Understanding Work Done: A Bicycle Puncture Example

In summary, the escaping air from a bicycle puncture doing work on the surrounding air and the air remaining in the tire doing work on the escaping air are both examples of work being done. Additionally, in this example, the temperature decreases as the air expands.
  • #1
Faiq
348
16
How is in a bicycle puncture, the air flowing out of the puncture constitutes as an example for work done?
 
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  • #2
Faiq said:
How is in a bicycle puncture, the air flowing out of the puncture constitutes as an example for work done?
Work done on what by what?
 
  • #3
I also don't know that. The question was stated as "In which example is work done?" and this option is the correct answer
 
  • #4
Faiq said:
I also don't know that. The question was stated as "In which example is work done?" and this option is the correct answer
Well the escaping air does work on the surrounding air in pushing it back. And the air still remaining in the tire does work on the air that it is pushing out ahead of it. So there are two places where work is being done.
 
  • #5
How is the still air pushing the escaping air out?
 
  • #6
Faiq said:
How is the still air pushing the escaping air out?
The air remaining in the tire is expanding as it pushes the air ahead of it out. So it is not still air.
 
  • #7
Temperature has increased in this example (since the air is expanding)?
 
  • #8
Faiq said:
Temperature has increased in this example (since the air is expanding)?
Temperature decreases in this example since the air is expanding.
 
  • #9
Oh yeah sorry got it
 

1. What is work done?

Work done is a measure of the energy transferred when a force is applied to an object and that object moves in the direction of the force. It is calculated by multiplying the force applied by the distance over which the force is applied.

2. How is work done related to a bicycle puncture?

In a bicycle puncture, work is done when the cyclist applies a force to the pump, causing it to compress the air and transfer energy into the bicycle tire. This energy is then used to push the tire outward, inflating it and fixing the puncture.

3. Can you give an example of calculating work done in a bicycle puncture scenario?

Sure, let's say a cyclist applies a force of 50 newtons to a pump to inflate a tire. The pump moves 0.5 meters in the direction of the force. The work done would be 50 newtons multiplied by 0.5 meters, which equals 25 joules.

4. How is work done affected by changes in force or distance?

According to the formula for work done (force x distance), an increase in either force or distance will result in an increase in work done. Similarly, a decrease in either force or distance will result in a decrease in work done.

5. Is work done a scalar or vector quantity?

Work done is a scalar quantity, meaning it has only magnitude and no direction. This is because it is the product of two scalar quantities (force and distance), and does not have a specific direction associated with it.

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