Work done along a path: how does velocity play into it?

[clueless_roboticist]

TL;DR

Work done along a path: how does velocity play into it?

To boil down the question, if you have a body at rest and apply a constant force, it will accelerate and the work done on it will be F*s (or the integral version of that statement). However, as the body accelerates due to the force, does that mean, per a given time unit, more and more work will be done to it as it will cover more and more distance in that time unit?

 

[PeroK]

TL;DR Summary: Work done along a path: how does velocity play into it?

To boil down the question, if you have a body at rest and apply a constant force, it will accelerate and the work done on it will be F*s (or the integral version of that statement). However, as the body accelerates due to the force, does that mean, per a given time unit, more and more work will be done to it as it will cover more and more distance in that time unit?

Yes!

 

[clueless_roboticist]

I guess that confirms my understanding of what the math indicates, but it really goes against my intuition that by applying a constant force, you are transferring an increasing amount of work over time.

 

[PeroK]

I guess that confirms my understanding of what the math indicates, but it really goes against my intuition that by applying a constant force, you are transferring an increasing amount of work over time.

That's the main reason that you can only go so fast on a bike! You need more and more power to maintain an accelerating force as you speed up. And, at about 10m/s you reach the point where the max force you can generate is only enough to equalize the retarding forces of wind and rolling resistance.

Consider, by contrast, cycling into a 10m/s headwind. You are not brought to a standstill by air resistance.

This is, in fact, a critical aspect of mechanics. It's the speed relative to the road that is the key factor.

 

[Dale]

I always think that the power formula is more useful and clear than the work formula. The power formula is $$P=\vec F \cdot \vec v$$ So as $v$ increases so does $P$ for a fixed force.

 

[A.T.]

but it really goes against my intuition that by applying a constant force, you are transferring an increasing amount of work over time.

Work/Energy are rather abstract concepts, so you cannot rely on intuition here.

However, it should be obvious that applying a force to a static object doesn't transfer any energy: You can lean something against a wall, or keep a book laying on your table indefinitely, without any energy input. So the energy/power transferred by a force must depend on displacement/velocity of the object.

 

[russ_watters]

I think it is important to recognize that more speed itself costs more power regardless of how resistance forces change with speed. I'll explain the bike example from that angle:

Bikes are basically constant power machines. The multiple gear ratios allow the rider to maintain a constant pedaling RPM and torque while speeds change. Bike riders will often select their power "setting" and accelerate fairly slowly. As speed increases the rider will gear up, losing mechanical advantage and trading more speed for lower propulsive force. Acceleration stops when resistance has increased and propulsive force has decreased to the point where they intersect.

 

[Dullard]

Your intuition got tricked. 'Apply a constant force' is an example of one of those things that are a lot easier to say than they are to do. Like: 'Hey, hand me that piano."