm{181}.\mathrm{5}\:{m}/{s}^{\mathrm{2}} \\ $$
Commented by Tawa11 last updated on 14/Apr/22
God bless you sir. I really appreciate.
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}.\:\mathrm{I}\:\mathrm{really}\:\mathrm{appreciate}. \\ $$

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