Question

in the arrangement shown, m1 = 1 kg, m2 = 2 kg, the pulleys and strings are ideal and all surfaces in contact are frictionless. the value of m for which the mass m1 moves with constant velocity isfig

Answer 1

The total force on m1 is m1g−m2g=−g, here '-' indicates force is in upward direction. So the acceleration of m1 is:

 −g/m1+m2=−g/3

Now acceleration of the total system is:

(m1+m2)g/M+m1+m2=3g/M+3

Now for constant velocity total acceleration should be zero so we have:

3g/M+3+(−g/3)=0

⇒9=M+3

⇒M=6

Answer 2

Answer: M = 6  Kg

Given:

= 1 kg,  = 2 kg  

Solution:

From the figure, total force acting on m1 is given by,

g – g = -g

negative sign indicates force is on upward direction

acceleration of  is given by,

-g/+ = -g/3

Acceleration of total system is given by,

(+)g/(M++) = 3g/M+3

In order to achieve the constant velocity, the acceleration should be zero

                 3g/M+3 + (-g/3) = 0

                 3g/M+3 = g/3

                 9 = M+3

                 M = 6

Hence the total mass should be 6kg to achieve constant velocity.