A) \[\frac{2{{\mu }^{2}}m{{g}^{2}}}{{{v}^{2}}}\]
B) \[\frac{4{{\mu }^{2}}m{{g}^{2}}}{{{v}^{2}}}\]
C) \[\frac{3{{\mu }^{2}}m{{g}^{2}}}{{{v}^{2}}}\]
D) \[\frac{16{{\mu }^{2}}m{{g}^{2}}}{{{v}^{2}}}\]
Correct Answer: B
Solution :
\[\frac{1}{2}m{{v}^{2}}=(\mu mg)x\]and\[\frac{1}{2}k{{x}^{2}}=\frac{1}{2}m{{v}^{2}}\] |
On solving, we get \[k=\frac{4m{{\mu }^{2}}{{g}^{2}}}{{{v}^{2}}}\] |
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