Answer:
(i) \[m=2kg,\,\,K=100N/m,\,\,x=?\]
we know, \[F=-kx\Rightarrow mg=-kx\]
\[\Rightarrow x=\frac{mg}{k}=\frac{2\times 10}{100}=\frac{20}{100}=0.2m\]
(ii) Let further elongation be 'x' then
\[F=-k(x+x')\Rightarrow mg=-k(x+x')\]
\[\Rightarrow 3\times 10=100(0.2+x')\]
\[\Rightarrow 30=100(0.5+x')\]
\[\frac{3}{10}-0.2=x'\Rightarrow x'=\frac{1}{10}=0.1m\]
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