The question is not a real difficult one, but i really need to clear my concept about the sign of the workdone when :-
1)A body falls towards earth due to gravity

2)A body is lifted against the gravitational force

Please tell is it negative or positive in the above cases?

 

Depends on one's point of reference. Either will suffice.

Example: I haul a box of comic books up to the attic. I'm doing positive work by diminishing my wife's opportunity to nag. I accidentally drop the box from the top of the ladder. This causes negative work!

Example: I haul down a box of seasonal cloths from the attic. This is positive work because we will be more comfortably dressed. I then must haul a box of non-seasonal cloths back up to the attic. This is negative work because I'll just have to haul it down again at the next Equinox.

 

As someone once said, "everything is relative".

Your point of reference will determine whether your calculations imply positive or negative work done. It is worth noting that the implication of positive and negative work comes from the fact that work done is an integral over a incremental component of a positional vector.

Dave

 

two cases where u might encounter use of -ve sign for a work done are.
1.when displacement is against the direction of applied force.
2.In thermodynamics...IIRC work done on the system is/was (dont remember, it was changed from positive to -ve or -ve to +ve..check this if possible) taken as -ve.



sen(japanese)=1000

 

Sorry I can't agree with you guys.

Work is an unsigned scalar quantity. Work is always done by something on something else.

Uzair

Case 1
(positive) work is done on the body by an outside agent. This work increases the overall energy of the body as an increase in potential energy

Case2
NO work is done untill the body hits something! The overall energy state of the body does not change. Some of its potential energy is converted to kinetic energy of falling motion.

Think about a pile driver.

Work is done by the winch and pulleys on the drop hammer raising it against gravity. After it is let go it falls and hits the top of the pile.
Work is done by the drop hammer on the pile pushing the pile another inch into the ground against friction.
In both instances the work done is positive, if you must have a sign. No work it done by gravity or friction at any stage.

 

Sorry I can't agree with you guys.

Work is an unsigned scalar quantity. Work is always done by something on something else.

Uzair

Case 1
(positive) work is done on the body by an outside agent. This work increases the overall energy of the body as an increase in potential energy

Case2
NO work is done untill the body hits something! The overall energy state of the body does not change. Some of its potential energy is converted to kinetic energy of falling motion.

The problem with not using positive and negative descriptions of work done is that you cannot conserve the energy used in the process. Think about it, if work is done to raise a box to a certain height, in other words it gains potential energy. When the box is returned from the height it looses the potential energy and work is done (positionally) counter to the initial work done.

You are right to say "Work is always done by something on something else" but that doesn't take away from the fact that work done can act against each other as in the above example.

Think about the definition of work done:

\(W = \)\(\int\)\(F.ds\)

Think about the above integral over a complete circle, actual work is done however the total work done is zero because S is a positional vector which is relative to point of reference.

Dave

 

Usually the earth is taken as the reference when considering P.E. of any object, right? So When the object is lifted against gravity, the lifting force(work done) is stored in it as its +P.E., so the energy is conserved.

 

Usually the earth is taken as the reference when considering P.E. of any object, right? So When the object is lifted against gravity, the lifting force(work done) is stored in it as its +P.E., so the energy is conserved.

Correct. So think about what if you were to take as a reference the point to which you are raising the object to.

Dave

 

Thanks for your replies guys!

 

Work is a mechanical concept and I would guess that uzair is studying at about GCSE level. So I have tried to avoid more advanced mathematics such as line integrals, taken over a closed loop or not.
Work is also about purely mechanical energy, there are other forms and energy conservation is about the idea that energy can neither be created nor destroyed, merely changed from one form to another.
To obtain useful information about work done you have to split a closed loop and here lies the trick that often trips people up.
You separate things into the system of interest and the rest of the universe. The trick is knowing where to draw the dividing line.

No system can do work on itself.......This is just a staement of conservation of energy.
Either
work is done by the rest of the universe on the system
or
work is done by the rest of the universe on the system
or
No work is done.

Now in my piledriver example the drophammer plus gravity form the system. The winch and pile form the rest of the universe. If you take gravity away it would not affect the rest of the universe this is the acid test. However taking gravity away from the hammer would prevent the system working.

Firstly the rest of the universe does work on the hammer. this could be from chemical or electrical energy. This energy is stored in the hammer as potential energy.

When the hammer is released this potential energy is converted to an exactly equal amout of kinetic energy.

The problem with not using positive and negative descriptions of work done is that you cannot conserve the energy used in the process. Think about it, if work is done to raise a box to a certain height, in other words it gains potential energy. When the box is returned from the height it looses the potential energy and work is done (positionally) counter to the initial work done.

If gravity did any work on the hammer it would now have more energy than it started with. Consequently gravity does no work on the hammer. In fact the hammer could fall forever in a gravitational field and no work be done. If any further work was done on it by gravity the hammer would have more energy than it started with as it passed its starting point. This would clearly violate the law of conservation of energy. This is also consistant with my statement no system can do work on itself when we are just considering the sytem of the hammer and gravity.

The hammer then hits the pile and does work on the rest of the universe, coming to a halt in the process. It now has lost (kinetic) energy so its total is still correctly in balance.

You cannot of course do work on gravity as it is not material.

All is correct.

 

I don't disagree with what you are saying. But gravity is a conservative force and therefore will behave in accordance with the integral equation as stated above. Remember this question is about work done being classified as either positive or negative relative to some frame of reference.

Lets bring this closer to home with a more familiar force; the electric-force, that acts on electrically charged particles. If we move a particle from point A to point B irrespective of which route or the time it takes to get there, the work done to move the particle to that point is the same - this is a fundamental feature of conservative forces. A point to note is that there is a distinct non-zero value for the work done which correlates directly to positional movement from point A to point B. If we now return the particle to its initial position, because the electric-force on the particle is conservative, the work done is 0. i.e:

\(W = \)\(\oint\)\(F.dS = 0\)

Therefore, positionally speaking the work done to move the particle from point A to point B is positive whereas the work done to return the particle from point B to point A from the original frame of reference must be negative in order to satisfy the conservative result as stated above - this is directly a consequence of the vectorial nature of the force and incremental distance over which the integral is calculated. Change the limits of the integral such that they are calculated from A-B and independently from B-A and you will find the scalar values will be positive - over a closed integral they are not independent in this way.

Dave

 

it makes life easier for engineers to use sign conventions for work done.
if some W amount of work is done by an external agency to compress a gas in a IC engine at this time what is the work done by the engine?..has to be taken as -ve W.
of course work is stored in it and for computing work o/p of the engine this work has to be
taken into account.