Solving Bug Sliding In Bowl: Friction & Potential Energy

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In summary, the problem involves a bug sliding back and forth in a frictionless bowl with a 1.5 cm wide sticky patch on its bottom. The coefficient of friction in the sticky patch is .61. The question is how many times the bug crosses the sticky region. The bug starts with some energy at the top and loses some in the sticky patch. It then goes to the other side with less energy and a smaller height. The force due to friction is equal to the net force, which is equal to the product of the coefficient of friction and the normal force (or opposite of weight). The work done on the bug can be calculated using the formula W=F*d, where d is the displacement (or distance) traveled. The bug
  • #1
xXmarkXx
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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
 
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  • #2
Hi, Mark, welcome to the forums!

What happens to the total energy of the bug when it crosses the sticky patch?
 
  • #3
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
xXmarkXx said:
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
The integral of Fnetdx?
 
  • #6
xXmarkXx said:
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
neutrino said:
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
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
neutrino said:
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
xXmarkXx said:
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
neutrino said:
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
Yes. I think you now have enough info to solve for the height on the other side of the bowl. :smile:
 
  • #13
neutrino said:
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
xXmarkXx said:
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
neutrino said:
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
Great!

Glad to help. :)
 

1. How does friction affect the movement of a bug sliding in a bowl?

Friction is a force that resists motion, so it can have a significant impact on the movement of a bug sliding in a bowl. The rough surface of the bowl will create resistance against the bug's movement, making it more difficult for the bug to slide.

2. What is potential energy and how does it relate to bug sliding in a bowl?

Potential energy is the stored energy an object has based on its position or state. In the case of a bug sliding in a bowl, the bug has potential energy at the top of the bowl because it has a higher position. As the bug slides down the bowl, its potential energy is converted into kinetic energy, causing it to move faster.

3. How can friction and potential energy be used to solve the bug sliding in a bowl problem?

By understanding the relationship between friction and potential energy, we can use this knowledge to solve the bug sliding in a bowl problem. By adjusting the amount of friction present in the bowl or manipulating the potential energy of the bug, we can control the bug's movement and determine how far it will slide.

4. Are there any other factors that can affect the bug's movement in the bowl?

Yes, there are other factors that can impact the bug's movement in the bowl. These include the shape and size of the bowl, the weight and size of the bug, and any external forces acting on the bug such as air resistance or gravity.

5. How can the understanding of friction and potential energy in this problem be applied to other real-world scenarios?

The principles of friction and potential energy are present in many real-world scenarios, such as the movement of objects on inclined planes or the sliding of cars on icy roads. By understanding these concepts and their relationship, we can better predict and control the movement of objects in various situations.

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