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 ($W_\text{ascend} = E_\text{pot} = m gh$).
However, when descending, you have to exert an equivalent force to stop yourself from accelerating and hitting the ground (with $v_\text{splat} = \sqrt{2 g h}$ ). If you arrive downstairs with :$$v_\text{vertical} \ll v_\text{splat}$$ you counteracted basically all of your potential energy, i.e. $$\int F(h) \cdot \mathrm dh = W_\text{descend} \approx E_\text{pot} = m g h$$
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?
Edit-1:
I felt 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 whether you ascended very slowly or very quickly, you would have invested the same amount of work and gained the same amount of Potential energy ($\delta W = m \cdot g \cdot \delta h_\text{step} = \delta E_\text{pot}$).
The same holds true while descending. After each step, you would have gained kinetic energy equivalent to $$E_\text{kin} = m \cdot g \cdot \delta h_\text{step}$$ 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 $$W_\text{step} = \int F(h) \cdot \mathrm dh = m \cdot g \cdot \delta h_\text{step}$$
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).
Edit-2: 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:
- 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.
- 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 $F_1$, 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).
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 roughly 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
$$\begin{align}
v(x) &= 1 – 2 \pi \cdot A \sin(2 \pi \cdot x)\\
a(x) &= -(2 \pi)^2 \cdot A \cos(2 \pi \cdot x)
\end{align}$$
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 $m = 80\ \mathrm{kg}$. I interpret the image as showing the following:
-
In order to gain height, you forcefully push with your lower leg,
a) counteracting gravity
b) gaining momentum in $y$ direction.
This corresponds to the maxima in the force plotted roughly in the center of each step.
- 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 $F(x)$.
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:
-
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 $$W_\text{desc} = \int F(x) \cdot \mathrm dx = W_\text{asc}$$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
b) decelerate your fall to near zero vertical velocity.
- 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: $$W_\text{muscle} \approx \int F(t) \cdot \mathrm dt$$ 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 $W_\text{muscle}$ 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.
Best Answer
Absolutely correct.
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.