Two blocks A and B of masses 2 kg and 1 kg respectively are interconnected by a pulley string
system as shown in the figure. The system is released from rest and after the 1 kg block had
descended through 1 m, it attains a speed 0.4 m/s. The coefficient of friction between block A and
the table is
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Given:
Mass of block A = 2 kg and
Mass of block B = 1 kg
They are interconnected by a pulley string
The system is released from rest and after the 1 kg block had descended through 1 m, it attains a speed 0.4 m/s.
To Find:
The coefficient of friction between block A and the table.
Solution:
Lets write the force equation.
Block B :
- Tension upwards , Weight downwards , accelerating downwards.
- mg - T = ma, m =1Kg
- g - T = a , --(1)
Block A :
- Tension towards left, Friction towards right, accelerating towards left.
- T - uMg = Ma
- T - 2ug = 2a --(2)
Given that when the block B descends through 1m, it attains a speed of 0.4m/s.
- Equation of motion gives us ,
- V² - U² = 2as
- 0.4² = 2a
- a = 0.16/2 = 0.08
Substituting T in (2) and value of a ,
- g - a -2ug = 2a
- 10 - 20u = 3 x 0.08
- 20u = 10 - 0.24
- u = 9.76/20
- u = 0.488
The coefficient of friction between block A and the table is 0.488.
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