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Question

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 :


A

m1m2(1 - μk)g/(m1 + m2)

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B

(m2 + μkm1)g/(m1 + m2)

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C

(m2 - μkm1)g/(m1 + m2)

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D

m1m2(1 + μk)g/(m1 + m2)

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Solution

The correct option is D

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,

m2gT=m2a …(1)

for block B.

And

Tμkm1g=m1a …(2)

for block A.

Adding equation (1) and (2) we get,

(m2μkm1)g=(m1+m2)a
a=(m2μkm1)g(m1+m2) …(3)

From equation (2) and (3) we get,

T=μkm1g+m1a

=μkm1g+m1g[m2μkm1(m1+m2)]=m1g[μk(m2μkm1)(m1+m2)]=m1g[μkm1+μkm2+m2μ1m1(m1+m2)]=m1g[m2(1+μk)(m1+m2)]=m1m2(1+μk)g(m1+m2)


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