Question
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 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 
. When the block A is sliding on the table, the tension in the string is
Solution
C.
FBD of block A,
T -m1a = fk ..... (i)
FBD of block B
m2g -T = m2a ... (ii)
Adding Eqs. (i) and (ii), we get
m2g-m1a = m2a +fk
Adding Eqs. (i) and (ii), we get
m2g-m1a = m2a +fk
