Question

in the arrangement shown in the figure neglect the mass of the pulley and string also friction the acceleration of block A and B are​

Answer 1

Answer:

g m/s^2 respectively.

Explanation:

For the block A the equation will be T - m1g = m1a here m1 is the mass of the body a and its acceleration is a while the tension in all the strings will be same T since all the strings are connected to single pulley.

For the body B the equation will be m2g - T = m2a.

For the pulley P1  we will have  2×T - T = 0 × aP1 .

For the pulley P2 we will have   T2 - 2×T =  0 × aP2  .

 

On solving all the equations simultaneously we will get that a1 = - g and a2= g.

Hence, both the blocks will be moving downwards with acceleration g m/s2.

Answer 2

On solving all the equations simultaneously, we will get that a1 = - g and a2= g. Answer:

g m/s^2 respectively.

Explanation:

For the block A the equation will be T - m1g = m1a here m1 is the mass of the body a and its acceleration is a while the tension in all the strings will be same T since all the strings are connected to single pulley.

For the body B the equation will be m2g - T = m2a.

For the pulley P1  we will have  2×T - T = 0 × aP1 .

For the pulley P2 we will have   T2 - 2×T =  0 × aP2  .

 

Hence, both the blocks will be moving downwards with acceleration g m/s2.