Calculating Work for Removing Water from Syringe

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    Syringe Work
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To calculate the work required to remove water from a syringe, the force exerted on the plunger can be expressed as F = p * S1, where p is the pressure and S1 is the cross-sectional area of the opening. The relationship between the areas and velocities at the opening and the plunger is described by A1v1 = A2v2. The volume of water removed relates to the distance the plunger moves, which can be expressed as volume = area x distance. The challenge lies in integrating the volume and time to determine the necessary distance for calculating work. Understanding these principles will help in deriving the correct formula for work in this scenario.
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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