A block A of mass m, 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, is
suspended. The coefficient of kinetic friction
between the block and table is Hy When the
block A is sliding on the table, the tension in the
string is
[AIPMT-2015)
mm, (1-x)g
(m, + m)g
(m + m)
(m, + m)
(m, -,m)g
mm (1+x)g
(4)
(m, + m)
(m, + m₂)
(1)
Answers
Answered by
0
Answer:
The blocks m
1
and m
2
will move with combined acceleration a:
From F.B.D. of block m
1
T−f
1
=m
1
a...(i)
as the block m
1
is sliding, kinetic friction will be acting:
T−μ
k
N=m
1
a...(ii)
N=mg...(iii)
From F.B.D. of block m
2
m
2
g−T=m
2
a...(iv)
adding (ii) and (iv)
a=
m
1
+m
2
m
2
g−μ
k
N
...(v)
T=m
2
g−m
2
a=m
2
(g−
m
1
+m
2
m
2
g−μ
k
N
)
=m
2
(
m
1
+m
2
m
1
g+μ
k
m
1
g
)
=m
1
m
2
(
m
1
+m
2
1+μ
k
g)
hence correct answer is option A
solution
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