Question Number 180157 by Tawa11 last updated on 08/Nov/22
As the force on a string increases from 100N to 180N, the string extends by 10cm. The work done in increasing the tension in the string is?
Answered by DvMc last updated on 08/Nov/22
$${W}={F}×{d} \\ $$$${W}=\left(\mathrm{180}{N}−\mathrm{100}{N}\right)×\mathrm{0}.\mathrm{1}{m}=\mathrm{8}{Nm} \\ $$
Answered by mr W last updated on 08/Nov/22
![W=∫Fds=((100+180)/2)×0.1=14 J or k=((180−100)/(0.1))=800 N/m W=E_2 −E_1 =((k(s_2 ^2 −s_1 ^2 ))/2) =(k/2)[((F_2 /k))^2 −((F_1 /k))^2 ]=((F_2 ^2 −F_1 ^2 )/(2k)) =((180^2 −100^2 )/(2×800))=14 J](https://www.tinkutara.com/question/Q180167.png)
$${W}=\int{Fds}=\frac{\mathrm{100}+\mathrm{180}}{\mathrm{2}}×\mathrm{0}.\mathrm{1}=\mathrm{14}\:{J} \\ $$$${or} \\ $$$${k}=\frac{\mathrm{180}−\mathrm{100}}{\mathrm{0}.\mathrm{1}}=\mathrm{800}\:{N}/{m} \\ $$$${W}={E}_{\mathrm{2}} −{E}_{\mathrm{1}} =\frac{{k}\left({s}_{\mathrm{2}} ^{\mathrm{2}} −{s}_{\mathrm{1}} ^{\mathrm{2}} \right)}{\mathrm{2}} \\ $$$$\:\:\:\:=\frac{{k}}{\mathrm{2}}\left[\left(\frac{{F}_{\mathrm{2}} }{{k}}\right)^{\mathrm{2}} −\left(\frac{{F}_{\mathrm{1}} }{{k}}\right)^{\mathrm{2}} \right]=\frac{{F}_{\mathrm{2}} ^{\mathrm{2}} −{F}_{\mathrm{1}} ^{\mathrm{2}} }{\mathrm{2}{k}} \\ $$$$\:\:\:\:=\frac{\mathrm{180}^{\mathrm{2}} −\mathrm{100}^{\mathrm{2}} }{\mathrm{2}×\mathrm{800}}=\mathrm{14}\:{J} \\ $$
Commented by Tawa11 last updated on 08/Nov/22
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir},\:\mathrm{I}\:\mathrm{appreciate}. \\ $$