Who exerts the power?

  • #1
Trying2Learn
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Work is the dot product of force and displacement (let me set aside the path integral).

If a rock of 2 Newton is lifted up 3 meters, the work is NEGATIVE 6
Gravity force is down, displacement is up, the dot product is negative and we can say: "The force of gravity does work = -6"

If a rock of 2 Newton is lowered by 3 meters, the work is POSITIVE 6
Gravity force is down, displacement is down, the dot product is positive and we can say: "The force of gravity does work = 6"

I am ignoring what "I" feel, and am focused on the role played by the force (perhaps that is my problem in the following question.)

Now we turn to power. Power is the dot product of the force and the velocity.

For the following, I would rather think in terms of a bicycle going up a hill at a constant velocity.

Gravity is down, your velocity component is up. The power is negative.

Now break as you go down

Gravity is down, your velocity component is down. The power is positive.

OK, so WHO is "doing" or "exerting" the power? In both cases, the bicyclist breaks a sweat (more on the way up, of course, but there is
power exerted to hold the breaks)

With regard to POWER, WHO or WHAT is exerting it?
 

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  • #3
Trying2Learn
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(emphasis added)
(emphasis added)

Gravity

OK, so far, go good. I was expecting that.

Can you now extend that and tell me: why, in BOTH cases, do I, the bicyclist, sweat?
A LOT in the first case, but not much in the second (still, it takes an effort -- power? -- to hold the breaks)
 
  • #4
anorlunda
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Can you now extend that and tell me: why, in BOTH cases, do I, the bicyclist, sweat?
Physiology has only a weak connection to physics. You use energy and sweat just sitting doing nothing.
 
  • #5
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OK, so far, go good. I was expecting that.

Can you now extend that and tell me: why, in BOTH cases, do I, the bicyclist, sweat?
A LOT in the first case, but not much in the second (still, it takes an effort -- power? -- to hold the breaks)
If you want to see what the bicyclist is doing then you need to analyze the bicyclist, not gravity.

Now, as @anorlunda mentioned, humans are terrible systems to analyze for understanding physics. We are far too complicated and inefficient. So instead I will replace the bicyclist with a perfectly efficient electric motor for pedaling and a perfectly efficient linear actuator for squeezing the brake. So to understand why the battery drains you simply look at the force at the pedals/brakes instead of looking at the gravitational force.

For the pedals, the force is forward on top and backward on bottom and the velocity is also forward on top and backward on bottom. So the motor does positive work and drains the battery. For the brakes the force is backwards and the motion is backwards. So the actuator also does positive work and drains the battery.

If you want to see the work that gravity does then you look at the force of gravity. If you want to see the work that the motor does then you look at the force of the motor.
 
  • #6
russ_watters
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Can you now extend that and tell me: why, in BOTH cases, do I, the bicyclist, sweat?
A LOT in the first case, but not much in the second (still, it takes an effort -- power? -- to hold the breaks)
Do you have bad brakes? Static forces do not require power to maintain, so if you are consuming a lot of biological/chemical energy to hold the bakes, that seems like a problem with the bike.

We are talking about hand brakes, right? The muscles in your forearm are tiny compared to your legs.
 
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  • #7
hutchphd
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For the brakes the force is backwards and the motion is backwards.
Sorry @Dale but I am not following your meaning here at all...the pad force is opposite the wheel direction.
 
  • #8
Trying2Learn
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OK, can we put this in terms of lifting a rock?

If I lift the rock up, in my hand, at a steady speed of e m/sec, and the rock weighs 3 Newtons, then the power generated by gravity is negative

However, the power generated by my hand is positive because the normal force is my hand.

So far, so good.

If I lower the rock (same speed, same weight), my hand is exerting negative power. Gravity's power is positive. But I still have to exert power to lower the rock.

I am almost not confused.
 
  • #9
hutchphd
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But I still have to exert power to lower the rock.
"Exerting power" is not a well defined term. In the standard physics jargon the phrase is "doing work on". When lowering the rock, the rock is "doing work on" your hand, while gravity is "doing work on" the rock. You may be working hard but that is not a physics term, nor is sweating like a pig or powering through.
 
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  • #10
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Sorry @Dale but I am not following your meaning here at all...the pad force is opposite the wheel direction.
I assumed he was asking about squeezing the hand control of the brake.
 
  • #11
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OK, can we put this in terms of lifting a rock?
It is kind of irritating to actually analyze the scenario you first proposed and then have you change the scenario without even commenting on the analysis. Did you even glance at it before deciding to ignore it?

If I lower the rock (same speed, same weight), my hand is exerting negative power. Gravity's power is positive. But I still have to exert power to lower the rock.
Again, this is just because humans are horribly inefficient machines. An ideal linear actuator can send power to recharge the battery in this case.

Anytime you are confused by energy in a scenario involving a human, get rid of the human.
 
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  • #12
Drakkith
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OK, can we put this in terms of lifting a rock?
Adding to what others have said, if you hold the rock still, you still use energy to hold rock up, even though no work is being done on the rock. The energy is used in your muscle cells to alternately contract different muscle fibers in your arm and hand to keep the rock up.

Place the rock on a table and the table uses no energy to hold the rock up.
 
  • #13
Trying2Learn
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If you want to see what the bicyclist is doing then you need to analyze the bicyclist, not gravity.

Now, as @anorlunda mentioned, humans are terrible systems to analyze for understanding physics. We are far too complicated and inefficient. So instead I will replace the bicyclist with a perfectly efficient electric motor for pedaling and a perfectly efficient linear actuator for squeezing the brake. So to understand why the battery drains you simply look at the force at the pedals/brakes instead of looking at the gravitational force.

For the pedals, the force is forward on top and backward on bottom and the velocity is also forward on top and backward on bottom. So the motor does positive work and drains the battery. For the brakes the force is backwards and the motion is backwards. So the actuator also does positive work and drains the battery.

If you want to see the work that gravity does then you look at the force of gravity. If you want to see the work that the motor does then you look at the force of the motor.
OK, sorry about the change in problem.

Can I go with your's then (I did not see it earlier)
If you want to see what the bicyclist is doing then you need to analyze the bicyclist, not gravity.

Now, as @anorlunda mentioned, humans are terrible systems to analyze for understanding physics. We are far too complicated and inefficient. So instead I will replace the bicyclist with a perfectly efficient electric motor for pedaling and a perfectly efficient linear actuator for squeezing the brake. So to understand why the battery drains you simply look at the force at the pedals/brakes instead of looking at the gravitational force.

For the pedals, the force is forward on top and backward on bottom and the velocity is also forward on top and backward on bottom. So the motor does positive work and drains the battery. For the brakes the force is backwards and the motion is backwards. So the actuator also does positive work and drains the battery.

If you want to see the work that gravity does then you look at the force of gravity. If you want to see the work that the motor does then you look at the force of the motor.
OK (regarding your other comment: sorry for the two different cases)

So, can I return to this one, but with a motor and actuator for brakes?

If the bicycle is going up a hill, then power supplied by gravity is negative. In that case, I assume the motor must supply that power. This makes sense to me.

If the bicyle is going down the hill, then the power supplied by gravity is positive. For the rider to go at the same speed, the breaks must drain that power.

However, in both cases, the power is the same.

Good. Now replace the motor and breaks with a person. The person must somehow supply the power when he or she replaces the motor or actuator.

Why is it that it FEELS more difficult to ride up hill, then down, if we just calculated that the power is the same.
 
  • #14
PeroK
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Good. Now replace the motor and breaks with a person. The person must somehow supply the power when he or she replaces the motor or actuator.

Why is it that it FEELS more difficult to ride up hill, then down, if we just calculated that the power is the same.
The braking power is provided by friction in the braking mechanism and by static friction of the tyres on the road.

It's not like a car is rolling down a hill and you have to stop it by getting in front of it and pushing against it. That would be very different from applying a foot-brake!
 
  • #15
Trying2Learn
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The braking power is provided by friction in the braking mechanism and by static friction of the tyres on the road.

It's not like a car is rolling down a hill and you have to stop it by getting in front of it and pushing against it. That would be very different from applying a foot-brake!

NOW I see.
Now everyone's response about this makes sense.

Sorry for the time waste.
 
  • #16
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So, can I return to this one, but with a motor and actuator for brakes?
Sure, let’s stick with the first example.

However, in both cases, the power is the same.
The power by gravity is the exact opposite in the two cases (same magnitude but different sign), but the power supplied by your motor is vastly different from the power supplied by the linear actuator.

The power that must be the same as the gravitational power for the downhill case is the mechanical power drained by the brakes, but this is not at all related to the power supplied by the actuator.

Good. Now replace the motor and breaks with a person.
No. If your goal is to learn the physics then you should never ever do that. Once you already understand the physics then you can do that to learn some biology. But never for learning physics. Never.
 
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  • #17
hutchphd
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Sorry for the time waste.
In some ways your question is very fundamental: why do frictional forces have a preferred direction in time? So no apology required.
 
  • #18
jbriggs444
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Why is it that it FEELS more difficult to ride up hill, then down, if we just calculated that the power is the same.
The power is not the same. Draining power and providing power are two different things.

Do not ignore the sign.

In addition, moving your legs to provide power is much more difficult than gently squeezing your hands to provide tension (but not significant motion) to the caliper brakes. Then too, there is wind resistance. You can literally coast down a hill with no hands.
 
  • #19
sophiecentaur
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Summary:: Positive vs negative power

Work is the dot product of force and displacement (let me set aside the path integral).
This first statement is correct but you seem to go on to doubt what it tells you. Up is a positive distance and g acts in a negative direction. All the rest follows.

Aamof, this sort of discussion usually considers Work, rather than Power because it's hard to compare Power in one direction (say climbing stairs) with Power in the other direction ( falling out of the window), when the energy involved is the same in both directions.
 
  • #20
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Good. Now replace the motor and breaks with a person.
Nit: the bicycle has brakes. It will have breaks only if the frame or other component is cracked. I realize that English is probably not your first language, so such a mistake is not a big deal, and is one that many native speakers of English get wrong as well.

You can literally coast down a hill with no hands.
And if you're like me, without using the brakes. I once rode a bicycle from the top of Haleakala (elev. 10,023') on the island of Maui, all the way down to almost sea level and applied the brakes only once.
this sort of discussion usually considers Work, rather than Power because it's hard to compare Power in one direction (say climbing stairs) with Power in the other direction ( falling out of the window), when the energy involved is the same in both directions.
I agree. It's better to be thinking about the forces involved, or the work done (in simplest form, work = force times distance) than power, which is defined as the time derivative of Work, or ##P = \frac{dW}{dt}##.
 
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