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A block A of mass m_1 rests on a horizontal table. A light string connected to it passes over a frictionless pully at the edge of the table and from its other end another block B of mass m2 is suspended. The coefficient of kinetic friction between the block and the table is H. When the block A is sliding on the table, the tension in the string is

Answer» <html><body><p>`(m_1 m_2(1+mu _k ) g)/( (m_1 +m_2))`<br/>`(m_1m_2 (1- mu_k)g)/((m_1) +m_2)`<br/>`((m_2+mu_k m_1)g)/((m_1 +m_2))`<br/>`((m_2 -mu_k m_1) g)/((m_1 +m_2))`</p>Solution :Givensituation is shownin thefigure<br/> here`,N= m_1 g ,f = mu_k N = mu_k m_1g `…(i) <br/>Leta be theacceleationon <a href="https://interviewquestions.tuteehub.com/tag/blocks-21562" style="font-weight:bold;" target="_blank" title="Click to know more about BLOCKS">BLOCKS</a> . <br/>equationof <a href="https://interviewquestions.tuteehub.com/tag/motionfor-2845459" style="font-weight:bold;" target="_blank" title="Click to know more about MOTIONFOR">MOTIONFOR</a> <a href="https://interviewquestions.tuteehub.com/tag/aand-844912" style="font-weight:bold;" target="_blank" title="Click to know more about AAND">AAND</a> B <br/><br/> ` T-f = m_1 a … (ii) ` <br/>`m_2 g - T =m_2 a…(<a href="https://interviewquestions.tuteehub.com/tag/iii-497983" style="font-weight:bold;" target="_blank" title="Click to know more about III">III</a>)` <br/>Addingequation(ii) and (iii) <br/>we get<br/> ` m_2 g- f = (m_1 +m_2) a` <br/> ` a =(m_2g-f)/( m_1 +m_2)` <br/>substitingvalueof ainequation(iii) <br/>` T = m_2 g -m_2((m_2 g -f))/(m_1 +m_2)` <br/> `=(m_1 m_2g + m_1 m_2mu_s g)/(m_1 +m_2) = (m_1 m_2 (1 + mu _s ) g)/( m_1 +m_2)` ( using(i)) <br/> <img src="https://doubtnut-static.s.llnwi.net/static/physics_images/MTG_NEET_GID_PHY_XI_C03_E03_015_S01.png" width="80%"/></body></html>


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