Question

please
refer picture​

Answer 1

Answer:

Correct option is

A

m

1

+m

2

m

1

m

2

(1+μ

k

)g

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)