How Do You Calculate the Unknown Mass in an Atwood's Machine?

[Quadratic]

I dislike simply having answers told to me, so I won't give the actual number values of the question. However, I need some insight on how I would solve this:

There are two masses on an atwood's machine, but only one of the masses is known (let's say 5 kilograms). I am told that the unknown mass is greater than the known mass, and that the system is accelerating at 2 m/s^2. I am also told that friction is negligible, and to assume that the string is weightless. How would I solve for the unknown mass?

edit:
I think it came to me:
2(5) + 2(x) = 9.81(x) -9.81(5)
2(5) + 9.81(5) = 9.81(x) - 2(x)
10 + 49.05 = 7.81(x)
59.05/7.81 = (x) = 7.56kg

Is that right?

 

[SN1987a]

Yup! You've got it.

 

[quicknote]

I would approach this problem by first analyzing the information given and identifying the known and unknown variables. In this case, the known variables are the mass of one object (5 kg) and the acceleration (2 m/s^2). The unknown variable is the mass of the other object.

Next, I would use the principles of Newton's Second Law of Motion, which states that the net force on an object is equal to its mass multiplied by its acceleration. In this case, the net force is due to the difference in the weights of the two masses.

Using this equation, we can set up the following equation:

(5 kg)(9.81 m/s^2) - (x kg)(9.81 m/s^2) = (x kg + 5 kg)(2 m/s^2)

Simplifying this equation, we get:

49.05 - 9.81x = 2x + 10

We can then solve for x by isolating it on one side of the equation:

49.05 - 10 = 2x + 9.81x

39.05 = 11.81x

x = 3.31 kg

Therefore, the unknown mass in this Atwood's Machine is approximately 3.31 kg.

Your solution of 7.56 kg is incorrect as it does not take into account the acceleration of the system. Also, it is important to note that this solution assumes ideal conditions, such as negligible friction and a weightless string. In real-world situations, these factors may affect the actual value of the unknown mass.