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Question Number 13727 by tawa tawa last updated on 22/May/17

A light wire which obeys hooke′s law hangs vertically on a fixed support. The wire has an unstretched lenght of 15cm. The lenght of the wire however increase to 20cm when a load of 0.5kg is attached to it lower end . What is the tension in the wire when the load is at rest ?. If the load is pulled down until the lenght of the wire is 22cm. What is the new tension in the wire (g = 9.8 m/s).

$$\mathrm{A}\:\mathrm{light}\:\mathrm{wire}\:\mathrm{which}\:\mathrm{obeys}\:\mathrm{hooke}'\mathrm{s}\:\mathrm{law}\:\mathrm{hangs}\:\mathrm{vertically}\:\mathrm{on}\:\mathrm{a}\:\mathrm{fixed}\:\mathrm{support}. \\ $$$$\mathrm{The}\:\mathrm{wire}\:\mathrm{has}\:\mathrm{an}\:\mathrm{unstretched}\:\mathrm{lenght}\:\mathrm{of}\:\mathrm{15cm}.\:\:\mathrm{The}\:\mathrm{lenght}\:\mathrm{of}\:\mathrm{the}\:\mathrm{wire}\:\mathrm{however} \\ $$$$\mathrm{increase}\:\mathrm{to}\:\mathrm{20cm}\:\mathrm{when}\:\mathrm{a}\:\mathrm{load}\:\mathrm{of}\:\mathrm{0}.\mathrm{5kg}\:\mathrm{is}\:\mathrm{attached}\:\mathrm{to}\:\mathrm{it}\:\mathrm{lower}\:\mathrm{end}\:.\:\mathrm{What}\:\mathrm{is}\:\mathrm{the} \\ $$$$\mathrm{tension}\:\mathrm{in}\:\mathrm{the}\:\mathrm{wire}\:\mathrm{when}\:\mathrm{the}\:\mathrm{load}\:\mathrm{is}\:\mathrm{at}\:\mathrm{rest}\:?.\: \\ $$$$\mathrm{If}\:\mathrm{the}\:\mathrm{load}\:\mathrm{is}\:\mathrm{pulled}\:\mathrm{down}\:\mathrm{until}\:\mathrm{the}\:\mathrm{lenght}\:\mathrm{of}\:\mathrm{the}\:\mathrm{wire}\:\mathrm{is}\:\mathrm{22cm}.\:\mathrm{What}\:\mathrm{is}\:\mathrm{the}\:\mathrm{new} \\ $$$$\mathrm{tension}\:\mathrm{in}\:\mathrm{the}\:\mathrm{wire}\:\left(\mathrm{g}\:=\:\mathrm{9}.\mathrm{8}\:\mathrm{m}/\mathrm{s}\right). \\ $$

Commented by tawa tawa last updated on 23/May/17

please help. Thanks in advance.

$$\mathrm{please}\:\mathrm{help}.\:\mathrm{Thanks}\:\mathrm{in}\:\mathrm{advance}.\: \\ $$

Answered by ajfour last updated on 23/May/17

T=kx=mg T_1 =k(5cm)=(0.5kg)(9.8m/s^2 ) T_1 =k((1/(20))m)=4.9N also⇒ k=98N/m T_2 =k(x+Δx)=(98(N/m))(0.07m) T_2 =6.86N .

$${T}={kx}={mg} \\ $$$${T}_{\mathrm{1}} ={k}\left(\mathrm{5}{cm}\right)=\left(\mathrm{0}.\mathrm{5}{kg}\right)\left(\mathrm{9}.\mathrm{8}{m}/{s}^{\mathrm{2}} \right) \\ $$$${T}_{\mathrm{1}} ={k}\left(\frac{\mathrm{1}}{\mathrm{20}}{m}\right)=\mathrm{4}.\mathrm{9}{N}\:\:{also}\Rightarrow\:{k}=\mathrm{98}{N}/{m} \\ $$$${T}_{\mathrm{2}} ={k}\left({x}+\Delta{x}\right)=\left(\mathrm{98}\frac{{N}}{{m}}\right)\left(\mathrm{0}.\mathrm{07}{m}\right) \\ $$$$\:\:\:{T}_{\mathrm{2}} \:\:=\mathrm{6}.\mathrm{86}{N}\:. \\ $$

Commented by tawa tawa last updated on 23/May/17

God bless you sir.

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$

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