Question

Two blocks of mass 6 kg and 12 kg are connected through an ideal spring of spring constant 200 N/m pulled by two horizontal force as shown. Find maximum elongation in spring. Initially blocks are in rest on smooth floor and spring is in natural length. K = 200 N/m

Answer 1

Maximum elongation = 0.15 m.

Answer 2

Let the spring elongation be $x$ at maximum extension. The effective force trying to stretch the spring is the net force on the system, which is

$$F_{\text{net}} = F_2 - F_1 = 60\,\text{N} - 30\,\text{N} = 30\,\text{N}.$$

In the center-of-mass frame, the relative coordinate $r$ (which is the extension $x$) obeys

$$\mu\, r'' = (F_2 - F_1) - k\,x,$$

with the reduced mass

$$\mu = \frac{m_1 m_2}{m_1 + m_2} = \frac{6 \times 12}{6 + 12} = 4\,\text{kg}.$$

At maximum elongation the relative velocity is zero, hence $r''=0$. Thus,

$$0 = 30 - k\,x.$$

Solving for $x$:

$$x = \frac{30}{200} = 0.15\,\text{m}.$$

Core Explanation:

• Net stretching force $= 60 - 30 = 30\,\text{N}$.
• At maximum extension, effective force $kx$ balances $30\,\text{N}$.
• So, $x = \frac{30}{200} = 0.15\,\text{m}$.