The pulley arrangements in the figure (A) and (B) are identical. The mass of the rope is negligible. In figure (A), the mass m is lifted up by attaching a mass 2m to the other end of the rope. In figure (B), m is lifted up by pulling the other end of the rope with a constant downward force F = 2 mg . The acceleration of m is the same in both cases.

Explanation

In figure (a), for the motion of mass m,
`T - mg = ma " "` …..(1)
For motion of mass 2m figure (b)
`2mg-T=(2m)a " "` …..(2)
Adding equation (1) and (2), we get
`2mg-mg=2ma+ma`
`mg=3ma rArr a=(g)/(3)`
Case II :
In figure (c ) `T^(1) -mg = ma^(1)`
But `T^(1)=2mg`
`therefore 2mg - mg = a^(1)`
`therefore a^(1)=g`