Three blocks of masses m₁, m₂ are connected by massless strings as shown on a frictionless table. They are pulled with a force T₃=40 ~N . If m₁=10 ~kg , m₂=6 ~kg and m₃=4 ~kg , the tension T₂ will be

Step-by-step solution

Given : (T₃=40 ~N , ~m ₁=10 ~kg , ~m ₂=6 ~kg ) and ( m ₃=4 ~kg ) Acceleration of complete system Total mass of system, ( m = m ₁+ m ₂+ m ₃=10+6+4=20 ~kg ) Acceleration of system =a Applying Newton's law, ( F = ma ) ( a = T ₃ ~m = 40 ~N 20 Kg =2 ~m / s ^2 ) Calculation of ( T ₂ ) From F.B.D OF ( (m₁+m₂ ) ) Force (T₂ ) is pulling 1 and 2 blocks. If we consider (m₁ ) and (m₂ ) as a system then (T₁ ) will get cancel as (T₁ ) acts as internal force. (As shown in figure) Therefore from Newton's second law- ( aligned F &