Question

Two blocks A and B are attached with two ends of a massless string which passes over a frictionless pulley vertically. If mass of block A is 6 kg find mass of body B which moves down with acceleration of 4 m/s?. (take g = 10m/s) (b) 14 kg (c) 10 kg (d) 7 kg​

Answer 1

Answer:

Since 6kg is heavier than 4 kg so it will accelerate down while 4 kg will accelerate up

So the equation of net force on 6kg is given as

6×10−T=6a

equation of net force on 4kg is given as

T−4×10=4a

now add the above two equations

6×10−(4×10)=(6+4)×a

60 - 40 = 10a

a = 2m/s

2

So 6 kg will accelerate downwards while 4 kg will accelerate upwards

with acceleration 2 m/s

2

now in order to find the tension force

T−4×10=4×2

T = 48N

So the tension in the string is 48 N

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Answer 2

Explanation:

number of factors of 110 is greater than the number of prime factors of 110 by