A block A of mass m1 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 m2 is suspended. The coefficient of kinetic friction between the block and the table is k. When the block A is sliding on the table, the tension in the string is
- m1m2(1+ k)g(m1+m2)
- m1m2(1- k)g(m1+m2)
- (m2+ km1)g(m1+m2)
- (m2- km1)g(m1+m2)
Answer
(1) m1m2(1+ k)g(m1+m2)
Solution
For B (moving down): m2g - T = m2a … (i)
For A (on the table): T - km1g = m1a … (ii)
Adding: a = (m2 - km1)gm1+m2
From (i): T = m2(g - a)
= m2g m1 + m2 - m2 + km1m1+m2
= m1m2(1+ k)g(m1+m2)
Answer: (1)