- #1

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Could you clarify this? Is the solution merely that the work W is not EXPLICITLY dependent on time?

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- Thread starter gentsagree
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- #1

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Could you clarify this? Is the solution merely that the work W is not EXPLICITLY dependent on time?

- #2

A.T.

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F is not the net force, but any force applied to an object.However, W=Fs, and F is proportional to the acceleration a,

It is nor defined via time. But of course you can compute the time for some energy transfer, given some extra information.Is the solution merely that the work W is not EXPLICITLY dependent on time?

- #3

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Regarding the second point: I know it is not defined via time, I am just pointing out (naively, I know, but I'd like a better understanding) that implicitly in the notion of work we have that of time. Not only that, it is changing time, so something like Delta t.

- #4

A.T.

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It can, but it's not in general. You can't make general statements based on special cases.F can be taken as the net force

- #5

Nugatory

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You're both right (or more accurately, correctly describing subtly different things).I don't agree with your first reply: F can be taken as the net force [check page 186 of Young and Freedman, for one].

The total amount of kinetic energy imparted when moving an object is the net force times times the distance. In many problems that's the useful work done and the only thing we care about.

However, if the net force is the sum of multiple opposing forces, the total amount of work done may be much greater. For example, if you are pushing an object at constant speed against the force of friction, the net force is zero and the kinetic energy of the object doesn't change; but you're clearly doing work and the energy ends up as waste heat.

I don't see this. Apply a force of 100N over a distance of 100 meters and you've done 10,000 Joules of work. It doesn't matter whether it took a millisecond or a century to cover that distance, the work done is the same.I am just pointing out (naively, I know, but I'd like a better understanding) that implicitly in the notion of work we have that of time. Not only that, it is changing time, so something like Delta t.

- #6

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I don't see this. Apply a force of 100N over a distance of 100 meters and you've done 10,000 Joules of work. It doesn't matter whether it took a millisecond or a century to cover that distance, the work done is the same.

I was thinking of this example: the velocity of a car which hits a wall goes from, say, 80 to 0 km/h almost instantaneously (quite disastrously). The force applied is then F=ma, where the acceleration is great since it happened over the space of a very short time interval. The work the brakes did is then W=Fs.

If the car gradually decelerates from 80 to 0, in say 10 seconds, the acceleration is much smaller. Hence the force and the work exerted by the brakes decrease.

This is why I was thinking that, implicitly, the work depends on time. Am I wrong?

On a different note, please disregard my original comment on power: that's a completely different thing, just being the obvious description of how work is spread over time, thus a different aspect of time of what I wonder about.

- #7

Doc Al

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Yes, you are wrong. The net work done in both cases is the same and is independent of the time it takes for the car to be brought to rest. (In the second case, while the force is reduced the distance increases.)I was thinking of this example: the velocity of a car which hits a wall goes from, say, 80 to 0 km/h almost instantaneously (quite disastrously). The force applied is then F=ma, where the acceleration is great since it happened over the space of a very short time interval. The work the brakes did is then W=Fs.

If the car gradually decelerates from 80 to 0, in say 10 seconds, the acceleration is much smaller. Hence the force and the work exerted by the brakes decrease.

This is why I was thinking that, implicitly, the work depends on time. Am I wrong?

- #8

A.T.

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You can construct some scenario where some quantity depends on some other quantity. All physical quantities can be related somehow.This is why I was thinking that, implicitly, the work depends on time. Am I wrong?

- #9

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Work = Force * Distance

for a constant force. For a variable force you use an integral.

In your example involving the car, the car has high Force and low Distance when it hits the wall. It has low Force and high Distance when it slows down over a longer distance. The work for the two operations must be equal.

- #10

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That's right. Thank you.

- #11

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