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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 pulley at the edge of table and from its other end another block B of mass m_(2) is suspended. The coefficient of kinetic friction between the block and the table ismu_(k). When the block A is sliding on the table the tension in the string is

Answer» <html><body><p>`((m_(<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a>)+mu_(k)m_(1))g)/((m_(1)+m_(2)))`<br/>`((m_(2)-mu_(k)m_(1))g)/((m_(1)+m_(2)))`<br/>`(m_(1)m_(2)(1+mu_(k))g)/((m_(1)+m_(2)))`<br/>`(m_(1)m_(2)(1-mu_(k))g)/((m_(1)+m_(2)))`</p>Solution :Frictional force acting on block A = `mu_(k)m_(1)g` <br/> Hence resultant force on the blocks = `m_(2)g-mu_(k)m_(1)`g<br/> Their <a href="https://interviewquestions.tuteehub.com/tag/acceleration-13745" style="font-weight:bold;" target="_blank" title="Click to know more about ACCELERATION">ACCELERATION</a>, a = `(m_(2)g-mu_(k)m_(1)g)/(m_(1)+m_(2))` <br/> If the <a href="https://interviewquestions.tuteehub.com/tag/tension-1241727" style="font-weight:bold;" target="_blank" title="Click to know more about TENSION">TENSION</a> in the string is T, then for block B, <br/> `m_(2)g-T=m_(2)a` <br/> `:. T=m_(2)(g-a) = m_(2)g(1-(m_(2)-mu_(k)m_(1))/(m_(1)+m_(2)))` <br/> =`m_(2)g (m_(1)+mu_(k)m_(1))/(m_(1)+m_(2))` <br/> `=(m_(1)m_(2)(1+mu_(k))g)/(m_(1)+m_(2))` <br/> <img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/CHY_DMB_PHY_XI_P1_U03_C02_E07_016_S01.png" width="80%"/></body></html>


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