A) 1.0 cm
B) 2.0 cm
C) 10 cm
D) 2.0 cm
Correct Answer: C
Solution :
[c] Let \[{{\operatorname{x}}_{A}} and {{x}_{B}}\]be the position of ends A and B at time (from the block, then stretched length of the spring will be \[{{l}_{2}}={{x}_{A}}-{{x}_{B}}\] and so the stretch \[\Delta \ell ={{\ell }_{2}}-{{\ell }_{1}}=({{x}_{A}}-{{x}_{B}})-{{\ell }_{1}}({{\ell }_{1}}\]natural length of the spring) So, \[U=\frac{1}{2}k\Delta {{\ell }^{2}}=\frac{1}{2}k{{[({{x}_{A}}-{{x}_{B}})-{{\ell }_{1}}]}^{2}}\] \[P=\frac{dU}{dt}=\frac{1}{2}k.2({{x}_{A}}-{{x}_{B}}-{{\ell }_{1}})\left( \frac{d{{x}_{A}}}{dt}-\frac{d{{x}_{B}}}{dt} \right)\]\[P=F\left( {{v}_{A}}-{{v}_{B}} \right)F=\frac{d{{x}_{A}}}{{{v}_{A}}-{{v}_{B}}}\] \[\Delta \ell =\frac{F}{k}=\frac{P}{({{v}_{A}}-{{v}_{B}})}=\frac{20}{4-2\times 100}\] \[\Delta \ell =0.1m=10\,cm\]
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