Calculating Work for Removing Water from Syringe

[DanicaK]

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

 

[Simon Bridge]

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.

 

[DanicaK]

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.

 

[Simon Bridge]

volume = area x distance

 

[asteriatic]

...Well, in order to calculate the work required to remove the water from the syringe, we would need to know the force applied to the clip, as well as the distance the clip moves while removing the water. This can be calculated by multiplying the force by the distance, which would give us the amount of work done in joules.

However, since the question mentions a constant force and a time frame, we can also use the formula for work, W = F x d, and rearrange it to find the force, F = W/d. This would give us the force required to remove the water in a specific amount of time.

Additionally, the volume of water and the cross-sectional area of the opening are irrelevant in this calculation since we are only concerned with the force and distance. The smaller area of the clip compared to the opening may affect the amount of force needed, but it would not drastically change the calculation.

Finally, it is important to note that this calculation does not take into consideration any external factors such as friction, which may affect the actual amount of work required to remove the water. In order to accurately calculate the work, we would need to take into account all external forces acting on the syringe.