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.

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

0.93xc=0.93x3x10^8m/s

2.54cm is 1 inch

360seconds i 1 hr