Two blocks of masses 6 kg and 4 kg connected by a rope of mass 2 kg are resting on frictionless floor as shown in fig. If a constant force of 60 N is applied to 6 kg block, tension in the rope at A, B, and C will respectively be -

Explanation

As the mass of the system is `(6+4+2)=12kg` and applied force is 60 N the acceleration of the system

`a=F/m=60/12=5m//s^(2)`
Now at point A as tension is pulling the rope and block B of mass 4 kg,
`T_(A)=(2+4)xx5=30N`
Similarly for B and C,
`T_(B)=(1+4)xx5=25N and T_(C)=(0+4)xx5=20N`
Note: In this problem as rope is not massless, tension is different at different points of the string, being maximum at the end closest to the applied force and minimum at the end farthest from the force.