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A simple atwood's machine uses two masses, m1 and m2. starting from rest, the speed of the two masses is 2 m/s at the end of 5 s. at that instant, the kinetic energy of the system is 87 j and each mass has moved a distance of 5 m. determine the values of m1 and m2.

Respuesta :

In Atwood machine system we will have

[tex]a = \frac{m_1 - m_2}{m_1+m_2}*g[/tex]

[tex]0.4 = \frac{m_1 - m_2}{m_1 + m_2}*10[/tex]

[tex]0.4m_1 + 0.4m_2 = 10 m_1 - 10m_2[/tex]

[tex]10.4m_2 = 9.6m_1[/tex]

[tex]1.08m_2 = m_1[/tex]

Also we know the kinetic energy

[tex]\frac{1}{2} m_1v_1^2 +\frac{1}{2}m_2v_2^2 = 87[/tex]

[tex]m_1 + m_2 = 43.5 J[/tex]

now solving the two equations we will have

[tex]m_2 = 21 kg[/tex]

[tex]m_1 = 22.6 kg[/tex]

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