A simple Atwood\'s machine uses two masses, m1 and m2. Starting from rest, the s

Question

A simple Atwood's machine uses two masses, m1 and m2. Starting from rest, the speed of the two masses is 1.2 m/s at the end of 9.8 s. At that time, the kinetic energy of the system is 82 J and each mass has moved a distance of 5.88 m. part a find the value of heavier mass. The acceleration due to gravity is 9.81 m/s^2 Answer in kg

Part B> Find the value of hte lighter mass answer in kg.

From rest, the speed of the two masses is 1.2 m/s at the end of 9.8 s. At that time, the kinetic energy of the system is 82 J and each mass has moved a distance of 5.88 m. Find the value of heavier mass. The acceleration due to gravity is 9.81 m/s2. Answer in units of kg

Explanation / Answer

from the statement of KE, we have that:

82J = 1/2m1v^2+1/2m2v^2 = 1/2 (1.2)^2(m1+m2) or that

82=0.72(m1+m2) or m1+m2=113.89

we can write Newton's second law for each mass, where T is the tension in the rope:

assuming m1 moves up, we have:

T-m1g=m1a
T-m2g=-m2a

subtracting equations gives us:

(m2-m1)g=(m1+m2)a or a=(m2-m1)/(m2+m1) g

we can find the acceleration of the system also knowing that

the speed has changed by 1.2 m/s in 9.8s for an accel of 0.1224 therefore:

0.1224 = [(m2-m1)/(m1+m2)] g, or after some algebra

m1= 0.975*m2

since we know from above that m1+m2 = 113.89, we have mow

therefore m2= 57.67 kg and m1= 56.22 kg

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