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# Examination of work done by a pulley system

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An Atwood machine consists of masses of 1.9 Kg and 1.995 Kg hanging from opposite sides of a pulley.

As the system accelerates 3.3 meters from rest, how much work is done by gravity on the system?
Assuming no friction or other dissipative forces, use the definition of KE to determine the velocity of the system after having moved through the 3.3 meters, assuming that the system was released from rest.

https://brainmass.com/physics/system-work/examining-pulley-sytem-395828

## SOLUTION This solution is FREE courtesy of BrainMass!

Problem: As the system accelerates 3.3 meters from rest, how much work is done by gravity on the system?
Assuming no friction or other dissipative forces, use the definition of KE to determine the velocity of the system after having moved through the 3.3 meters, assuming that the system was released from rest.

Solution:
Given, two masses of m1 = 1.9 kg and m2 = 1.995 kg

- Net force (F) is

F = m2*g - m1*g = g(m2 - m1)

=> F = 9.81(1.995 - 1.900) = 9.81 x 0.095 N

=> F = 0.93195 N

- Work done (W) is given by

W = F*d {where F is net force and d is the distance traveled}

=> W = 0.93195 x 3.3 = 3.075 J

- Now KE of the system is given by

KE = ½*M*v^2

Where v is the velocity of the system and the total mass is given as M.

M = m1 + m2 = 1.9 + 1.995 = 3.895 kg

Thus KE of the system is

KE = ½*3.895*v^2 = 1.9475*v^2

Now after 3.3m we can equate the KE of the system to the Work done.

1.9475*v^2 = 3.075

=> v^2 = 3.075/1.9475 = 1.579 (m/s)^2

Thus velocity achieved is

v = Sqrt {1.579} = 1.257 m/s

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This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!