Question-142361

Question Number 142361 by BHOOPENDRA last updated on 30/May/21

Answered by mr W last updated on 30/May/21

δ=0.2m l=4.8+0.2=5m a=1m mglsin θ−μmgcos θl=(1/2)kδ^2 =mgasin θ+μmgcos θa l(sin θ−μcos θ)=a(sin θ+μcos θ) ⇒μ=(((l−a)tan θ)/(l+a))=((5−1)/(5+1))×(3/4)=0.5 ⇒k=((2mg(sin θ−μ cos θ)l)/δ^2 )=((2×2×10(0.6−0.5×0.8)5)/(0.2^2 ))=1000 N/m

$$\delta=\mathrm{0}.\mathrm{2}{m} \\ $$$${l}=\mathrm{4}.\mathrm{8}+\mathrm{0}.\mathrm{2}=\mathrm{5}{m} \\ $$$${a}=\mathrm{1}{m} \\ $$$${mgl}\mathrm{sin}\:\theta−\mu{mg}\mathrm{cos}\:\theta{l}=\frac{\mathrm{1}}{\mathrm{2}}{k}\delta^{\mathrm{2}} ={mga}\mathrm{sin}\:\theta+\mu{mg}\mathrm{cos}\:\theta{a} \\ $$$${l}\left(\mathrm{sin}\:\theta−\mu\mathrm{cos}\:\theta\right)={a}\left(\mathrm{sin}\:\theta+\mu\mathrm{cos}\:\theta\right) \\ $$$$\Rightarrow\mu=\frac{\left({l}−{a}\right)\mathrm{tan}\:\theta}{{l}+{a}}=\frac{\mathrm{5}−\mathrm{1}}{\mathrm{5}+\mathrm{1}}×\frac{\mathrm{3}}{\mathrm{4}}=\mathrm{0}.\mathrm{5} \\ $$$$\Rightarrow{k}=\frac{\mathrm{2}{mg}\left(\mathrm{sin}\:\theta−\mu\:\mathrm{cos}\:\theta\right){l}}{\delta^{\mathrm{2}} }=\frac{\mathrm{2}×\mathrm{2}×\mathrm{10}\left(\mathrm{0}.\mathrm{6}−\mathrm{0}.\mathrm{5}×\mathrm{0}.\mathrm{8}\right)\mathrm{5}}{\mathrm{0}.\mathrm{2}^{\mathrm{2}} }=\mathrm{1000}\:{N}/{m} \\ $$

Commented by BHOOPENDRA last updated on 30/May/21