Question

The blocks of masses $m_1$ and $m_2$ are connected by an ideal spring of force constant k. The blocks are placed on a smooth horizontal surface. A horizontal force F acts on the block $m_1$. Initially, spring is relaxed, both the blocks are at rest.

What is the maximum elongation of spring?

• A. $\frac{2m_{1}F}{(m_1+m_2)k}$
• B. $\frac{m_1^{2}F}{2(m_1+m_2{)}^{2}k}$
• C. $\frac{m_{2}F}{k(m_1+m_2)}$
• D. $\frac{m_2^{2}F}{2(m_1+m_2{)}^{2}k}$

Answer 1

Answer: (C)

$\frac{m_{2}F}{k(m_1+m_2)}$

The acceleration of the system is:

$a=F/(m_1+m_2)$.

The tension on the spring is:

$T=m_2\times a=\frac{m_{2}F}{m_1+m_2}$.

This will be equal to the force applied.

Now if the maximum elongation of the spring is $x$ then one can write:

$F=Kx=T$

or, $x=\frac{m_{2}F}{K(m_1+m_2)}$