Calculate the speed of the Cylinder in the pulley system

[Northbysouth]

Homework Statement

Each of the two systems is released from rest. Calculate the speed v of each 47-lb cylinder after the 38-lb cylinder has dropped 5.2 ft. The 17-lb cylinder of case (a) is replaced by a 17-lb force in case (b).

I have attached an image of the question.

Homework Equations

The Attempt at a Solution

I think I should use U₁ + K₁ = U₂ + K₂

But I'm not quite sure how I should interpret the situation with this formula. I had thought:

47lbf*5.2ft - 55lbf*5.2ft = 0.5*m*v²

But I'm not sure if this is the right way to look at it.

Any help is appreciated

 

[Northbysouth]

I've managed to figure out part A

m₁gh = m₂gh + .5m₁v²+.5m₂v²

(47lb)(5.2ft) = (55lb)(5.2ft) + .5(47lb/32.2)v² + .5(55/32.2)v²

solving for v gives me:

v = 5.1187 ft/sec

I had thought that for part B I would be able to use the same equation but use a reduced mass, 38 instead of 55 but it doesn't work.

Input on this would be greatly appreciated.

 

[haruspex]

(47lb)(5.2ft) = (55lb)(5.2ft) + .5(47lb/32.2)v² + .5(55/32.2)v²

Something crossed over there - that would give a negative value for the gain in KE. I guess that was just an error in the post.

I had thought that for part B I would be able to use the same equation but use a reduced mass, 38 instead of 55 but it doesn't work.

That should work. Pls post the details.