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I have been asked this question by school kids, colleagues and family (usually less formally):

When ascending a flight of stairs, you exchange mechanical work to attain potential energy ($W_{ascend} = E_{pot} = m \cdot g \cdot h$).

However, when descending, you have to exert an equivalent force to stop yourself from accelerating and hitting the ground (with $v_{splat} = \sqrt{2 \cdot g \cdot h}$ ). If you arrive downstairs with $v_{vertical} << v_{splat}$, you counteracted basically all of your potential energy, i.e. $\int F(h) \cdot dh = W_{descend} \approx E_{pot} = m \cdot g \cdot h$.

So is the fact that ascending stairs is commonly perceived as significantly more exhausting than descending the same stairs purely a biomechanical thing, e.g. having joints instead of muscles absorb/counteract kinetic energy? Or is there a physical component I am missing?

edit1:

I feel I need to clarify some points in reaction to the first answers.

A) The only reason I introduced velocity into the question was to show that you actually have to expend energy going downstairs to prevent ending up as a wet spot on the floor at the bottom of the steps.

The speed with which you ascend or descend doesn't make a difference when talking about the energy, which is why I formulated the question primarily using energy and mechanical work. Imagine that while ascending you pause for a tiny moment after each step ($v = 0$). Regardless of wether you ascended very slowly or very quickly, you would have invested the same amount of work and gained the same amount of potential energy ($\delta W = m \cdot g \cdot \delta h_{step} = \delta E_{pot}$).

The same holds true while descending. After each step, you would have gained kinetic energy equivalent to $E_{kin} = m \cdot g \cdot \delta h_{step}$, but again, imagine you take a tiny pause after each step. For each step, you will have to exert a force with your legs such that you come to a complete stop (at least in y direction). However fast or slow you do it, you mathematically will end up expending $W_{step} = \int F(h) \cdot dh = m \cdot g \cdot \delta h_{step}$.

If you expended any less "brake" work, some of your kinetic energy in y direction would remain for each step, and adding that up over a number of steps would result in an arbitrarily high terminal velocity at the bottom of the stairs. Since we usually survive descending stairs, my argument is that you will have to expend approximately the same amount of energy going down as going up, in order to reach the bottom of arbitrarily long flights of stairs safely (i.e. with $v_y \approx 0$).

B) I am quite positive fairly sure that friction does not play a significant role in this thought experiment. Air friction as well as friction between your shoes and the stairs should be pretty much the same while ascending and descending. In both cases, it would be basically the same amount of additional energy expenditure, still yielding identical total energy emounts for ascending and descending. Anna v is of course right in pointing out that you need the friction between your shoes and the stairs to be able to exert any force at all without slipping (such as on ice), but in the case of static friction without slippage, no significant amount of energy should be dissipated, since said friction exerts force mainly in x direction, but the deceleration of your body has a mostly y component, since the x component is roughly constant while moving on the stair (~orthogonal directions of frictional force and movement, so no energy lost to friction work).

edit2: Reactions to some more comments and replies, added some emphasis to provide structure to wall of text

C) No, I am not arguing that descending is subjectively less exhausting, I am asking why it is less exhausting when the mechanics seem to indicate it shouldn't be.

D) There is no "free" or "automatic" normal force emanating from the stairs that stops you from accelerating.

The normal force provided by the mechanic stability of the stairs stops the stairs from giving in when you step on them, alright, but you have to provide an equal and opposite force (i.e. from your legs) to decelerate your center of gravity, otherwise you will feel the constraining force of the steps in a very inconveniencing manner. Try not using your leg muscles when descending stairs if you are not convinced (please use short stairs for your own safety).

E) Also, as several people pointed out, we as humans have no way of using or reconverting our stored potential energy to decelerate ourselves. We do not have a built-in dynamo or similar device that allows us to do anything with it - while descending the stairs we actually have to "get rid of it" in order to not accelerate uncontrollably. I am well aware that energy is never truly lost, but also the "energy diversion instead of expenditure" process some commenters suggested is flawed (most answers use some variation of the argument I'm discussing in C, or "you just need to relax/let go to go downhill", which is true, but you still have to decelerate, which leads to my original argument that decelerating mathematically costs exactly as much energy as ascending).

F) Some of the better points so far were first brought up by dmckee and Yakk:

  1. Your muscles have to continually expend chemical energy to sustain a force, even if the force is not acting in the sense of $W = F \cdot s$. Holding up a heavy object is one example of that. This point merits more discussion, I will post about that later today.
  2. You might use different muscle groups in your legs while ascending and descending, making ascending more exhausting for the body (while not really being harder energetically). This is right up the alley of what I meant by biomechanical effects in my original post.

edit 3: In order to address E as well as F1, let's try and convert the process to explicit kinematics and equations of motion. I will try to argue that the force you need to exert is the same during ascent and descent both over y direction (amount of work) and over time (since your muscles expend energy per time to be able to exert a force).

Image 1: Center of gravity movement

When ascending (or descending stairs), you bounce a little to not trip over the stairs. Your center of gravity moves along the x axis of the image with two components: your rougly linear ascent/descent (depends on steepness of stairs, here 1 for simplicity) and a component that models the bounce in your step (also, alternating of legs). The image assumes $$h(x) = x + a \cdot \cos(2 \pi \cdot x) + c$$ Here, $c$ is the height of your CoG over the stairs (depends on body height and weight distribution, but is ultimately without consequence) and $A$ is the amplitude of the bounce in your step.

By derivation, we obtain velocity and acceleration in y direction $$ v(x) = 1 - 2 \pi \cdot A \sin(2 \pi \cdot x)\\ a(x) = -(2 \pi)^2 \cdot A \cos(2 \pi \cdot x) $$ The total force your legs have to exert has two parts: counteracting gravity, and making you move according to $a(x)$, so $$F(x) = m \cdot g + m \cdot a(x)$$ The next image shows F(x) for $A = 0.25$, and $m = 80 kg$. I interpret the image as showing the following:

  1. In order to gain height, you forcefully push with your lower leg, a) counteracting gravity and b) gaining momentum in y direction. This corresponds to the maxima in the force plotted roughly in the center of each step.
  2. Your momentum carries you to the next step. Gravity slows your ascent, such that on arriving on the next step your velocity in y direction is roughly zero (not plotted $v(x)$). During this period of time right after completely straightening the pushing lower leg, your leg exerts less force (remaining force depending on the bounciness of your stride, $A$) and you land with your upper foot, getting ready for the next step. This corresponds to the minima in F(x). Image 2: force for a = 0.25, m = 80 kg

The exact shape of h(x) and hence F(x) can be debated, but they should look qualitatively similar to what I outlined. My main points are:

  1. Walking down the stairs, you read the images right-to-left instead of left-to-right. Your h(x) will be the same and hence F(x) will be the same. So $W_{desc} = \int F(x) \cdot dx = W_{asc}$. The spent amounts of energy should be equal. In this case, the minima in F(x) correspond to letting yourself fall to the next step (as many answers pointed out), but crucially, the maxima correspond to exerting a large force on landing with your lower leg in order to a) hold your weight up against gravity and b) decelerate your fall to near zero vertical velocity.
  2. If you move with roughly constant x velocity, $F(x)$ is proportional to $F(t)$. This is important for the argument that your muscles consume energy based on the time they are required to exert a force, $W_{muscle} \approx \int F(t) \cdot dt$. Reading the image right-to-left, F(t) is read right-to-left, but keeps its shape. Since the time required for each segment of the ascent is equal to the equivalent "falling" descent portion (time symmetry of classical mechanics), the integral $W_{muscle}$ remains constant as well. This result carries over to non-linear muscle energy consumption functions that depend on higher orders of F(t) to model strength limits, muscle exhaustion over time and so on.
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    $\begingroup$ Your assumption "For each step, you will have to exert a force with your legs such that you come to a complete stop" is false. You land on each step, and the normal force provides the vertical deceleration necessary. Picture a Segway on (large) stairs - it's obvious that going up takes work and effort, but going down you can just coast and land on each step with a slight bump that requires no effort on your part beyond standing still. $\endgroup$ – SirTechSpec Sep 18 '17 at 15:45
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    $\begingroup$ Stand on one leg with your knee slight bent, as you would when going up- or downstairs, for 2 minutes and it'll become clear energy is being spent. :) The amount of energy does depend on the speed: it wouldn't for blocks or balls moving, but a tense muscle is spending energy even while there's no (macroscopic) displacement and no (macroscopic/visible) work is being done. $\endgroup$ – stafusa Sep 18 '17 at 15:51
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    $\begingroup$ As many hikers know from bitter experience long descents are actually worst (in a feeling the effort way) than long ascents. The bio-mechanics of walking uphill or climbing stairs use larger muscles than those of walking downhill or descending stairs. $\endgroup$ – dmckee Sep 18 '17 at 17:02
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    $\begingroup$ This is a physics forum but the question is much more about biology. You can go down a stair spending more or less or next or even no energy than when ascending it, depending on the grace and fitnes of your movements. When I was young I could go down stairs without ever slowing on any of the steps. Today I can't and descding stairs it is a real effort.. $\endgroup$ – TaW Sep 18 '17 at 19:55
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    $\begingroup$ Please do not let posts look like revision histories. This is not a discussion forum, if you find yourself in need of having to reply to answers at length three times in edits, you're doing something wrong. In particular, coming to this question without having read the answers and comments first, I have no idea what you are replying to in these edits, and all I see is a giant wall of text of a question at least half of which makes no sense without the context below. Questions must be questions, standing on their own. $\endgroup$ – ACuriousMind Sep 20 '17 at 10:07

15 Answers 15

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However, when descending, you have to exert an equivalent force to stop yourself from accelerating and hitting the ground...

Absolutely correct.

So is the fact that ascending stairs is commonly perceived as significantly more exhausting than descending the same stairs purely a biomechanical thing, e.g. having joints instead of muscles absorb/counteract kinetic energy?

Right. When going up the stairs, you must exert large forces by your large muscles. When your legs raise your torso, your muscles are supplying sufficient forces (with an energy cost) to do so.

When you go down the stairs, it is not the reverse of ascending. Instead of using your large muscles to decelerate, most people will take a straightened leg and plant it on the lower step. The deceleration is accomplished by plastic deformation in joints, fluid displacement in your foot, and the materials in your shoes and the floor. There is still some energy demand on the muscles for coordination and moving the legs, but it is significantly less than if the muscles were doing the deceleration job.

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    $\begingroup$ most of the "plastic deformation etc" uses scattered energy from the recovered from the potential. It is the skilled, directional use of the muscles to stop rolling that needs new caloric input, which is not very much. $\endgroup$ – anna v Sep 19 '17 at 3:56
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    $\begingroup$ "plastic deformation in joints" Really? That seems like it would be really quick to damage the body. $\endgroup$ – JMac Sep 19 '17 at 11:25
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    $\begingroup$ @JMac, "Plastic" as opposed to "elastic". The idea is that energy is dissipated, so any tendons or structures that act like springs and return energy don't help. But cartilage that deforms does. It's not meant to imply that damage is occuring. $\endgroup$ – BowlOfRed Sep 19 '17 at 15:32
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    $\begingroup$ As a perhaps illustrative "thought experiment", note that even a dead or unconscious human body is still capable of tumbling down a staircase (and coming to rest at the bottom). Thus, clearly, active muscle work is not needed to dissipate the potential energy gained from descending the stairs. (For a more practical experiment, substitute a sack of potatoes or whatever; almost any non-elastic material will have the same qualitative behavior.) $\endgroup$ – Ilmari Karonen Sep 19 '17 at 23:10
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    $\begingroup$ @DavidScarlett: Make sure it's a very small step, or just a horizontal walking stride. I worry that people could hurt themselves landing with joints locked even on a single step of a normal staircase if they really drop (not using the back leg to slow them before impact.) $\endgroup$ – Peter Cordes Sep 21 '17 at 6:14
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you have to exert an equivalent force to stop yourself from accelerating and hitting the ground

As animals, we spend calories to go up and gain potential energy. Tiredness is a measure of spent calories. Ideally going down needs no calories, and we have not evolved to the point of taking them back. Except a few calories are needed in the interaction with frictional forces and some skill to transfer the excess energy to the steps.

Think of skiing. To get up the hill on foot needs a lot of calories, (believe it or not back in 1958 I was taught to walk up with skis) to control the speed sliding down a few, and some skill (that is why at that course, I was fine going up, but ended in a splat at the bottom of the hill, having no skills). The energy is the return of the calories spent to go up (well partially, friction takes up a part of it).

Edit after edit of question:

The only reason I introduced velocity into the question was to show that you actually have to expend energy going downstairs

You are ab initio assuming that the velocity takes energy from your muscles. The going down velocity is sustained by the diminishing of the potential energy incrementally by going down a step. That turns into a velocity of your body, hitting the step a normal force bounces a ball back, you have to spend some muscle energy in order not to bounce, but in no way equal to the energy needed to carry your weight up one step.

I am quite positive that friction does not play a significant role in this thought experiment.

Wrong. Friction plays a very significant role in walking, climbing up or down. Have you tried walking on ice?

No, I am not arguing that descending is subjectively less exhausting, I am asking why it is less exhausting

It is less exhausting because less energy is needed from the muscles of the body, needed in directing the way of descent to control the release of energy from the incremental lowering of the potential energy of the body. Directing is much less energy absorbing than lifting.

There is no "free" or "automatic" normal force emanating from the stairs that stop you from accelerating.

You paid for it going up the stairs. The incremental velocity of lowering the body a step at a time hits the step and a normal force is created from the impact, not from the muscles. The muscles need to control against it so you do not bounce like a ball, but that is less energy than the potential step, because of friction taking up most of it.

Also, as several people pointed out, we as humans have no way of using or reconverting our stored potential energy to decelerate ourselves.

No, but our body is smart enough when in a velocity situation to spend a bit of muscle energy to direct where that velocity goes. The velocity coming from the acceleration of falling from step to step is transformed to friction (no slide shoes help) and a bounce of the body due to the normal force, all eaten up in friction and radiation. The new energy input is small with respect to the energy spent to get at a high potential. See the ski example above.

After the third edit, here is a simple example:

1) Take a half inflated ball that would bounce a few times and stop on a flat floor.

2) Lift it upstairs, next to the edge. Potential energy acquired.

3) Give it a small push just to fall on the next step: a tiny bit of energy expended.

It will bounce down the steps without any extra energy and, depending on how deflated it is, may reach the ground, or stop in between due to the normal force being larger than the gain of kinetic from potential energy from the fall of one step.

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    $\begingroup$ I think the skiing example shows exactly why you need to brake - otherwise you convert basically all of your potential to kinetic energy. My argument is that braking takes mathematically exactly the same amount of energy. $\endgroup$ – Daniel Sep 18 '17 at 15:24
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    $\begingroup$ @Daniel and my argument is that you recover the energy spent and use it to brake , as you would do with skiing skills, just spending a little new energy by turning your ankle to dig in deeper and brake the velocity. it is the original energy spent to reach the higher potential, given to the braking friction,. $\endgroup$ – anna v Sep 18 '17 at 16:47
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    $\begingroup$ a simpler example : a bucket of water has potential energy at the top of the hill, pour it and it spends the potential energy running down hill. In the case of the stairs we do not roll down, but use up the potential energy incrementally step by step with friction and back scattering of feet on the step. Some calories are spent in muscles stepping down, but no way as much as in going up. $\endgroup$ – anna v Sep 18 '17 at 17:30
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    $\begingroup$ @Daniel no it doesn't, friction will take care of it for you $\endgroup$ – PyRulez Sep 19 '17 at 1:14
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    $\begingroup$ Wouldn't most of the friction when walking stairs both up and down (at least the friction between your feet and the steps) be static friction, meaning it only transfers potential and kinetic energy between the earth and the person, and not into heat? On the other hand, I have no idea whether there friction internally in the legs. $\endgroup$ – Arthur Sep 19 '17 at 11:25
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  • Your muscles exert more force when ascending than descending:

When going downstairs they have to exert a force smaller than gravity's in order to control your speed, while when going upstairs, the force they exert must be at least equal to your weight, in order for you to ascend. So your muscles are doing more work ascending than descending, the movements are typically not symmetrical.

That's especially true because the braking force (for the "fall" from step to step, provided by the step's normal force) is not a reaction from a force exerted by your leg muscles - you can hit the step straight-legged, and let the impact energy dissipate passively through your body, spending very little energy in the process, as well explained in BowOfRed answer.

  • The natural energy losses help you keeping a comfortable speed going downstairs, while that's a loss you have to compensate for when going upstairs.

  • And, yes, there are also certainly some biomechanical aspects at play too. Consider, e.g., how much more tiresome it is to descend in slow motion: going downstairs very slowly is hardly any easier than going upstairs at the same speed - it increases the symmetry between both movements.

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  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$ – ACuriousMind Sep 20 '17 at 10:09
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It is biomechanical.

Well, it is enthropic.

Gravitational potential energy is really high quality (low entropy) energy. Converting it to doing nearly arbitrary work is really easy.

When we go down, we convert that gravitational potential energy into heat by soaking it with our elastic bones and ligaments. This is an easy conversion, as we are going from low entropy energy to high entropy energy.

Now, some muscle work is done beyond simply absorbing the shocks; this keeps us balanced and in control as we descend.

Going up, energy-wise there is nothing stopping us from cooling our muscles and ligaments and bones and using it to spring up the steps, generating gravitational potential energy. But that would violate the laws of thermodynamics, namely converting high-entropy energy to low-enthropy energy.

Instead, we are forced to convert our stored chemical energy -- ATP and others -- into kinetic energy, which we then turn into gravitational potential energy.

Our ATP (and other stored chemical) energy reserves are depleted, and we feel tired.

The biomechanical way this is realized involves how we climb and descend; you could probably make some creature who isn't very efficient at descending and uses muscles the entire way.

There are people who go "down stairs" by sliding down a bannister and only burn energy to generate friction against the bannister. This is probably the most efficient way for someone to go down stairs.

Fundamentally one cannot climb as efficiently as one can descend.

Energy isn't used, it is transferred and converted. "Available" energy is high-quality low-entropy energy. You never "spend" energy on something (other than creating rest mass if you don't talk about mass-energy equivalence), instead you convert low-entropy energy into a mixture of low-entropy energy of a different form, and higher entropy-energy "loss".

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It might be the case that what I have to say has already been said in other answers implicitly but I am posting this answer as I don't see an answer that is manifestly similar to the one that I have in mind.

While going upstairs, the earth-person system gains potential energy. This increment in potential energy must come from the biochemical energy of the person. Thus, while going upstairs, the person needs to work at least of the amount of the increment in the potential energy of the earth-person system.

Now, while going downstairs, the earth-person system loses potential energy. Thus, this lost potential energy ought to go somewhere. The first place for it to go is in the macroscopic kinetic energy of the person. Upto this point, it is completely clear that the person doesn't spend a penny out of her biochemical energy. But, we require that the person shouldn't gain any macroscopic energy. So, we ought to redistribute the energy that is released from the earth-person system into some other forms. This redistribution is done by the normal reaction forces between the legs of the person and the stairs. They redistribute the energy into vibrational motion of the steps and partially into the vibrational motion of the molecules of the person's legs etc. But this is just the redisribution of the energy. The person doesn't need to spend any of her biochemical energy at all. Actually, if the person were to spend some energy then there would be an additional requirement of redistributing this additionally spent energy.

I have ignored the inefficiency losses etc. which can be reasonably assumed to be the same while going upstairs or downstairs.

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The answer is simple:

-> Going up is done by muscle work.

-> Going down is (mostly) done by shock absorbing.

Explaining:

When going up, one bends its knees and then must use a considerable amount of force (depending on one’s weight) to get his leg straight and lift himself to the next step.

Going down (ideal, simplified case), first, one uses gravity to straight up his leg and then relaxes the muscles in the other leg and starts falling. Before he gains a dangerous falling speed (depending on the height of the steps) the straight leg hits the next step and all energy is dissipated by your body shock absorbing systems.

In other words, going down is made of little jumps. I call it ideal case, since this configuration results in the least use of muscle power to go downstairs. In reality, however, one still uses some muscle energy to straight up his leg, keeping it stiff, etc. still, that is considerably less than the energy required to lift yourself up.

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Work done is equal to force exerted times distance moved in the direction of the force.

You're right that (to a first approximation) the forces exerted when going up and down are the same: in both cases (again, to a first approximation) you have a body moving at constant velocity – whether up or down – subject to gravity, so there must be an upward force matching the force of gravity.

The problem is that, when ascending, the muscles (tendons, ligaments, bones etc - the entire "machine" of the body) are exerting a downwards force while moving upwards, so they are losing/expending energy; when descending, the force is still downwards, but now the movement is downwards too, so the muscles (etc) are receiving/gaining energy.

Now, as you know, muscles can't work in reverse: they're good at converting chemical energy to mechanical energy, but you can't put in mechanical energy and get chemical energy back out. But that doesn't mean they can't take in energy: they can, and they do this by getting warm.

It's also true that muscles require energy to operate, whether they're doing any useful work or not. But it's not true that the energy required for a muscle to exert a specific force is constant: very roughly, there will be an overhead of "wasted" energy $W(F)t$ for a given force over a given time, plus any work done by the muscle through movement $F\cdot x$. If the muscle isn't moving (think pushing against a brick wall), you use just $W(F)t$; if you are doing actual work (so the muscle is moving by contracting) it's $W(F)t+F\cdot x$. The wastage will likely be similar going up and down stairs, but the work done by the muscles will not.

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  • $\begingroup$ This is the right answer. OP is confused by ignoring the sign of the energy. Basically OP is saying "up or down: same force, same distance, so same work". But it should be "same force, opposite distance, so opposite work". $\endgroup$ – Marc van Leeuwen Sep 22 '17 at 4:48
  • $\begingroup$ Very true: When walking downstairs one needs to shed excess energy! (Or one would, in the OP's words, end up as a wet spot at the bottom.) That can be a significant challenge, for example when running down a steep slope. Some survivors of the 1996 Mt Everest disaster basically slid down the slopes to the camp. They could never ever have climbed the same way up: They were too exhausted. $\endgroup$ – Peter A. Schneider Sep 22 '17 at 10:11
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For the stairs example I was thinking in terms of torque.

When you want to ascend, yo place your leg bent in the upper step and then you take impulse to rise the other leg to that very palce or even the next step. When you do that, you must compensate the torque that gravity is producing on your former knee.

However, when desceinding, the gravity does help that torque for reaching the lower step.

I don't know if this is right but this is what came to my mind.

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When you descend, you transfer the energy, you do not need to supply (almost) anything. What little energy you do need to spend is that required to handle and control the transfer (and the descent); the rest comes from the gravitational potential energy, and will be transferred as mechanical energy and/or dissipated as heat. Mechanical transfer in joints and muscles may lead to trauma which can be perceived as kin to tiredness or fatigue.

You could in theory recover energy when descending, but really you don't. The best you can do is recycle some elastic energy from one step to propel the next step (there are several climbing-down techniques that teach how to move to do this as gracefully, safely, quickly, or cheaply as possible. Braking by flexing and extending with the upper leg - the latter against gravity - is more expensive than absorbing the shock with the foot and the lower leg and slinking down from one step to the next).

A lot of energy gets dissipated in the soles of the shoes (try going down a long stair with wooden slats instead of running shoes, with your leg muscles having to pick up the slack), in whatever covers the steps themselves, in the steps themselves if they're elastic enough, etc.

So while you can descend efficiently or not so efficiently, and get tired and/or aching going down too, the energy you spend going down is but a fraction of what you need to go up, when you have to supply that gravitational potential energy from your own chemical stores.

If you were a perfectly rigid body on perfectly rigid stairs, with a piston damper in both knees, you'd expend very little energy to slide forward and fall on the next step, and then you'd fall on it, the dampers absorbing the shock and dissipating it as heat.

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  • $\begingroup$ Energy is not created or destroyed. All energy use is transfer. $\endgroup$ – Yakk Sep 18 '17 at 18:12
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I think the question can be simplified by asking considering the difference between the upwards and downwards part when doing squats.

Let's first consider a very simple model: A vertical spring hanging from the ceiling and a mass hanging from the spring which pullsthe spring downwards. When the mass goes downwards, the potential energy on the spring increases. When the mass goes upwards, the potential energy on the spring decreases. That's even though in both cases, the spring exerts the same forces. Force isn't work. The dot product of force and displacement is work.

In other words: When a spring (or a muscle) exerts a force, it doesn't necessarily mean it does any work. It does work on an outside object only if the force moves something.

Now go back to a real muscle. Like the spring in our example, a human muscle does work when it shortens, and the work is positive because the force exerted by the muscle is in the direction of the displacement.

Your legs are wired in such a way that when doing squats upwards, you can shorten certain muscles and your legs straighten. So, as I explained, when going upwards, muscles do mechanical work.

When going down, the force is in the same direction but the displacement is opposite. Therefore, when going downwards, mechanical work is done on the muscles. This might be hard to grasp, but now comes the biomedical part: Unlike the spring, the human muscle can't store the energy it gains this way and the energy just turns to heat. In addition, because of how the cells in the muscle actually work, tensed muscles need to generate heat even when they are static or lenghtening. That's why going downwards needs energy.

You can try this at home. (It might be easier to observe if you use a huge extra weight you're not used to, but I don't recommend that for medical reasons.) If you do squats very slowly, the energy needed to generate heat because biomechanical reasons dominates, and going down feels almost as hard as going up. If you do squats very fast, the energy needed to generate mechanical work dominates, and going down feels a lot easier.

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    $\begingroup$ Might the downvoter explain the reason? $\endgroup$ – JiK Sep 19 '17 at 19:58
  • $\begingroup$ As someone who squats, I think this is the best answer to the question, but due to the biological nature of the problem, Physics Stack Exchange may not have been the best venue for the question. $\endgroup$ – David Scarlett Sep 20 '17 at 2:47
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There are significant edits to your post, so I have to make significant edits to address them because the question is interesting.

Let's cut to the heart of your question.

To go from point A, bottom of stairs to point B, top of stairs we must exert $mg\Delta h = mg(B-A)$ energy to do so. In climbing we will convert chemical energy of the body/muscles to do this. To descend, as you've noted, there is no way to descend the stairs without dissipating $mg(B-A)$ energy. It is physically impossible to translate some height without discharging at least $mg\Delta h$ in energy. The question is, how much of $mg(B-A)$ does my body need to provide in the form of chemical energy?

Say I take a rope and pulley with a brake to limit my rate of descent. The friction between the brake will dissipate at least $mg\Delta h = mg(B-A)$ worth of energy. Turning mechanical friction into heat.

Say you jump (off a cliff of equal height). Then your body will absorb $mgh$ and you'll probably break things or die.

In the above two examples, your energy output was negligible. The key here is something else dissipated the energy and it was necessary for at least $mgh$ to be dissipated even on descent where "gravity works in your favor." The thing I'm trying to illustrate is that you can descend without exerting much of your own energy. So how does this happen without a pulley, or jumping?

The energy you want to dissipate will be dissipated in each step by using the mechanics of your body tissues. A proportion of the normal force exerted by the stair on your joints, bones, muscles, tendons, etc... will all compress and rebound dissipating the energy as heat. Which your body will then radiate away. If you think this is not significant, drop a brick or a piece of wood, and see how long it bounces. If it doesn't keep bouncing forever, then that means the energy is dissipated by the material itself, via compression and rebound. This energy is dissipated by inter-molecular and atomic forces.


Original analysis (pre edits)

Energy (Non-rigorous Analysis)

Climbing

In order to climb the stairs, 100% of the energy provided to vertically climb must be provided by your body.

$E_{\text{climb}} = E_{\text{pe}} = mgh$

Descending

In order to descend the stairs, you only need to provide a small vertical climb (to break your foot free from friction) and then a small amount of energy to pivot your leg forward. Gravity takes over from there. Let's assume you "step" 1/100th the height of the stairs to initiate a descending step, then:

$E_{\text{descend}}\approx \frac{1}{100}mgh$

Clearly, from the simplified mechanics described above $E_{\text{climb}} >> E_{\text{descend}}$.

Naturally, other forces are involved. You will use your leg muscles to resist falling down the stairs, however, you can see that you are taking advantage of the stored up potential energy of your vertical height to use to descend.

Net Energy (More Rigorous Analysis)

Based on the paragraph above we can see that we've made assumptions and not really made a rigorous model to account for all factors. It was a simple mental experiment to quickly show we are probably on the right line of thinking. Therefore, a better analysis will look at the entire system such that conservation laws hold.

$E_{\text{net}} = 0$

Net energy for climb

The following net energy equation of the system will show better how human energy relates to climbing energy. Let's break the model into four parts: net energy ($0$), potential energy ($mgh$), energy output of the human, and any gravitational energy ($E_{\text{extra}}$) that we can use to help us.

$E_{\text{net,climbing}}= E_{\text{human}} - E_{\text{pe}} + E_{\text{extra}}$

In a climb, our $E_{\text{extra}}=0$ as we cannot use gravitational energy to help us (that is, nothing "pushes" us up).

(1) $E_{\text{human,climb}} = E_{\text{pe}}$

Net energy for descent

Clearly, in descent, we can convert part of the potential energy to do work for us. We can use gravitational energy to help us as it pulls us where we want to go.

$E_{\text{net,descending}}= E_{\text{human}} - E_{\text{pe}} + E_{\text{extra}}$

Here, our $E_{\text{extra}}\gt0$ as some gravitational energy can be converted/harnessed to help us descend.

$ (2) E_{\text{human,descend}}=E_{\text{pe}}-E_{\text{extra}}$

Clearly, $(2) \lt (1)$ because $E_{\text{extra}}\gt0$.

Power vs Energy

Speaking about velocity does certainly change the model. Primarily, introducing the rate at which you descend or climb the stairs means we are now talking about power which is:

$P_{\text{stairs}} = \frac{E}{t}=\frac{mgh}{t}$

If we cut our climb time in half, then we double the required power.

$P_{2}=\frac{mgh}{0.5t_{1}}\rightarrow P_{2}=2P_{1}=2\left(\frac{mgh}{t_{1}}\right)$

This is why running up the stairs will be more exhausting than a leisurely walk.

(And, interestingly, power is the reason why you will go splat if you try to take a shortcut down really tall stairs. While $\Delta E$ is constant, as $\Delta t$ approaches zero, you will find that you have serious problems.)

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Think about the energy, ascending stairs require energy from you to push yourself up. this energy is stored in potential energy form. However, in case of descending, each step you took is all about transferring your potential energy to the stair (not back to your body).

In summary, you loss energy (calories from food) when you ascend. And you loss (almost)nothing when you descend.

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  • $\begingroup$ No, the energy largely does go back to your body when you descend - just not in a useful form (you don't get more energetic, just hotter!). $\endgroup$ – psmears Sep 19 '17 at 10:21
  • $\begingroup$ @psmears : in what form? it doesnot go back and stored as bodily energy(calories). neglecting muscular friction and other losses, the energy largely transfer to the earth(through the force exerted on stairs). $\endgroup$ – someone_ smiley Sep 20 '17 at 8:02
  • $\begingroup$ In the form of heat - you get (marginally) warmer. There's not really any energy transferred to the Earth via the force exerted on the stairs, because energy transfer = (force x distance moved in the direction of the force), and the stairs don't really move. $\endgroup$ – psmears Sep 20 '17 at 8:22
  • $\begingroup$ Stair does move, only it is negligible due to gigantic size of the earth compare to our body. if our size and mass were comparable, it will be observable $\endgroup$ – someone_ smiley Sep 20 '17 at 8:46
  • $\begingroup$ energy transformation from PE can be in form of work done on moving earth, vibration wave energy when come in contact, heat due to friction as we move down the stair, but not back to body. $\endgroup$ – someone_ smiley Sep 20 '17 at 8:49
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To exert a force and to strain your muscles is not the same thing. Even if you completely relax, work will be needed to move your limps around. This work is exactly what creates the force slowing you down when you're going down the stairs.

Of course, you still need to strain your muscles when going down, in order to control your trajectory and speed. But when you're going up, that work needs to be done in addition to the work required to lift your weight.

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I'm going to provide another answer, because none of the existing answers seem to succinctly address energy efficiency.

Let's say your muscles are 25% efficient. This seems to be on the generous side as cycling and rowing are probably more efficient uses of your muscles compared to walking, where you have to exert more effort to retain your balance and absorb shock.

So going up a hill, you are actually going to expend 4 times the energy climbing with your legs as the actual amount of potential energy you gain. Three parts of that is in the 75% inefficiency, generating heat in your body, and the last part is the 25% which goes into actual potential energy.

Now let's consider going down. If you walk backwards down the hill you are using all the same muscles and you're going to make roughly the same motion. I walked up and down and nearby steep hill like this to confirm this. Now walking down the hill we know that you have to generate at least the amount of potential energy at the top in order to wind up at the bottom with no increase in speed. But that's all the energy you need to generate for a simple backwards walk down the hill! All your muscle energy is specifically for the purpose of dumping potential energy and converting it to heat.

So walking up is going to take at least 4 times as much energy from your body's stores as walking down. It may be more, because there are ways you can dissipate the potential energy more efficiently - it's called being less efficient at using your muscles! If your muscles are only 16% efficient (the low end cites in the linked page) that going uphill will take 6.25 times as much energy. If you slide part of the way down the hill it will take even less energy as you are dissipating the energy as heat from friction and not in your muscles.

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    $\begingroup$ You missed a very big point about biological inefficiency: muscles burn energy even when no physics work is done. They burn energy even when negative work is done. Sometimes a lot (trying doing negatives at the gym!). You're not looking at a proportionality. The claim here "So walking up is going to take at least 4 times as much energy from your body's stores as walking down." is built on a misunderstanding. That's the basic problem with this question: you can't understand the situation without understanding more biology than most physicist ever learn. $\endgroup$ – dmckee Sep 20 '17 at 5:09
  • $\begingroup$ @dmckee I disagree. The efficiency numbers quoted were measured via actual oxygen consumption. This would only be done during the actual exercise. Yes, some of that inefficiency is due to base metabolism, but it's in the context of exertion, not at rest. We aren't concerned about energy flows outside the window of time that the actual work is being done. $\endgroup$ – Michael Sep 20 '17 at 16:07
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Simple. There's a constant force of 1g pulling you down.

(yeah, it depends on distance from Earth etc etc... but a simplified example is enough for an explanation)

So if you want to ascend at, let's say half a g, you need to produce 1.5 g of force, 1g of which goes just to cancel the gravity pull.

If you need to descend at the same acceleration (half a g) you need to produce just half a g of force - to cancel half a g of gravity.

So, 0.5 g to descend, 1.5 go to ascend.

For other desired accelerations (say, 0.1 g, 0.05g etc) you can do the math.

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  • $\begingroup$ I don't think that's it, see point A) in my edit... if I kept overcompensating g (by whatever tiny amount) each step, I would accelerate indefinitely. $\endgroup$ – Daniel Sep 18 '17 at 15:21
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    $\begingroup$ "1.5g of force" probably shouldn't be written on physics.stackexchange. $\endgroup$ – Eric Duminil Sep 18 '17 at 16:58
  • $\begingroup$ -1 Descending stairs is easier than ascending even during a long distance with constant speed (that is, no acceleration). $\endgroup$ – JiK Sep 18 '17 at 17:32
  • $\begingroup$ g is a unit of acceleration, not force and definitely not speed. If you're traveling at a constant speed, then you have acceleration of zero g. There will be a small acceleration (but nowhere near .5) at the beginning, and a small acceleration in the other direction at the end. If you start at rest and end at rest, your average acceleration must be zero. $\endgroup$ – Acccumulation Sep 18 '17 at 18:06
  • $\begingroup$ If I walk up or down a flight of steps, then most of the time, I'm not actually accelerating or decelerating - at least, not by much. If it were the acceleration that caused exhaustion, it would be no more tiring to walk up 50 flights of steps than to walk up just one. $\endgroup$ – Dawood ibn Kareem Sep 19 '17 at 2:12

protected by Qmechanic Sep 18 '17 at 14:48

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