The pulley arrangements of figures (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=2mg . Calculate the accelerations in the two cases.

Explanation

In figure (a), for the motion of mass m , T-mg=ma

For motion of mass 2m figure (b)
2 mg -T (2m)a …(2)
Adding equation (1) and (2) , we get
`2mg-mg= 2ma+ma, mg=3ma, a = g//3`
(b)In figure `T^(1)-mg=ma^(1)`
But `T^(1)=2mg , :. 2mg-mg =a^(1), ` Therefore `a^(1)=g`