Question

In figure (5-E11) m₁, = 5 kg, m₂, = 2 kg and F = 1N . Find the acceleration of either block. Describe the motion of m₁, if the string breaks but F continues to act.

Answer 1

Let the acceleration of blocks be 'a' and tension in the string T. Resultant downward force on block m₁ is  

=m₁g+F-T =5g+1 -T, 

Downward acceleration is a. so we get,   , 5g+1 -T = 5a, → T= 5g+1-5a    

Consider the block m2, Resultant upward forces = T- 2g -1, upward acceleration assumed 'a', we get,   

T- 2g -1=2a         (Put the value of T)  

5g+1-5a -2g -1=2a  → 7a = 3g →a =3g/7 =3 x9.8/7 =3x1.4 =4.2 m/s².  

If the string breaks, the upward pull of string due to tension becomes zero and the resultant downward force on the block m₁ is 

=5g+1  N, Mass = 5 kg,  hence Acceleration

 = Force/Mass   = (5g+1)/5  m/s² =g+0.2  m/s²    (dowmwards)

Answer 2

This is the correct answer