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Work Done On or By The System

by brentd49
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brentd49
#1
Nov1-05, 10:02 PM
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Will someone help solidy this concept of Work 'On' a system and Work 'Done' by a system.

If I lift an object from some defined zero potential. Then the work BY the system is -mgh, because gravity 'g' is in the reverse direction as the displacement 'h', but work done ON the system is +mgh because it is the opposite of work BY the system.

But what do I do if I don't know the work done by the system, i.e. there is not a given 'g' vector pointing down? How can I tell if the work is plus or minus?
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James R
#2
Nov1-05, 10:46 PM
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Work is always associated with a particular force. The work done by the force is:

[tex]W = \int \vec{F}.d\vec{r}[/tex]

In your example of lifting an object, the work done by the gravitational force is negative, while the work done by the lifting force is positive.
Doc Al
#3
Nov2-05, 08:03 AM
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Quote Quote by brentd49
But what do I do if I don't know the work done by the system, i.e. there is not a given 'g' vector pointing down? How can I tell if the work is plus or minus?
Can you give an example of what you mean? As James R stated, work is always associated with a particular force acting through a displacement. If the displacement is in the direction of the force, the work is positive; if opposite to the force, negative.

tashi
#4
Nov2-05, 02:15 PM
P: 7
Work Done On or By The System

It's sometimes easier to find the change in energy first (whether +ve or -ve) and decide whether it is positive or negative afterwards.

energy lost ===> ΔE is negative ==> work done by system
energy gained => ΔE is positive ==> work done on system

A good example is the first law of thermodynamics:

change in internal energy = heat added to system - work done by system; OR, equivalently:
change in internal energy = heat added to system + work done on system

E.g. in a four-stroke engine, if you consider a single cylinder,
work is done BY the system during the power stroke (i.e. BY the gas expanding on the piston to move it outwards and thus transferring energy outsdie the system to the other cylinders) and work is done ON the system during the exhaust stroke (i.e. by the piston (the force is supplied by the power stroke of a different cylinder) ON the exhaust gas as it is pushed out of the cylinder).
russ_watters
#5
Nov2-05, 02:52 PM
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For me it's always easiest to draw a picture. Start with a big circle that defines the boundary of your system, then place objects in it. And if often doesn't matter whether something is placed inside or outside, as long as you are consistent.

Since gravity seems like an all-encompassing force, it can get confusing, but if the system is just the box, then both you lifting and gravity acting against you are outside the system. And if you put you in the system, you need to put gravity in the system as well. Otherwise, you could accidentally create a perpetual motion machine (no inputs, only outputs).


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