How is work calculated for a conservative force?

[waynexk8]

Conservative force ?

Hi all,

Work or mechanical work in physics and kinesiology, is more narrowly defined as the product of force applied over a distance. If the distance traveled is zero, than any force times zero is zero (U = F x 0 = 0).

So if I move a weight overhead 1m with the force/strength of my muscle work has been done, and if I lower the weight back down 1m with the force/strength on my muscles work has been done, I have moved the weight 2m, and if we knew the kgs of the weight and speed it was traveling we could work out the power {work energy} it used.

However, according to conservative force, conservative force is a force with the property that the work done in moving a particle between two points is independent of the path taken. Equivalently, if a particle travels in a closed loop, the net work done (the sum of the force acting along the path multiplied by the distance travelled) by a conservative force is zero.

The question,
So how if I move a weight 1m up and 1m down with the force/strength of my muscles, and am able to calculate the power {work energy} used, how come conservative force says the work done is zero, this seems contradictory to me.

Wayne

 

[f95toli]

There are a few issues here.
Firstly, muscles are a bit "weird" compared to say an ideal spring in that they can't absorb the energy gained when you lower the weight (which is what would happen when a spring is being compressed: it "stores" the energy); muscles always use energy.

The second issue is that you have moved the weight 1+(-1)=0 meters (you need to keep track of the sign). Hence, if you were to calculate the energy used for an ideal closed system (and the human body is not an ideal, closed, system) the distance involved and hence the energy would be zero.

So, there is no contradiction.

 

[pgardn]

I might add you also have to be very careful about identifying what is doing the work, and what is having the work done on it. Most of the time if you can identify this, the problem makes more sense.

 

[waynexk8]

Hi there,

And thanks for the reply.

There are a few issues here.
Firstly, muscles are a bit "weird" compared to say an ideal spring in that they can't absorb the energy gained when you lower the weight (which is what would happen when a spring is being compressed: it "stores" the energy); muscles always use energy.

I get the spring thing.

The second issue is that you have moved the weight 1+(-1)=0 meters (you need to keep track of the sign). Hence, if you were to calculate the energy used for an ideal closed system (and the human body is not an ideal, closed, system) the distance involved and hence the energy would be zero.

So, there is no contradiction.

I am not sure why you are saying I have moved the weight 1+ and then 1- ? As I have moved the weight up 1m and this takes force and power {work energy} and then moved the weight down 1m and this takes force and power {work energy} so would not there be 1+ and 1+ ? As the weight has moved 1m up and 1m down, this = 2m.

mechanical work In biomechanics and physics, work is more narrowly defined as the product of force applied over a distance, and I have moved the weight by force over a distance of 2m, so work has taken place.

As what if I moved the weight up on an angle 1m, and back down on the opposite angle, thus the weight would not be in the same position it started ?

What if I lifted the weight, walked a mile {ouch my aching arm ROL} and then lowed the weight ?

Wayne

 

[waynexk8]

I might add you also have to be very careful about identifying what is doing the work, and what is having the work done on it. Most of the time if you can identify this, the problem makes more sense.

Yep see you point.

Wayne

 

[arunma]

My friends who are familiar with physiology tell me that muscles actually do move when they are contracted. Apparently, small muscle fiber's which connect to each other during muscular contraction are constantly being detached and must reconnect. These fibers must do work against a frictional force (which is non-conservative, and will contribute positive work during detachment and reconnection). Someone correct me if my physiological information is wrong. But if not, this would be where the displacement comes into play.

 

[arunma]

mechanical work In biomechanics and physics, work is more narrowly defined as the product of force applied over a distance, and I have moved the weight by force over a distance of 2m, so work has taken place.

Actually the net work done by you is still zero. You're doing positive work by lifting the weight up, and negative work by putting the weight down.

As what if I moved the weight up on an angle 1m, and back down on the opposite angle, thus the weight would not be in the same position it started ?

We're being a bit fudgy when we talk about "closed loops." Gravity acts only in the downward direction, so the only coordinate that matters here is height. You know from basic physics that gravitational potential near the Earth is $U = mgh$. So as long as the object is returned to the original height, it could be a mile away, and the work done against gravity would still be zero.

All of this actually makes perfect logical sense when we talk about energy in the language of multivariable calculus. But I'm trying to avoid the math here, so I have to ask you to take what I just said on faith.

What if I lifted the weight, walked a mile {ouch my aching arm ROL} and then lowed the weight ?

In this case you'd have done no work on it, because it takes no energy to move an object perpendicular to the force. This would actually be the same thing as simply holding the weight in place for the same amount of time that it takes you to walk a mile. Here the only work being done is the "physiological" energy required to contract your muscles, e.g. dephosphorylation of ATP by hexokinase (which is fundamentally electric potential energy), the mechanical work of the muscle fibers, etc.

 

[Yeti08]

I am not sure why you are saying I have moved the weight 1+ and then 1- ? As I have moved the weight up 1m and this takes force and power {work energy} and then moved the weight down 1m and this takes force and power {work energy} so would not there be 1+ and 1+ ? As the weight has moved 1m up and 1m down, this = 2m.

mechanical work In biomechanics and physics, work is more narrowly defined as the product of force applied over a distance, and I have moved the weight by force over a distance of 2m, so work has taken place.

As what if I moved the weight up on an angle 1m, and back down on the opposite angle, thus the weight would not be in the same position it started ?

I think where you're going wrong is in saying that Work = Force times Distance, which is the simplified version of the vector dot product:

$$W = \textbf{F} \cdot \textbf{d}$$

Or more generally for a curve

$$W =\int \textbf{F} \cdot d \textbf{s}$$

where s is the position vector. When using vectors the direction obviously matters, so when going straight up and then straight down these equations simplify to the 1+(-1) as f95toli stated

 

[waynexk8]

Actually the net work done by you is still zero. You're doing positive work by lifting the weight up, and negative work by putting the weight down.

Hmm, but as you say, I am doing positive work and negative work, and both need energy to do so, and say it takes 3E to lift the weight and {because the muscles are more efficient at lowering the weight, but let's not get into that} 2E to lower the weight = 5E, and in both cases 1m has the object moved = 2m.

So I just do not get/see how zero work has been done, too me its like saying if you move something up and then down, that it has not traveled and distance or used and energy, but it has.

We're being a bit fudgy when we talk about "closed loops." Gravity acts only in the downward direction, so the only coordinate that matters here is height. You know from basic physics that gravitational potential near the Earth is $U = mgh$. So as long as the object is returned to the original height, it could be a mile away, and the work done against gravity would still be zero.

So are you saying that if I lift something up gravity takes it down ? Well yes, but I am meaning if I lift the weight using my force/strength power, {work energy} and then lower the weight using my force/strength power {work energy}, and on both accounts I am resisting gravity.

All of this actually makes perfect logical sense when we talk about energy in the language of multivariable calculus. But I'm trying to avoid the math here, so I have to ask you to take what I just said on faith.

I will take your words for it, as the maths would be over my head. But can you see the point I am trying to make ?

In this case you'd have done no work on it, because it takes no energy to move an object perpendicular to the force. This would actually be the same thing as simply holding the weight in place for the same amount of time that it takes you to walk a mile. Here the only work being done is the "physiological" energy required to contract your muscles, e.g. dephosphorylation of ATP by hexokinase (which is fundamentally electric potential energy), the mechanical work of the muscle fibers, etc.

Work would be done by the body in waling a mile.

Wayne

 

[waynexk8]

Actually the net work done by you is still zero. You're doing positive work by lifting the weight up, and negative work by putting the weight down.



We're being a bit fudgy when we talk about "closed loops." Gravity acts only in the downward direction, so the only coordinate that matters here is height. You know from basic physics that gravitational potential near the Earth is $U = mgh$. So as long as the object is returned to the original height, it could be a mile away, and the work done against gravity would still be zero.

All of this actually makes perfect logical sense when we talk about energy in the language of multivariable calculus. But I'm trying to avoid the math here, so I have to ask you to take what I just said on faith.



In this case you'd have done no work on it, because it takes no energy to move an object perpendicular to the force. This would actually be the same thing as simply holding the weight in place for the same amount of time that it takes you to walk a mile. Here the only work being done is the "physiological" energy required to contract your muscles, e.g. dephosphorylation of ATP by hexokinase (which is fundamentally electric potential energy), the mechanical work of the muscle fibers, etc.

I think where you're going wrong is in saying that Work = Force times Distance, which is the simplified version of the vector dot product:

$$W = \textbf{F} \cdot \textbf{d}$$

Or more generally for a curve

$$W =\int \textbf{F} \cdot d \textbf{s}$$

where s is the position vector. When using vectors the direction obviously matters, so when going straight up and then straight down these equations simplify to the 1+(-1) as f95toli stated

Hmm I think I see a little more here.

However, the above zero work does not seem to work in real life situations like the one I am chatting about ?

As you have lifted the weight up 1m and lowered it down 1m and in both the positive and negative direction, you have moved 2m and in both directions you have used strength/force and power .{work energy}

Wayne

 

[Yeti08]

However, the above zero work does not seem to work in real life situations like the one I am chatting about ?

Yes, what I stated was for a perfect mechanical system - i.e. no losses. A person lifting a weight is a system using chemical energy to do mechanical work and inefficiencies abound. For example, for a person sitting in a chair, the chair is exerting a force on the person, but no work is being done since there is no distance moved. But for a person doing a wall sit there is obviously energy being used, just not mechanical energy - the person is expending chemical energy in order to maintain a force via muscles.

Work would be done by the body in waling a mile.

The human body is a very complex system. If you simplify by holding the weight by some structure on a frictionless surface and then moving a mile, you can better see that there is no work being done on the the suspended object other than accelerating/decelerating it - just imagine a bicycle carrying a weight. So I think the easiest way to explain your scenario is that there human body is inefficient and thereby expends energy both when lifting and lowering a weight, whereas a perfect system will not.

 

[pgardn]

My friends who are familiar with physiology tell me that muscles actually do move when they are contracted. Apparently, small muscle fiber's which connect to each other during muscular contraction are constantly being detached and must reconnect. These fibers must do work against a frictional force (which is non-conservative, and will contribute positive work during detachment and reconnection). Someone correct me if my physiological information is wrong. But if not, this would be where the displacement comes into play.

yes they do, but usually in pairs, one set of muscles contract and the other set relaxes. Each individual Sarcomere (the unit of proteins in muscle cells) actually does get shorter when contraction occurs. But underneath it all, the energy to allow all of this to occur, is chemical in nature. Which really is a type of positional energy due to electron configuration changing. So it all comes back to ideas represented in physics.

 

[waynexk8]

Yes, what I stated was for a perfect mechanical system - i.e. no losses. A person lifting a weight is a system using chemical energy to do mechanical work and inefficiencies abound. For example, for a person sitting in a chair, the chair is exerting a force on the person, but no work is being done since there is no distance moved. But for a person doing a wall sit there is obviously energy being used, just not mechanical energy - the person is expending chemical energy in order to maintain a force via muscles.

The human body is a very complex system. If you simplify by holding the weight by some structure on a frictionless surface and then moving a mile, you can better see that there is no work being done on the the suspended object other than accelerating/decelerating it - just imagine a bicycle carrying a weight. So I think the easiest way to explain your scenario is that there human body is inefficient and thereby expends energy both when lifting and lowering a weight, whereas a perfect system will not.

Therefore, if you lift a barbell up and then down, it takes force power {work energy} to go up, and to go down, this is also the same with a rocket/helicopter. I understand that if you pull a spring that it goes back using its own force, but a muscle rocket/helicopter does not, it must use force power {work energy}to go back down as well as up.

Wayne

 

[Born2bwire]

Therefore, if you lift a barbell up and then down, it takes force power {work energy} to go up, and to go down, this is also the same with a rocket/helicopter. I understand that if you pull a spring that it goes back using its own force, but a muscle rocket/helicopter does not, it must use force power {work energy}to go back down as well as up.

Wayne

You do not need to expend any power to go back down. The gravitational field does all the work. Any energy that you are expending in lowering an object is extraneous, it is spent in reconfiguring your muscles or in checking the object's descent for example which have nothing to do with actually lowering the object.

 

[waynexk8]

You do not need to expend any power to go back down. The gravitational field does all the work. Any energy that you are expending in lowering an object is extraneous, it is spent in reconfiguring your muscles or in checking the object's descent for example which have nothing to do with actually lowering the object.

If I was to let it drop, then yes the gravitational field would do all the work, but I am not letting it drop, I am controlling the decent, thus I must be using force and power.

The concentric action and eccentric action are complete different actions, but both need force and power to do. Thus there is work on the concentric and work on the eccentric.

Wayne