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Question

A block of 2 kg is suspended from a ceiling by a massless spring of spring constant k = 100 N/m. What is the elongation of the spring? If another 1 kg is added to the block, what would be the further elongation?

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Solution

Given,
mass of the first block, m = 2 kg
k = 100 N/m
Let elongation in the spring be x.

From the free-body diagram,
kx = mg
x=mgk=2×9.8100 =19.6100=0.196≈0.2 m

Suppose, further elongation, when the 1 kg block is added, is ∆x.
Then,kx+∆x=m'g
⇒ k∆x= 3g − 2g = g
⇒∆x=gk=9.8100=0.098≈0.1 m

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