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Question Number 21388 by Tinkutara last updated on 22/Sep/17

A block of mass m is connected with another block of mass 2m by a light spring. 2m is connected with a hanging mass 3m by an inextensible light string. At the time of release of block 3m, find tension in the string and acceleration of all the masses.

$$\mathrm{A}\:\mathrm{block}\:\mathrm{of}\:\mathrm{mass}\:{m}\:\mathrm{is}\:\mathrm{connected}\:\mathrm{with} \\ $$$$\mathrm{another}\:\mathrm{block}\:\mathrm{of}\:\mathrm{mass}\:\mathrm{2}{m}\:\mathrm{by}\:\mathrm{a}\:\mathrm{light} \\ $$$$\mathrm{spring}.\:\mathrm{2}{m}\:\mathrm{is}\:\mathrm{connected}\:\mathrm{with}\:\mathrm{a}\:\mathrm{hanging} \\ $$$$\mathrm{mass}\:\mathrm{3}{m}\:\mathrm{by}\:\mathrm{an}\:\mathrm{inextensible}\:\mathrm{light}\:\mathrm{string}. \\ $$$$\mathrm{At}\:\mathrm{the}\:\mathrm{time}\:\mathrm{of}\:\mathrm{release}\:\mathrm{of}\:\mathrm{block}\:\mathrm{3}{m},\:\mathrm{find} \\ $$$$\mathrm{tension}\:\mathrm{in}\:\mathrm{the}\:\mathrm{string}\:\mathrm{and}\:\mathrm{acceleration} \\ $$$$\mathrm{of}\:\mathrm{all}\:\mathrm{the}\:\mathrm{masses}. \\ $$

Commented by Tinkutara last updated on 22/Sep/17

Commented by mrW1 last updated on 22/Sep/17

since the string is inextensible, “2m” and “3m” have the same velocity and acceleration. let T=tension in the string let a=acceleration of “2m” and “3m” 3mg−T=3ma T=2ma ⇒a=(3/5)g=0.6g ⇒T=(6/5)mg=1.2mg the elongation of the spring is zero, therefore the force in it is also zero. ⇒the acceleration of “m” is zero.

$$\mathrm{since}\:\mathrm{the}\:\mathrm{string}\:\mathrm{is}\:\mathrm{inextensible},\: \\ $$$$``\mathrm{2m}''\:\mathrm{and}\:``\mathrm{3m}''\:\mathrm{have}\:\mathrm{the}\:\mathrm{same} \\ $$$$\mathrm{velocity}\:\mathrm{and}\:\mathrm{acceleration}. \\ $$$$ \\ $$$$\mathrm{let}\:\mathrm{T}=\mathrm{tension}\:\mathrm{in}\:\mathrm{the}\:\mathrm{string} \\ $$$$\mathrm{let}\:\mathrm{a}=\mathrm{acceleration}\:\mathrm{of}\:``\mathrm{2m}''\:\mathrm{and}\:``\mathrm{3m}'' \\ $$$$ \\ $$$$\mathrm{3mg}−\mathrm{T}=\mathrm{3ma} \\ $$$$\mathrm{T}=\mathrm{2ma} \\ $$$$\Rightarrow\mathrm{a}=\frac{\mathrm{3}}{\mathrm{5}}\mathrm{g}=\mathrm{0}.\mathrm{6g} \\ $$$$\Rightarrow\mathrm{T}=\frac{\mathrm{6}}{\mathrm{5}}\mathrm{mg}=\mathrm{1}.\mathrm{2mg} \\ $$$$ \\ $$$$\mathrm{the}\:\mathrm{elongation}\:\mathrm{of}\:\mathrm{the}\:\mathrm{spring}\:\mathrm{is}\:\mathrm{zero}, \\ $$$$\mathrm{therefore}\:\mathrm{the}\:\mathrm{force}\:\mathrm{in}\:\mathrm{it}\:\mathrm{is}\:\mathrm{also}\:\mathrm{zero}. \\ $$$$\Rightarrow\mathrm{the}\:\mathrm{acceleration}\:\mathrm{of}\:``\mathrm{m}''\:\mathrm{is}\:\mathrm{zero}. \\ $$

Commented by Tinkutara last updated on 23/Sep/17

Thank you very much Sir!

$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much}\:\mathrm{Sir}! \\ $$

Commented by mrW1 last updated on 23/Sep/17

at t=0 the displacement of “2m”is zero, therefore the force in the spring between “m” and “2m” is zero. ⇒acceleration of “m” is zero at t=0. this is independent from which mass it has.

$$\mathrm{at}\:\mathrm{t}=\mathrm{0}\:\mathrm{the}\:\mathrm{displacement}\:\mathrm{of}\:``\mathrm{2m}''\mathrm{is} \\ $$$$\mathrm{zero},\:\mathrm{therefore}\:\mathrm{the}\:\mathrm{force}\:\mathrm{in}\:\mathrm{the}\:\mathrm{spring} \\ $$$$\mathrm{between}\:``\mathrm{m}''\:\mathrm{and}\:``\mathrm{2m}''\:\mathrm{is}\:\mathrm{zero}.\: \\ $$$$\Rightarrow\mathrm{acceleration}\:\mathrm{of}\:``\mathrm{m}''\:\mathrm{is}\:\mathrm{zero}\:\mathrm{at}\:\mathrm{t}=\mathrm{0}. \\ $$$$\mathrm{this}\:\mathrm{is}\:\mathrm{independent}\:\mathrm{from}\:\mathrm{which}\:\mathrm{mass} \\ $$$$\mathrm{it}\:\mathrm{has}. \\ $$

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