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The two masses in the Atwood's machine shown in the figure are initially at rest at the same height. After they are released, the large mass, m2, falls through a height h and hits the floor, and the small mass, m2, rises through a height h.
Part A: Find the speed of the masses just before m2 lands, giving your answer in terms of m1, m2, g, and h. Assume the ropes and pulley have negligible mass and that friction can be ignored.
Part B: Evaluate your answer to part A for the case h = 1.7 m , m1 = 3.5 kg , and m2 = 4.3 kg.

Answer :

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According to the description given in the photo, the attached figure represents the problem graphically for the Atwood machine.

To solve this problem we must apply the concept related to the conservation of energy theorem.

PART A ) For energy conservation the initial kinetic and potential energy will be the same as the final kinetic and potential energy, so

[tex]E_i = E_f[/tex]

[tex]0 = \frac{1}{2} (m_1+m_2)v_f^2-m_2gh+m_1gh[/tex]

[tex]v_f = \sqrt{2gh(\frac{m_2-m_1}{m_1+m_2})}[/tex]

PART B) Replacing the values given as,

[tex]h= 1.7m\\m_1 = 3.5kg\\m_2 = 4.3kg \\g = 9.8m/s^2 \\[/tex]

[tex]v_f = \sqrt{2gh(\frac{m_2-m_1}{m_1+m_2})}[/tex]

[tex]v_f = \sqrt{2(9.8)(1.7)(\frac{4.3-3.5}{3.5+4.3})}[/tex]

[tex]v_f = 1.8486m/s[/tex]

Therefore the speed of the masses would be 1.8486m/s

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