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Alonzo19
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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?
 one year ago
 one year ago
Alonzo19 Group Title
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?
 one year ago
 one year ago

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Qasim_Ahmed Group TitleBest ResponseYou've already chosen the best response.1
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
 one year ago

Alonzo19 Group TitleBest ResponseYou've already chosen the best response.0
Thanks Ahmed! Let me try this way and see. Also, good luck on your exam :)
 one year ago

VVALLEY Group TitleBest ResponseYou've already chosen the best response.0
this may help
 one year ago
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