(Atwood’s Machine): Two masses, 9 kg and 12 kg, are attached by a lightweight cord and suspended over a frictionless pulley. When released, find the acceleration of the system and the tension in the cord.

The acceleration of the system is 1.401 m/s² and

The tension in the cord is 100.902 N

Explanation:

Let the 9 kg mass be m

Let the 12 kg mass be M

By Newton's second law of motion we have

For the 9 kg mass, T - mg = ma and for the 12 kg mass we have T - Mg  = -Ma

Here we took the upward acceleration as positive a of the 9 kg mass and the downward acceleration of the 12 kg mass as -a

Solving for T for the 9 kg mass we have

T = mg + ma

Substituting  the value of T in to the 12 kg mass equation, we have

mg + ma - Mg = -Ma or  a = therefore the acceleration is

1.401 m/s²

and the tension is T = mg + ma = 9×(9.81+1.401) = 100.902 N

get the total number of pieces of fruit and divide it by 4 to get the average. the result is the answer to the problem. 14 total number of fruits and there are 4 members. 14 divided by 4 is equals to 3.5.