Question

Two blocks are connected by a massless string through an ideal pulley as shown. A force of 22N is applied on block B when initially the blocks are at rest. Then speed of centre of mass of block A and block B, 2 sec, after the application of force is (masses of A and B are 4 kg and 6 kg respectively and surfaces are smooth) –

Answer 1

Answer:

the speed of centre.........4,6kg.....

Answer 2

Given:

Force exerted on the block B when the blocks are at rest = 22N

Mass of A = 4kg

Mass of B = 6kg

To find:

The speed of centre of mass of blocks A and B after 2 seconds from the application of force.

Solution:

Let the tension in the string be T, then;

F - 2T = 6a

T = 4*2a = 8a

Putting the value of T, we get:

F - 16a = 6a

a = F/22

Putting the value of F, we get:

a = 1 m/s^2

Then the acceleration of the centre of mass will be:

= (6*acceleration of B) + (4*acceleration of A)/ (6+4)

=(6*1) + (4*2)/10

=1.4 m/s²

Speed of centre of mass = acceleration*time =1.4*2 = 02.8 m/s

Therefore, the speed of centre of mass will be 2.8 m/s