A) h/2
B) h
C) \[\sqrt{2}h\]
D) 2h
Correct Answer: A
Solution :
\[W=k\delta \] \[U=\frac{1}{2}k\delta \times \delta =\frac{1}{2}=k{{\delta }^{2}}=mgh\] \[m=\frac{k{{\delta }^{2}}}{gh}\] For two springs in series, \[{{k}_{e}}=\frac{k}{2}\]\[U=\frac{1}{2}\times \frac{k}{2}\times {{\delta }^{2}}=\frac{k{{\delta }^{2}}}{4}=mg{{h}_{1}}\] \[{{h}_{1}}=\frac{k{{\delta }^{2}}}{4mg}=\frac{k{{\delta }^{2}}}{4\times \frac{k{{\delta }^{2}}}{2h}}=\frac{h}{2}\]
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