A) 0.6 m
B) 0.5 m
C) 0. 7 m
D) 0.2 m
Correct Answer: A
Solution :
By law of conservation of energy loss in KE = gain in PE of spring \[\frac{1}{2}m{{v}^{2}}+\frac{1}{2}l{{\omega }^{2}}=\frac{1}{2}k{{x}^{2}}\] \[\frac{1}{2}m{{v}^{2}}+\frac{1}{2}\left( \frac{m{{R}^{2}}}{2} \right){{\left( \frac{v}{R} \right)}^{2}}\,=\,\,\frac{1}{2}k{{x}^{2}}\] \[\left[ l=\frac{m{{R}^{2}}}{2};\,\,\,v=R\omega \right]\] \[\Rightarrow \,\,\,\,\,\,\,\,\frac{1}{2}m{{v}^{2}}+\frac{1}{4}m{{v}^{2}}=\frac{1}{2}k{{x}^{2}}\] \[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{{x}^{2}}=\frac{3}{2}\,\frac{m{{v}^{2}}}{k}\] Substituting \[\operatorname{m}=3\,\,kg,\,\,v=4m/s,\,\,k=200\,\,N/m\] \[{{x}^{2}}=\frac{3}{2}\times \frac{3\times 4\times 4}{200}\] \[{{x}^{2}}=\frac{36}{100}\,\,\,\Rightarrow \,\,\,\,x=0.6\,\,m\]
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