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Question Number 39649 by rahul 19 last updated on 09/Jul/18

Commented by rahul 19 last updated on 09/Jul/18

$$\mathrm{Given}\mathrm{mass}\mathrm{of}\mathrm{block}\mathrm{A}\mathrm{\&}\mathrm{C}=\mathrm{m}$$$$\mathrm{mass}\mathrm{of}\mathrm{block}\mathrm{B}=2\mathrm{m}.$$$$\mathrm{Spring}\mathrm{constant}=\mathrm{k}.$$$$\mathrm{Find}{\mathrm{acc}}^{\mathrm{n}}\mathrm{of}\mathrm{the}\mathrm{blocks}\mathrm{A},\mathrm{B}\mathrm{\&}\mathrm{C}\mathrm{at}\mathrm{the}$$$$\mathrm{instant}\mathrm{when}\mathrm{the}\mathrm{string}\mathrm{between}\mathrm{B}\mathrm{\&}\mathrm{C}$$$$\mathrm{is}\mathrm{cut}?$$

Answered by MrW3 last updated on 09/Jul/18

$$Beforethecut:$$$$T_2=m_{c}g=mg$$$$T_1=m_{b}g+T_2=3mg$$$$T_k+m_{a}g=T_1$$$$\Rightarrow T_k=2mg$$$$$$$$Afterthecut:$$$$T_2=0$$$$T_1-m_{b}g=m_{b}a$$$$\Rightarrow T_1=2mg+2ma$$$$T_k+m_{a}g-T_1=m_{a}a$$$$\Rightarrow 2mg+mg-2mg-2ma=ma$$$$\Rightarrow a=\frac{g}{3}$$

Commented by rahul 19 last updated on 09/Jul/18

$$\mathrm{Sir},\mathrm{i}\mathrm{want}\mathrm{to}\mathrm{know}\mathrm{why}\mathrm{after}\mathrm{the}\mathrm{cut},$$$${\mathrm{T}}_1\mathrm{and}{\mathrm{T}}_{\mathrm{k}}\mathrm{remains}\mathrm{same}?\mathrm{Why}\mathrm{not}$$$$\mathrm{after}\mathrm{the}\mathrm{cut}\mathrm{new}\mathrm{tension}\mathrm{develops}?$$

Commented by ajfour last updated on 09/Jul/18

$$sincetilljustaftercutspringextension$$$$hasnotchanged.$$

Commented by rahul 19 last updated on 09/Jul/18

Thank you sir !����

Commented by MrW3 last updated on 09/Jul/18

$$forcesinstringsandinsprings$$$$willchangewhentheirdeformation$$$$changesorwhentheaccelerationof$$$$objectchanges.$$

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