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  4. A 2 kg block is dropped from a height of 0.4m on a spring of force constant k=1960 text Nm-1 . The maximum compression of the spring is

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Q. A 2 kg block is dropped from a height of 0.4m on a spring of force constant $ k=1960\text{ }N{{m}^{-1}} $ . The maximum compression of the spring is

JamiaJamia 2009

Solution:

By law of conservation of energy, we have loss in gravitational potential energy = gain in elastic potential energy $ \Rightarrow $ $ mg(h+x)=\frac{1}{2}k{{x}^{2}} $ $ 2\times 10(0.4+x)=\frac{1}{2}1960\times {{x}^{2}} $ $ \Rightarrow $ $ 8+20x=980{{x}^{2}} $ $ \Rightarrow $ $ 980{{x}^{2}}-20x-8=0 $ Solving, we get $ x=0.1\text{ }m $