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A block of 200 g mass is dropped from a height of 2 m on to a spring and compresses the spring to a
distance of 50 cm. The force constant of the spring is [Kerala CET 2013]

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$60 {Nm}^{- 1}$

$20 {Nm}^{- 1}$

$40 {Nm}^{- 1}$

$30 {Nm}^{- 1}$

answer is B.

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Detailed Solution

Given, $m = 200 g = 0.2 kg, g = 10 {ms}^{- 2} and$

$H_{eff} = 2 + 0.5 = 2.5 m$

Here, $\frac{1}{2} k x^{2} = m g H_{eff} \Rightarrow k = \frac{2 m g H_{eff}}{x^{2}}$

$\Rightarrow k = \frac{2 \times 0.2 \times 10 \times 2.5}{(0.5)^{2}} \Rightarrow k = \frac{10}{0.25} = 40 {Nm}^{- 1}$

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