CBSE-XI-Physics
05: Newton's Laws of Motion
- #17A 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?Ans : 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=\frac{mg}{k}=\frac{2\times 9.8}{100}``
`` =\frac{19.6}{100}=0.196\approx 0.2\,\mathrm{\,m \,}``
Suppose, further elongation, when the 1 kg block is added, is `` ∆x``.
Then,`` k\left(x+∆x\right)=m\text{ ' }g``
⇒ k`` ∆x``= 3g - 2g = g
`` \Rightarrow ∆x=\frac{g}{k}=\frac{9.8}{100}=0.098\approx 0.1\,\mathrm{\,m \,}``
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