Insights Blog
-- Browse All Articles --
Physics Articles
Physics Tutorials
Physics Guides
Physics FAQ
Math Articles
Math Tutorials
Math Guides
Math FAQ
Education Articles
Education Guides
Bio/Chem Articles
Technology Guides
Computer Science Tutorials
Forums
Intro Physics Homework Help
Advanced Physics Homework Help
Precalculus Homework Help
Calculus Homework Help
Bio/Chem Homework Help
Engineering Homework Help
Trending
Featured Threads
Log in
Register
What's new
Search
Search
Search titles only
By:
Intro Physics Homework Help
Advanced Physics Homework Help
Precalculus Homework Help
Calculus Homework Help
Bio/Chem Homework Help
Engineering Homework Help
Menu
Log in
Register
Navigation
More options
Contact us
Close Menu
JavaScript is disabled. For a better experience, please enable JavaScript in your browser before proceeding.
You are using an out of date browser. It may not display this or other websites correctly.
You should upgrade or use an
alternative browser
.
Forums
Homework Help
Introductory Physics Homework Help
Solving for different masses in pulley system with KE
Reply to thread
Message
[QUOTE="jb007, post: 4994668, member: 541005"] [h2]Homework Statement [/h2] So I am stuck on this homework problem. I understand the general direction I have to take, but my algebra and physics aren't good. Here's the problem: A simple Atwood's machine uses two masses, m1 and m2. Starting from rest, the speed of the two masses is 10.0 m/s at the end of 6.0 s. At that instant, the kinetic energy of the system is 90 J and each mass has moved a distance of 30.0 m. Determine the values of m1 and m2.[h2]Homework Equations[/h2] Wnet = change in KE total KE of the system = 0.5(m1)v^2 + 0.5(m2)v^2 W = force * distance [h2]The Attempt at a Solution[/h2] My work so far has not really gotten me anywhere, any tips would be very helpful. First, I saw that the KE of the system would be the KE equation but with m 1 and m 2 , like this: KE = 0.5(m1+m2 )v^2, or 90=0.5(m1+m2 )v^2, or 90=0.5(m1+m2 )100 So then then m1+m2 would be equal to 9/5. I'm not sure about this next part I've done: The net work is equal to the change in KE. Since the system starts from rest, the initial KE is 0, and the final KE is 90. So I did: W=0.5(m1+m2 )100. Then using the work equation: FΔX=0.5(m1+m2)100, and since ΔX is equal to 30m: 30F=0.5(m1+m2)100 I tried substituting in the acceleration in order to get the force of the masses, 10/6=5/3m/s/s, so weight 1 = m1*5/3, and weight 2 = m2*5/3. With these, I know there is supposed to be 2 equations for 2 unknowns, but I can't seem to figure them out. I'm kind of lost at his point. I've uploaded a picture of the problem. I believe I'm headed in the right direction. Any tips? [/QUOTE]
Insert quotes…
Post reply
Forums
Homework Help
Introductory Physics Homework Help
Solving for different masses in pulley system with KE
Back
Top