Question

two blocks of mass 20 kg and 10kg are connected through light inextensible cord and lying on a smooth horizontal surface the system is pulled by a horizontal force of 600 Newton find the tension in the cord in common acceleration of blocks of the pulling force applied o (a) 20 kg block (b) 10 kg block

Answer 1

Answer:

Solution

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Horizontal force,F=600 N

Mass of body A, m

1

=10 kg

Mass of body B, m2=20 kg

Total mass of the system, m=m1+m2=30 kg

Using Newtons second law of motion, the acceleration (a) produced in the system can be calculated as:

F=ma

a=

m

F

=

30

600

=20m/s

2

When force F is applied on body A:

The equation of motion can be written as: F−T=m

1

a

T=F−(m

1

)a

=600−10×20=400 N (i)

When force F is applied on body B:

The equation of motion can be written as:

F−T=m

2

a

T=F−(m

2

)a

T=600−20×20=200 N (ii)

which is different from value of T in case (i). Hence our answer depends on which mass end, the force is applied.