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Mechanics Question Help

Hey guys!

I've come across this example in my mechanics textbook... And I'm so confused... I really don't get the working, not even the first line! Could someone possibly point out to me what's going on at each step? Many thanks in advance (HAHA LOL, geddit? ;p) Seriously though, stressing out if this is a very simple concept that I just can't seem to see !

Reply 1

TheConfusedMedic

Post moved to new thread.

Reply 2

old_engineer

The working in the textbook is misleading, in my opinion, as it uses kg where the units are actually kg per second, and J where the units are really J per second. But leaving that aside, I would tend to do the working in the opposite order by first considering the work done on 1kg of water in the course of lifting it by 2.5m and accelerating it from rest to 5m per second, then having done that I would work out how many kg are being pumped per second.

A few clues.

a) The work done (in Joules) in lifting 1kg of water through 2.5m is given by mgh;

b) The work done (in Joules) in accelerating 1kg of water from rest to 5ms^-1 is given by 0.5mv^2;

c) The total work down - i.e. the sum of (a) and (b) should come to 37J;

d) Now for for the number of kg pumped per second. To get that we must first work out the volume pumped per second. Imagine for a moment that the pipe is longer than 2.5m. Imagine also that you can mark a particle of water so that it is visible, and that you can run the pump for exactly one second. Imagine further that you note the position of the marked particle of water, then run the pump for exactly one second, then note the new position of the marked particle. We can infer from the information given in the question that the pump is moving water at a linear rate of 5ms^-1, so it's reasonable to assume that the two marked positions will be 5m apart. The volume of water in the pipe between the two marked positions is the volume of water pumped in one second. The volume of a cylinder (which is what we're talking about here) is given by length times cross sectional area. This will give us the flow rate in m^3s^-1;

e) We can used mass = density x volume to convert the rate of flow in m^3s^-1 to an alternative rate of flow in terms of kgs^-1. Finally, now we know how many kg of water are being pumped per second, we can multiply that by the number of Joules of work done on each kg - from (c) - to give us the rate of work being done in Js^-1. 1Js^-1 = 1W.