A body of 200 g was hung from the lower end of a spring which obeys Hooke’s law. Given that the spring extended by...

A body of 200 g was hung from the lower end of a spring which obeys Hooke’s law. Given that the spring extended by 100 mm, what is the spring constant for this spring?

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F = k e. F = 200 × 10^-3 kg × 10 N /kg = 2 N. Extension = 100 × 10^-3
m = 0.1 m.
Spring constant (k) = 2 / 0.1 = 20 N/m.

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Mutiso answered the question on October 12, 2018 at 10:54

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