A block A of mass m1 rests on a horizontal table. A lights 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 . When the block A is sliding on the table, the tension in the string is :
m1m2(1 + μk)g/(m1 + m2)
See figure alongside
Let T be the tension in the string.
Let a be the acceleration of the combination. We have,
m2g−T=m2a …(1)
for block B.
And
T−μkm1g=m1a …(2)
for block A.
Adding equation (1) and (2) we get,
(m2−μkm1)g=(m1+m2)a
∴a=(m2−μkm1)g(m1+m2) …(3)
From equation (2) and (3) we get,
T=μkm1g+m1a
=μkm1g+m1g[m2−μkm1(m1+m2)]=m1g[μk(m2−μkm1)(m1+m2)]=m1g[μkm1+μkm2+m2−μ1m1(m1+m2)]=m1g[m2(1+μk)(m1+m2)]=m1m2(1+μk)g(m1+m2)