Physics, 31.08.2020 01:01 kholman

Find the tension between the 4 kg and the 6 kg blocks once the system is released from rest and the 9 kg block accelerates downward.

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Answer from: asandors9617

Tension on the string is 48 N

Explanation:

Mass of A = 6 kg

mass of B = 4 kg

If the 9 kg accelerates downwards then the equation for the 6 kg mass side of the pulley will be

6g - T = 6a 1

where

g is acceleration due to gravity = 10 m/s^2

a is the acceleration of the body

T is the tension in the string.

For the 4 kg mass side of the string, we have

T - 4g = 4a 2

transposing, we have

T = 4a + 4g 3

substitute equation 3 in equation 1, we have

6g - (4a + 4g) = 6a

6g - 4a - 4g = 6a

6g - 4g = 6a + 4a

2g = 10a

but g = 10 m/s^

2 x 10 = 10a

20 = 10a

acceleration a = 20/10 = 2 m/s^2

substitute value of a in equation 1

6g - T = 6a

6(10) - T = 6(2)

60 -T = 12

60 - 12 = T

T = 48 N

Answer from: Quest

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Answer from: Quest

Distance^2= yellow line ^2 (see picture) = 30^2 + 49^2so distance = squareroot of (300^2 +49^2)i dont know if this is how to solve it, but i cnt think of another way

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Find the tension between the 4 kg and the 6 kg blocks once the system is released from rest and the...

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