Two masses m_(1) = 4.8 kg and m_(2) = 5 kg tied to a string are hanging over a light frictionless pully. What is the acceleration of masses when they are free to move ?

Two masses m_(1) = 4.8 kg and m_(2) = 5 kg tied to a string are hangin

1.	Two masses m_(1) = 4.8 kg and m_(2) = 5 kg tied to a string are hanging over a light frictionless pully. What is the acceleration of masses when they are free to move ?
Answer» <html><body><p>0.<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a> `m//s^(2)`<br/>9.8 `m//s^(2)`<br/><a href="https://interviewquestions.tuteehub.com/tag/5-319454" style="font-weight:bold;" target="_blank" title="Click to know more about 5">5</a> `m//s^(2)`<br/>4.8 `m//s^(2)`</p><a href="https://interviewquestions.tuteehub.com/tag/solution-25781" style="font-weight:bold;" target="_blank" title="Click to know more about SOLUTION">SOLUTION</a> :Supposethe accelerationof compositesystemis a<br/> `m_(2) g - T =m_(2)a ` and <br/> Addingthesetwoequations <br/> `(m_(2)-m_(1)) g =(m_(1) =m_(2)) a` <br/> `a= ((m_(2) - m_(1))/(m_(1) + m_(2))) g ` <br/> <img src="https://doubtnut-static.s.llnwi.net/static/physics_images/KPK_AIO_PHY_XI_P1_C05_E05_061_S01.png" width="80%"/> <br/> `=((5-4.8)/( 5+4.8)) 9.8` <br/> `=(0.2)/(9.8) xx 9.8` <br/> `a= 0.2 m//s^(2)`</body></html>