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Question

The block of mass $$m_1$$ in shown in figure is fastened to the spring and the block of mass $$m_2$$ is placed against it. (a) Find the compression of the spring in the equilibrium position. (b) The blocks are pushed a further distance $$(2/k)(m_1+m_2)g\sin\theta$$ against the spring and released. Find the position where the two blocks separate. (c) What is the common speed of blocks at the time of separation ?

Solution
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(a) Due to the gravitational force on the blocks that's will equal the spring force

So,
$$kx = (m_1+m_2)g sin \theta$$

So,
$$x = \dfrac{(m_1+m_2)gsin \theta}{k}$$

Now,
(b) The blocks are further pushed $$\dfrac{2}{k}(m_1+m_2)g sin \theta$$
So, Total compression = $$\dfrac{2}{k}(m_1+m_2)g sin \theta$$
That means that The amplitude of the SHM followed will be $$\dfrac{2}{k}(m_1+m_2)g sin \theta$$

And, after the total compression in the spring will become zero that is the spring comes to it's natural length , then the blocks will leave each other's contact.

(c) By work energy theorem,

Change in Kinetic Energy = Work done by all forces,

So,
$$\dfrac{1}{2} \times (m_1+m_2) \times v^2 = \dfrac{1}{2}kx^2 - (m_1+m_2)gsin\ theta \times x$$
Therefore,
$$\dfrac{1}{2}(m_1+m_2)v^2 = \dfrac{1}{2}k{\dfrac{3}{k}(m_1+m_2)g sin \theta}^2 - (m_1+m_2)g sin \theta \times \dfrac{3}{k} (m_1+m_2) g sin \theta$$
So,
$$\Rightarrow$$ $$v = g sin \theta \sqrt{\dfrac{3(m_1+m_2)}{k}}$$



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