
#1
Feb2507, 12:51 PM

P: 69

1. The problem statement, all variables and given/known data
A bug slides back and forth in a bowl 11 cm ddep, starting from rest at the top. The bowl is frictionless except for a 1.5 cm wide sticky patch on its flat bottom, where the coefficient of frictino is .61. How many times does the bug cross the sticky region??? Anybody have any ideas. I can't find an equation that uses the coefficient of friction and potential energy... thanks 



#2
Feb2507, 12:56 PM

P: 2,048

Hi, Mark, welcome to the forums!
What happens to the total energy of the bug when it crosses the sticky patch? 



#3
Feb2507, 01:01 PM

P: 69

Thanks for the welcome.
It will decrease, the kenetic energy is at its peek when it reaches the patch, but the total energy will decrease after it passes the 1.5 in sticky area. 



#4
Feb2507, 01:05 PM

P: 2,048

A bug in a bowl, lol 



#5
Feb2507, 01:07 PM

P: 69

The integral of Fnetdx???




#6
Feb2507, 01:10 PM

P: 2,048

The force is constant (what is it?), and (presumably) it's going to travel in a straight line of 1.5 cm. So it boils to down to the simpler version, W = F(net)d. 



#7
Feb2507, 01:15 PM

P: 69

Well, we have a gravitational force, weight...and we also have the kenetic friction. It's weight is constant, but we don't know its weight because we don't have a mass. 



#8
Feb2507, 01:20 PM

P: 2,048

In the patch, the gravitational force is balanced by the normal force. The net force is due to friction.
Actually, you don't need to know the mass. Once you look at the problem in its "entirety," the mass will vanish, so to speak. The bug starts with some energy at the top, then loses some in the sticky patch, and then it goes to the other side with less energy than it started with, and hence to a smaller height. All you need to know is that second height, and then repeat the process again (and again). 



#9
Feb2507, 01:28 PM

P: 69

I'm confused, so i take the work from the friction, and subtract that from the total energy..and then solve for height2? My Fnet is equal to (ma+Fsubk) right? And you said W=(Fnet)d. What is d? 



#10
Feb2507, 01:33 PM

P: 2,048

F_{net} = F_{k}. What's the force due to friction? 



#11
Feb2507, 01:39 PM

P: 69

So the displacement is going to be 1.5cm? 



#12
Feb2507, 01:41 PM

P: 2,048

Yes. I think you now have enough info to solve for the height on the other side of the bowl.




#13
Feb2507, 01:50 PM

P: 69

Ok, so i solved for work. W=.915mg. Can i use U=mgh and then substitute U/h in for mg?? And i don't know my initial total energy. I'm still a little confused. 



#14
Feb2507, 01:58 PM

P: 2,048

After that, it repeats the same thing again, but with "initial" energy decreased every time. 



#15
Feb2507, 02:12 PM

P: 69

Ok, so i start off with 11mg. Now i minus .915mg until i reach 0? I got it. Because it loses .915mg of energy every time it passes the sticky spot. So i minus that from the total each time it passes until the sticky spot completely stops the bug. Thanks for your help! 



#16
Feb2507, 02:18 PM

P: 2,048

Great!
Glad to help. :) 


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