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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
2m1F(m1+m2)k
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B
m21F2(m1+m2)2k
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C
m2Fk(m1+m2)
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D
m22F2(m1+m2)2k
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Solution

The correct option is D $$\displaystyle \frac{m_2 F}{k (m_1 + m_2)}$$

The acceleration of the system is:
$$a=F/(m_1+m_2)$$.
The tension on the spring is:

$$T=m_2 \times a =\dfrac{m_2F}{m_1+m_2}$$.

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, $$x= \dfrac{m_2F}{K(m_1+m_2)}$$

567274_169871_ans_8bee80f4cdd64322a47ff04bf4cb5a46.png

Physics

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