Total work/effort/energy spent to climb a hill

[Stella.Physics]

https://scontent.fath3-2.fna.fbcdn.net/hphotos-xtf1/v/t1.0-0/s526x296/12804814_10208727068471849_3639991170291956856_n.jpg?oh=a60b7b49070602e57e99d4dfdf3545dd&oe=57640241
Yesterday my bf asked me to calculate the calories he burns to go up a hill.
So of course the first thing I thought was to take the difference of the potential energy to estimate that work and so I said that work equals W=mgh, so for a person of 60kgr, assuming that g=10 m/s^2 and h=200m I said that work is W=120.000 J. So then I thought that this calculation is only for vertical lifting of his weight but still is correct since I estimated it with the potential energy from level zero (Point A) to point B which has potential energy. But still is this calculation enough? Or should I add the work from point A to C and then add the vertical work?
I want to know the total work needed to climb the hill and the energy consumption to move his weight from A to B.

 

[hackhard]

total work needed to climb the hil

in physics we deal with-
work done by a force ,on a body, from initial to final pos
mgh = work done by gravity , on bf, in climbing hill

Yesterday my bf asked me to calculate the calories he burns to go up a hill.

theres no way equations be framed for calories burnt
suppose ur bf wears 10 sweaters on midday June
or just a vest on midnight december ?

 

[Stella.Physics]

theres no way equations be framed for calories burnt
suppose ur bf wears 10 sweaters on midday June
or just a vest on midnight december ?

i never mentioned clothes, weather or ground because I take them a negligible. I just want to know the work and effort to just move the weight from A to B. Please, consider metabolism activity like sweating negligible as well :P

 

[hackhard]

if u want to reduce problem to -
work done by minimum force required to push a block (60 kg) up an inclined plane (vertical height 200m , hori length 3000m)
then work by that force is (mg.sinx + umg.cosx) S
s is slant height u coeff of friction & x is incline angle

 

[Drakkith]

I want to know the total work needed to climb the hill and the energy consumption to move his weight from A to B.

You're right in that just calculating the work required to lift his mass up the vertical distance is insufficient. The human body is an inefficient machine, and a lot of energy is lost in a variety of ways during movement. Using a random calories burned calculator I found on google, I get that he would burn somewhere around 200 calories (actually kcals) just by running 3 km horizontally. 200 kcals = 840 kJ.

 

[hackhard]

work done by minimum force required to push a block (60 kg) up an inclined plane (vertical height 200m , hori length 3000m)
then work by that force is (mg.sinx + umg.cosx) S
s is slant height u coeff of friction & x is incline angle

but in this case it takes forever for block to reach top

 

[Drakkith]

if u want to reduce problem to -
work done by minimum force required to push a block (60 kg) up an inclined plane (vertical height 200m , hori length 3000m)
then work by that force is (mg.sinx + umg.cosx) S
s is slant height u coeff of friction & x is incline angle

Unfortunately people aren't blocks so you can't really use this method.

 

[hackhard]

consider metabolism activity like sweating negligible as well :P

loss in chemical energy is not negligible as compared to loss in total energy
and negligible loss in chemical energy(sweating etc) would mean 0 calories burnt

 

[nasu]

You did calculate the work done against gravity correctly. The horizontal motion does not contribute to work done by (or against) gravity.
But, as other posts mentioned already, this work will give just a minimum amount of energy spent, assuming 100% efficiency of the body "machine".
We burn calories even when doing nothing, sleeping in the bed, not mentioning when we walk on a horizontal road.
So the actual number of calorie burnt may be several times higher than the minimum you get from just mechanical considerations.

 

[hackhard]

this work will give just a minimum amount of energy spent, assuming 100% efficiency of the body "machine".

even this loss in potential energy can occur without any need of burning calories for instance in elevator

 

[russ_watters]

There are websites that calculate calories burned when hiking that you can find via google.

 

[nasu]

even this loss in potential energy can occur without any need of burning calories for instance in elevator

And do you think this is relevant for the OP?

 

[hackhard]

probably op must consult a heath forum not a physics forum

 

[Stella.Physics]

You did calculate the work done against gravity correctly. The horizontal motion does not contribute to work done by (or against) gravity.

Yeah that's right about the work against gravity. So I guess the horizontal motion only contains effort and of course energy loss. I wanted to know about work vs calories and if there is a difference. But now I see that work is not the same as effort. Because work depends only from the difference of point A and B whereas effort counts all the way (like if I go around circles and then go to point B). That's why I assumed metabolism etc negligible.
I was just curious if I needed to know if the horizontal motion will contribute to the work from W=F dx and then the total work would be W=Fdx+mgh or if W=mgh is enough.

 

[CWatters]

I was just curious if I needed to know if the horizontal motion will contribute to the work from W=F dx and then the total work would be W=Fdx+mgh or if W=mgh is enough.

What would F be ?

Presumably F is some notional or equivalent "force required to walk horizontally". Thats mentioned in here but I haven't read it all...

http://www.sigmadewe.com/fileadmin/user_upload/pdf-Dateien/Bergaufgehen_engl.pdf

 

[Stella.Physics]

What would F be ?

I guess F is created by our bodies to move. So F is the effort we put to walk and is linked with the energy we consume :)

 

[PeroK]

I guess F is created by our bodies to move. So F is the effort we put to walk and is linked with the energy we consume :)

What about the energy consumed walking downhill?

 

[Stella.Physics]

What about the energy consumed walking downhill?

I think it would be a force of the same kind, against grtavity to hold your body from falling while in the previous case was the force to lift your weight against gravity. In a theoretical model it would be the same force but in different direction.
Now, as far as calories are concerned, i don't know which would cause more calories consumption but let's stick to the physics problem and not the "sports" aspect :P

 

[CWatters]

That's also mentioned in the article I posted above.

 

[Merlin3189]

... but let's stick to the physics problem and not the "sports" aspect :P

But it still depends on what you think is the physics problem.
If ur bf were a particle being pushed up a smooth incline, the physics work is just the mgh as you first calculated and there is nothing more to consider.
You have already raised the question of sideways movement, so you are not content with mgh. Once you start down this path, you are getting into the "sports" aspect, even for old codgers like me tottering to the shops at 2km/hr. Russ Watters' link mentions the work done, when walking or running on the flat, by raising and lowering your CG with each step. In any normal gait you must lift your leg & foot from the ground and lower it again to make a pace.
IMO this is the crux of biomechanical work. There is no mechanism for reclaiming energy that has been expended by a muscle (though some energy can be stored elastically), so every necessary movement does work and requires energy. Even relaxing a muscle causes some work to be done by the antagonist muscle. The question is, which movements you ignore.
By avoiding the "sports" aspects, presumably you ignore work done by muscles to operate heart, lungs and balance. Probably also work done by muscles purely in reconfiguring the body during different phases of gait. Ultimately you're back to mgh, but you have ignored the majority of work done by muscles in the body.

So I would say you need to consider these "sports" aspects, because that is where most of the energy goes and he could not move his mass from A to B without them. But they are the most difficult to calculate - and won't even be the same for him when he does it on different occasions. AFAIK the way this is usually calculated is by inference from the measured CO₂ expired during and after exercise compared with resting levels.

Incidentally, in your equation W=Fdx+mgh if Fdx is the work required to travel horizintally for a distance dx, then I would say it should be W=Fds+mgh where ds is the distance traveled along the route chosen.
Even if you climbed a ladder, you would have Fds for the height of the ladder, ds = h, and F = (the vertical force applied) - mg . The vertical force here must be greater than mg since you have to lift your leg on each step. Generally F is part of the "sports" aspect involved in bodily reconfiguration and applies throughout the distance actually traveled and of course also changes in magnitude depending on the direction of travel.

 

[Stella.Physics]

Because I want to avoid this F coming from the body and that it's difficult to define i just assumed a body (dimensionless for simplicity) that needs a known value force to move up an inclined floor with no friction. so this F should be greater in magnitude than the projection of weight on x-axis which should be B(x)=mg sin(θ) where θ is the degree of the inclination.
And then the work would be W=F dx where dx now is the hypotinuse.

So F=mg sin(θ) where sin(θ)=200/3006, where 3006m is the hypotinuse so sin(θ)=0.07
F=600*0.07=42 N
so the work would be W=Fdx=42 * 3006 = 126252 J ≅ 30.15 kcal

(use this site for conversion:
https://www.unitjuggler.com/convert-energy-from-J-to-kcal.html?val=126252 )

This result is again without friction but with friction it would be F=mg sin(θ) + μNcos(θ), where μ is the friction coefficient and N is the reaction to the surface due to the weight.And then the work would be W=F dx⇔W=mg sin(θ) dx+ μNcos(θ)dx.
With this approach we do not need the unknown F from the horizontal motion.

Assuming a human body with all metabolism and muscles working this F that I found would be multiplied with a human-body chemical-to-usable kinetic energy conversion coefficient (with obvious max value 1 for 100% efficiency) that would tell us the efficiency of human body energy conversion so we can calculate the total calories needed to move up the hill.

Taking data from:

http://web.applied.com/assets/attachments/492ACC9E-E5C2-2D43-0B8CCDA72ACE3361.pdf

for Rubber tire on pavement which would be our shoe of course the friction coefficient is around 0.85
sooooo W=mg sin(θ) dx+ μNcos(θ)dx⇔W=126252(from previously) + 0.85*600*(3000/3006)⇔ W=126252+509⇔W=126761 J≅30.27 kcal.

Now according to Wikipedia the energy conversion efficiency for the muscles is 14–27%.
So If I say that it is about 20% that means that this 126761 J is the 0.2 so the total work that the body needed in the first place was W=126761/0.2=633805 J≅151 kcal, which sounds right.

idk anymore I've been thinking over this matter for days now, I even had a dream about it :P

 

[Merlin3189]

This link http://www.sigmadewe.com/fileadmin/user_upload/pdf-Dateien/Bergaufgehen_engl.pdf[/PLAIN] from CWatters is really rather good and discusses your question in a lot of detail. I think you will find it helpful. It takes empirical data and makes equations to describe it, but also discusses possible underlying mechanisms for some of it.
I think its approach will give you more realistic results than these unrealistic physics models. Your addition of friction above seems more like the work done by someone dragging ur bf up the hill. If the coefft of friction were very small, say the road is covered in ice, you calculate that he will do less work running up the hill. Maybe? But I suspect the opposite is the case, if he can even get up it.
And how can you be doing work against friction when the direction of the frictional force is in the direction and sense that you are going?

 

[Stella.Physics]

And how can you be doing work against friction when the direction of the frictional force is in the direction and sense that you are going?

I assumed that friction and the weight component face at the same direction or "downwards" while we are going "upwards".

 

[Merlin3189]

If the friction force on BF is down hill, what is pushing him uphill?
If I stand still on a slope, gravity is pulling me down the hill, but what is balancing that?
If I stand on an icy slope so that there is very little friction, which way do I slide? Then when I slide off the ice onto a rough part, which way is the friction force on me?

I don't know the answers, I just ask what is on my mind. I'm always puzzled when an object pushes itself along, but friction is supposed to oppose its motion. Cars obviously do work against friction as they need petrol even moving on the flat, but if you remove friction by putting oil or ice on the road, they don't move. The ultimate example seems to be a baby sitting on the floor pushing itself backwards: friction is clearly pushing against the movement, as you can see its trousers or nappy being pushed down, but nevertheless it moves.)

 

[Stella.Physics]

If the friction force on BF is down hill, what is pushing him uphill?
If I stand still on a slope, gravity is pulling me down the hill, but what is balancing that?
If I stand on an icy slope so that there is very little friction, which way do I slide? Then when I slide off the ice onto a rough part, which way is the friction force on me?

the force that he creates with his muscles pulls him up.
what balance you is a force also coming from your body that is the same magnitude with your weight so that ΣF=0.
with the friction pointing downwards i guess you would start to accelerate downwards unless you oppose it with a force from your body. even if you change terrain friction will continue to face to the opposite direction from the one you are heading.

 

[Kaura]

Work is commonly referred to as a force applied over a distance

Right off the bat we have the force acting upon the climbing which is 600N

The distance they must climb can easily be solved using the Pythagorean Theorem

(200²+3000²)^1/2

This solves to a distance of about 3006.66 meters

We can solve for the amount of work done in joules by multiplying the two

(600N)(3006.66m) = 1803996Nm

The theoretical answer should be that about 1803996 Joules of energy are required to climb the mountain

This translates roughly to 431200 Calories

The issue is that there are a multitude of other variables that are unaccounted for

Humans tend to waste a lot of energy on other body process so finding an accurate answer would be somewhat difficult

Take my answer with a grain of salt because it is likely very far off from the actual value

 

[Stella.Physics]

Work is commonly referred to as a force applied over a distance

Right off the bat we have the force acting upon the climbing which is 600N

The distance they must climb can easily be solved using the Pythagorean Theorem

(200²+3000²)^1/2

This solves to a distance of about 3006.66 meters

We can solve for the amount of work done in joules by multiplying the two

(600N)(3006.66m) = 1803996Nm

The answer should be that about 1803996 Joules of energy are required to climb the mountain

I might be wrong on this so any feedback would be great

I'm afraid you can't do that because you have to analyze the weight to a vertical and a horizontal component. and the horizontal component will affect your movement "upwards".
You cannot put a force and a distance that are not in the same direction in the formula of work.

 

[Kaura]

I'm afraid you can't do that because you have to analyze the weight to a vertical and a horizontal component. and the horizontal component will affect your movement "upwards".
You cannot put a force and a distance that are not in the same direction in the formula of work.

That is one of the things I was wondering about

Would one be able to solve for the normal force after taking the angle of the mountain into account and then proceed to solve it?

Quick edit normal force is not technically the correct term I meant to refer to but I think you guys know what I mean

 

[Merlin3189]

I'm not sure how his muscles pull him up, or how forces come from your body. I can see how you apply forces to other objects, but not how you apply a net force to yourself.
Have you come across Newton's third law?

Anyhow, if I understand your answers, I should be able to run up an icy hill much easier than up a rough hill:

 

[Stella.Physics]

That is one of the things I was wondering about

Would one be able to solve for the normal force after taking the angle of the mountain into account and then proceed to solve it?

yes if you take the inclination into account. that's what I did above :)

 

[Stella.Physics]

I'm not sure how his muscles pull him up, or how forces come from your body. I can see how you apply forces to other objects, but not how you apply a net force to yourself.
Have you come across Newton's third law?

Anyhow, if I understand your answers, I should be able to run up an icy hill much easier than up a rough hill:
View attachment 97267

the third law says about action and reaction but of course these two are applied to different bodies. like i create and push the ground and the ground pushes back.
also in a theoretical model yes, because in reality you would slide down and the go back and try again so the effort would be much more. on ice there is less friction but then you have the slippery aspect :P and as we know we walk and move around thanks to friction, that's why you can't move up an ice hill because the friction is minimum

 

[Merlin3189]

the third law says about action and reaction but of course these two are applied to different bodies. like i create and push the ground and the ground pushes back.

And I think, the ground pushing back is the forces of friction and normal reaction. You pushing down on the ground is opposed by the ground pushing up on you, and you pushing backwards on the ground is opposed by friction pushing forwards on you. Which is where I start to get concerned about the work we do against friction when walking.

Anyhow, this is getting a bit away from ur original Q. My point was simply that the work done going horizontally was not work against ground friction, but work lifting legs and feet over the ground.