Calculating Work for Removing Water from Syringe

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Homework Help Overview

The discussion revolves around calculating the work required to remove water from a syringe using a constant force. The problem involves understanding the relationship between force, pressure, volume, and time in a fluid dynamics context.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore the use of the work formula W=Fd and question how to determine the distance d. There is discussion about the pressure exerted on the water and how the size of the opening affects the motion of the plunger. Some participants attempt to apply fluid dynamics equations and relate volume to area and distance.

Discussion Status

The discussion is ongoing, with participants sharing various equations and concepts. Some guidance has been offered regarding the relationships between pressure, area, and velocity, but there is still uncertainty about how to incorporate volume and time into the calculations.

Contextual Notes

Participants are working under the assumption that friction is negligible and are considering the implications of the opening size in relation to the syringe's mechanics. There is a lack of clarity on how to connect the volume of water and the time factor to the work calculation.

DanicaK
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A syringe is filled with water and placed in horizontal position. How much work should we do pressuring the clip with a constant force in order to remove the water from the syringe in a time t. The volume of the water in the syringe is V, the cross-sectional area of the opening is S1 and is very smaller than the area of the cross-section of the clip. Don't take into consideration the friction.

hmm
 
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hmm ... indeed.
So what have you tried?

W=Fd seems a good start - what would d be then?
Pressure exerted on the plunger-end of the water would be F/S1
If the opening were the same as S1 what sort of motion would you expect from the plunger?
How does this change since the opening is "very small"?
Stuff like that? Show me.

Start the hmm problems by playing around with the concepts and something will usually strike a chord with a recent lesson or a related concept.
 
p1+ρgh+ρv1^2/2=p2+ρgh+ρv2^2/2
p1=F/A1
we cancel ρgh
p2=p(atm)
ρv1^2/2=0 because the opening is very smal
so, F/A1=p(atm)+ρv2^2/2

A1v1=A2v2

Now I don't know how to use the volume and the time.
And how to find d in order to calculate the work.
 
volume = area x distance
 

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