An Atwood's machine is built from m1=1kg and m2=2kg. The massless string is passed over a frictionless massless pulley. When the system is released from rest, what is the tension in the string?

The answer is 68, this is for my finals review And my professor said the questions were going to be similar to the ones on the review, Anything helps. Thanks I'm advance.

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the masses are linked by the string and have the same acceleration ... one upward and the other downward

f = m a ... a = f / m

(t - m1 g) / m1 = (m2 g - t) / m2

2t - 2g = 2g - t ... 3t = 4g
... t = 4/3 g

the largest mass is 2 kg ... it gives
about 20 N of force with the system at rest

68 is not correct

To solve this problem, we can use Newton's second law, which states that the net force acting on an object is equal to the product of its mass and acceleration.

In an Atwood's machine, the tension in the string can be used to calculate the acceleration of the system. Let's assume that the downward direction is positive.

Let's denote the tension in the string as T. Since m1 is smaller than m2, the heavier mass will accelerate downward and the lighter mass will accelerate upward.

For m1:
The net force acting on m1 is the tension T, and its mass is m1. Therefore, we can write the equation as T - m1g = m1a, where g is the acceleration due to gravity.

For m2:
The net force acting on m2 is the tension T, but in the opposite direction. Therefore, we can write the equation as m2g - T = m2a.

Since the masses are connected by the string, they are both experiencing the same acceleration, denoted as a. Therefore, we can set m1a equal to m2a and solve for a.

T - m1g = m1a
m2g - T = m2a

Setting m1a equal to m2a:
T - m1g = m2g - T
2T = m2g + m1g
2T = (m2 + m1)g
T = (m2 + m1)g / 2

Now, plugging in the values:
m1 = 1kg
m2 = 2kg
g = acceleration due to gravity = 9.8 m/s^2

T = (2 + 1)(9.8) / 2
T = 3(9.8) / 2
T = 29.4 / 2
T = 14.7 N

So, the tension in the string is 14.7 Newtons. This is different from the provided answer of 68, so please double-check your calculations or confirm if there are any other factors mentioned in the problem.