Mechanical energy of sliding beads

AI Thread Summary
The discussion revolves around calculating the change in mechanical energy of a bead sliding on a curved wire, specifically addressing the transition from a frictionless segment to a rough segment. The initial calculations for the change in mechanical energy due to friction were incorrect, as the height at point C was misunderstood. Participants clarified that the change in mechanical energy is negative, indicating energy loss due to friction. Another problem involving a bead on a loop-the-loop was also discussed, where the participant initially miscalculated the speed at point A due to rounding errors and sign mismanagement. The conversation emphasizes the importance of understanding energy conservation principles and careful calculation in physics problems.
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Homework Statement


A 0.5 kg bead slides on a curved wire, starting
from rest at point A as shown in the figure.
The segment from A to B is frictionless, and
the segment from B to C is rough. The point
A is at height 6.9 m and the point C is at
height 1.6 m with respect to point B.
The acceleration of gravity is 9.8 m/s^2 :
http://img376.imageshack.us/img376/2720/motion8pj.gif
the image for this problem is fig.2
If the bead comes to rest at C, find the change
in mechanical energy due to friction as it
moves from B to C. Answer in units of J.

Homework Equations


mgh=energy of grav.
Force*displacement=diss. energy

The Attempt at a Solution


First, I set mgh at point A to mgh at C + diss energy:
(.5)(9.8)(6.9)=.5(9.8)+diss.
diss=26J
the answer above is not right, am i misunderstanding the question?
 
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Well, isn't the height at C 1.6m above B??

Remember, though that the change of mechanical energy is NEGATIVE.
 
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oh, thanks alot!
 
since i have fig.1 as well, I might as well post another one of my problem:
1. Homework Statement
A bead slides without friction around a loop-
the-loop. The bead is released from a height
of 9.4 m from the bottom of the loop-the-loop
which has a radius 2 m.
The acceleration of gravity is 9.8 m/s2 :
What is its speed at point A? Answer in
units of m/s.
3.Work
this is basically the same as the first problem, and mass isn't needed because it ends up canceling.
so:
mgh=.5mv^2+mgh <-there is also grav. energy, because A is not the lowest
gh=.5v^2+gh
(9.8)(9.4)=.5v^2+9.8(4)
v=10.3m/s
but that answer is wrong. I'm sure that i have the equation set out right, because at point A kinetic energy is transferred to grav. potential energy.
 
you did some thing that does not make logical sense. take a look at your plus/minus signs.
 
so you are saying that one of the value that i plugged in should've been negative?
 
yep. But don't just change it without understanding why. after it falls, it goes up 4 meters--not down.
 
i figured out what i did wrong... I rounded wrong. I was off by more than 1% from the real answer, so Utexa's homework service rejected it.
 
Those drawings are down in Latex?
 
  • #10
no, I copied and edited them on paint :)
 
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