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

A block A of mass m_(1) rests on a horizontal table. A light string co

1.	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» `((m_(2)+mu_(k)m_(1))g)/((m_(1)+m_(2)))` `((m_(2)-mu_(k)m_(1))g)/((m_(1)+m_(2)))` `(m_(1)m_(2)(1+mu_(k))g)/((m_(1)+m_(2)))` `(m_(1)m_(2)(1-mu_(k))g)/((m_(1)+m_(2)))` Solution :Frictional force acting on block A = `mu_(k)m_(1)g` Hence resultant force on the blocks = `m_(2)g-mu_(k)m_(1)`g Their ACCELERATION, a = `(m_(2)g-mu_(k)m_(1)g)/(m_(1)+m_(2))` If the TENSION in the string is T, then for block B, `m_(2)g-T=m_(2)a` `:. T=m_(2)(g-a) = m_(2)g(1-(m_(2)-mu_(k)m_(1))/(m_(1)+m_(2)))` =`m_(2)g (m_(1)+mu_(k)m_(1))/(m_(1)+m_(2))` `=(m_(1)m_(2)(1+mu_(k))g)/(m_(1)+m_(2))`