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A bug in a bowl, lol

  • Thread starter xXmarkXx
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  • #1
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Homework Statement



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
 

Answers and Replies

  • #2
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Hi, Mark, welcome to the forums!

What happens to the total energy of the bug when it crosses the sticky patch?
 
  • #3
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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
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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.
Exactly! Some work is being done on the bug to decrease its energy. How would you calculate this work done?
 
  • #5
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The integral of Fnetdx???
 
  • #6
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The integral of Fnetdx???
Mmm...you wouldn't actually need an integral here, but the principle's the same.
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
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Mmm...you wouldn't actually need an integral here, but the principle's the same.
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.

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
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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
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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).

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
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so i take the work from the friction, and subtract that from the total energy..and then solve for height2?
That's right.

My Fnet is equal to (ma+Fsubk) right?
I assume Fsubk refers to the force due to kinetic friction.

Fnet = Fk. What's the force due to friction?

And you said W=(Fnet)d. What is d?
Dispalcement (or distance). Work = Force.Displacement. Remember, the force and displacement are in opposite directions.
 
  • #11
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That's right.


I assume Fsubk refers to the force due to kinetic friction.

Fnet = Fk. What's the force due to friction?



Dispalcement (or distance). Work = Force.Displacement. Remember, the force and displacement are in opposite directions.
Fk=ukn where n is the normal force or the opposite of weight in this case.

So the displacement is going to be 1.5cm?
 
  • #12
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Yes. I think you now have enough info to solve for the height on the other side of the bowl. :smile:
 
  • #13
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Yes. I think you now have enough info to solve for the height on the other side of the bowl. :smile:

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
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Ok, so i solved for work. W=-.915mg.
Okay, that's how much it lost to friction.

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.
Let's say it started off at height h (which is eqaul to 11 cm, in this case). Therfore initial energy is mgh; it loses .915mg, and with the remaining energy goes up the other side to some height, say h'. Can you solve for h'?

After that, it repeats the same thing again, but with "initial" energy decreased every time.
 
  • #15
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Okay, that's how much it lost to friction.



Let's say it started off at height h (which is eqaul to 11 cm, in this case). Therfore initial energy is mgh; it loses .915mg, and with the remaining energy goes up the other side to some height, say h'. Can you solve for h'?

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

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
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Great!

Glad to help. :)
 

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