Question

A block of mass $3kg$is suspended from a light spring of spring constant $100N{m}^{-1}$ as shown in figure. Determine the elongation in spring when the block is in equilibrium. (Take $g=10m{s}^{-2}$)

Answer 1

Step 1: Given

1. Mass of the block, $m=3kg$
2. Spring constant, $k=100N{m}^{-1}$
3. Acceleration due to gravity, $g=10m{s}^{-2}$

Step 2: Draw the free body diagram

1. The restoring force of the spring is $kx$ and the weight of the block is $mg$

Step 3: Determine the elongation

1. In equilibrium condition,
2. $$\begin{gathered} kx=mg \\ 100x=3\times 10 \\ x=0.3m \end{gathered}$$

Hence, the elongation in the spring when the block is in equilibrium is $0.3m$