A 26.3 kg block (m1) is on a horizontal surface, connected to a 5.70 kg block (m2) by a massless string as shown in the figure below. The frictionless pulley has a radius R = 0.057 m and a moment of inertia I=0.090 kgm2. A force F = 192.9 N acts on m1 at an angle theta = 30.3°. There is no friction between m1 and the surface. What is the upward acceleration of m2?

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A 26.3 kg block (m1) is on a horizontal surface, connected to a 5.70 kg block (m2) by a massless string as shown in the figure below. The frictionless pulley has a radius R = 0.057 m and a moment of inertia I=0.090 kgm2. A force F = 192.9 N acts on m1 at an angle theta = 30.3°. There is no friction between m1 and the surface. What is the upward acceleration of m2?

Physics
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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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look buddy i dont know the absolute solution for i dont recall the equation but i'll tell you what to do .. first you should analyse F to calculate the force component that acts on the system.. in this case this component equals \[F * \cos \theta \] then you can calculate the m1's acceleration by newtons 2nd low \[F = m_{1} a\] now comes my erased part of my memory ! you can write down the equation of inertia and put in your variables to find the up going acceleration i know i didn't made lot of help .. so please forgive me .. i have an important exam tomorrow in Plasma Physics and i am so distracted .. Good Luck
Thanks Ahmed! Let me try this way and see. Also, good luck on your exam :)

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