Question

VICUL ANSWer:
Two blocks of masses 2 kg and 4 kg are connected
by a massless string which is just taut. Now two
forces 3 N and 14 N are applied on blocks. The
tension in the string is
14 N
3 N
14 kg
u= 0.4
u = 0.3​

Answer 1

Answer:

Maximum Frictional Force experienced by the block of mass 4 kg.

⇒ Frictional Force = μN

⇒ N = mg

⇒ Frictional Force = 0.4 × 4 kg × 10 m/s²

⇒ Frictional Force = 16 N

Here we can see that Frictional Force of 16 N is greater than applied force of 14 N.

Therefore the block wont move and hence tension wont change with respect to 4 kg block.

In the case of 2 kg block.

Maximum Frictional Force = μN

⇒ Frictional Force = 0.3 × 2 × 10

⇒ Frictional Force = 6 N

Here, we can see that the frictional force of 6 N is greater than the applied force of 3 N. Therefore the block wont move.

Since the net movement of both the blocks is 0, the tension in the string remains constant ( taut ) which therefore is 0 N.

Hence Option ( D ) is the right answer