Question #104973

Two blocks are con-

nected by a light string that

passes over two frictionless

pulleys.

The block of mass m 2 is

attached to a spring of force m1 constant k and m1 . m2.

If the system is released from rest, and the spring is initially not stretched or compressed, find an expression for the maximum dis- placement d of m2.

Expert's answer

As per the given question,

the masses of the blocks are m1m_1 and m2m_2 , which is connected with the string and spring system and string is passing over the pulley, let the blocks are going x distance downwards, so the spring will elongate 2x.

Hence applying the conservation of the energy

⇒k(2x)22=m1gx+m2gx\Rightarrow \dfrac{k(2x)^2}{2}=m_1gx+m_2gx

⇒2Kx2=(m1+m2)gx\Rightarrow 2Kx^2=(m_1+m_2)gx

So, xx=(m+m2)g2Kx=\dfrac{(m_+m_2)g}{2K}

Hence the maximum elongation in the spring will be x=(m1+m2)g2Kx=\dfrac{(m_1+m_2)g}{2K}


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