Question

A block A of 2kg on the incline is connected by a flexible cord to a block B of 9 kg suspended in the air. The cord pass by a pulley of 5 kg having a shape of a disc of 30 cm radius. In addtion the surface of the incline has a coeffcient of friction uk=0.25 and the incline has an angle of 30. The system of the 2 block is maintained at rest with the block B kept at 4 meters above the ground.

The system is released from rest and the bloc B falls on the ground. What will be the velocity of bloc B when it reaches the ground.

Answer 1

We can find out the acceleration of the system

Assuming T1 tension between block B and pulley and T2 tension between block A and pulley

For block B

MBg - T1 = MBa

9g - T1 = 9a

88.2 - T1 = 9a

For block A

T2 - MAgSin30 - f = MAa

f = ukmgCos30 = 0.25×2×9.8×Cos30 = 4.24 N

T2 - 2×9.8×Sin30 - 4.24 = 2a

T2 - 14.04 = 2a

Rotational equation for pulley

(T1 - T2)×R = I×α

α = a/R

(T1 - T2) ×R =( mR2/2)×a/R

T1 - T2 = ma/2 = 5a/2 = 2.5a

Adding all three equations

88.2 - 14.04 = (9a +2a +2.5a)

a = 5.5 m/s2​​​​​​

Now using equation of motion for B

V^2 = U^2 + 2as

V^2 = 0 + 2×5.5×4

V = 6.6 m/s

This is the velocity of block B when reaches ground.

This question can also be solved by using work energy theorem.