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?

Answer 1

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
$$\begin{gathered} x=\frac{mg}{k}=\frac{2\times 9.8}{100} \\ =\frac{19.6}{100}=0.196\approx 0.2\mathrm{m} \end{gathered}$$

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
$\Rightarrow ∆x=\frac{g}{k}=\frac{9.8}{100}=0.098\approx 0.1\mathrm{m}$