Haven't yet begun to dig into the guts of this one, but I do see a sticking point: The author if this CQ is looking for an equation for time t for both the ascent and descent. The descent portion should be easy considering the acceleration will be provided by gravity. The ascent portion depends on the force applied by the person giving the sled an uphill shove.

We do have a starting point for our equation: t = t_ascent + t_descent.

t(accelerate)+t(ascent)+t(descent)?

Regards JD

The question only asked the ascent time.

If t starts at the end of the quick push, (we'll call that point a) the the time to reach the top of the trajectory is 1/2t. Obviously, the time to come back down to point a would also be 1/2t. Its velocity at point a on the way back down will be the same as that imparted by the quick push.

We are assuming that the runners have equal forward and reverse friction coefficients.

This is gibberish.

Actually it's not entirely gibberish, but it answers the question for only for the frictionless condition. If the sled/snow interface is frictionless, then after you release the sled on its way up, it will decelerate according to the component of gravity along the slope (G x cos Θ). Eventually, it will stop, and then start coming back down, with the same acceleration (and therefore it will return at the velocity with which it left).

Without actually doing the math, and with befuddled mind, I assumed that the friction, being the same for going up as for going down, would simply cancel out. But, of course, that's wrong. The friction converts some of the sled's kinetic energy to heat on the way up (and on the way back down). The friction increases the deceleration on the way up and decreases the acceleration on the way down. Given that the up distance and down distance are equal, the time going up must be less than the time going down.

Tkot's explanation is very good. Coincidentally, his typo relates directly to my brain fade. If:

• Moreover, accelerations (using Newton's law a=∑F/m) are obviously:
• a_up = g (sinθ + μ cosθ) (3) and
• a_down = g (sinθ + μ cosθ) (4)

were true, then a_up and a_down would be equal, and my original answer would not be so nonsensical.

The aggregate of tkot's answers is correct and mathematically complete (subject to the limiting assumptions specified in the challenge itself) - but I suspect more explanation may be needed.
I'm hoping someone else will provide this.

fyz

I'm sure there's a μmgCos(θ) in there somewhere, but I can't stop my sled from sliding.. Somebody make the nasty F and sliding friction coefficients go away. My impulse force and static resistance is all jumbled up. I'm not even sure if force was applied horizontally or along the slope, and even worse I don't know if it matters.

a_down = g (sinθ + μ cosθ) (4)

It is obvious I actually meant:

a_down = g (sinθ - μ cosθ) (4)

How do you get those sexy mu & theta symbols? ffeJ

Hi, tkot!

You stated "If there is no friction (μ=0) then t_up = t_tot / 2, as expected. Otherwise, t_up < t_tot / 2 because β > α for any logical θ (i.e. 0 ≤ θ < π/2)"

I respectfully submit that it would be impossible for t_total to be anything divided by 2 since we don't know the velocity of the upward trip (pushed), and the downward trip is extremely likely to be faster, multiplied by 9.8 m/s/s.

Mark

Without friction, the situation is the same as the basic trajectory of a ball thrown upward -- it returns to your hand with the same velocity with which it departed (assuming you don't move your hand after release). Gravity decelerates the ball (on the way up) with exactly the same magnitude that is accelerates it (on the way down).

It that were not the case, then energy mysteriously disappears. (With friction, however, some energy is converted to heat -- and thus the unequal travel times when friction is accounted for.)

BTW, my post 16 is a correction to my post 3, in which I proposed something fairly close to your logic.

I like jdretired's response [#2], becaue t_total includes the push or initial acceleration phase of the sled's t_up.

Mark

Of course, a problem with jdretired's response is that it does not answer the question, which asks for the time to the top of its trajectory, not total time.

However, should we include the initial acceleration time? My feeling is no, because we cannot know what that time might be. Also, it makes the end of the thought experiment indeterminate: if you included the push, then you should probably include the catch and deceleration, so the the sled ends up stationary at start and finish.

I prefer the simpler approach of tkot's solution, which highlights the essential nature of the problem (the part which I failed to appreciate in my first response), which is that the friction causes the problem to be asymmetric (whereas without friction, the time up and time down are the same).

If one wanted to include acceleration before release and deceleration after catch, it could be a footnote to tkot's solution -- but I don't think it adds anything other than distraction.

Hi, Blink!

"if you included the push, then you should probably include the catch and deceleration, so the the sled ends up stationary at start and finish."

The initial question did include a push, but it didn't include deceleration and a catch phase, only asked for the time taken to return to the initial point. I can see though, that the riddler may not have intended to include the initial push as a part of the calculations: that's murky to me; and my clue that it might not have been intended to be included was being given a slope at a constant angle as a determining factor. I think the constant angle thing is an interesting inclusion for consideration though, since regardless of the angle or curve of the slope the answer might still be the same.

Mark

I can see though, that the riddler may not have intended to include the initial push as a part of the calculations: that's murky to me;

It's murky to me too. As in so many of these challenges, the "answer" depends heavily on one's interpretation. Tkot's and my interpretations were the same, but his answer benefited from his actually thinking things through. (I consider that cheating!)

We'll all be up a creak if the official answer is something like "4.2 seconds".

I don't agree "for any logical θ (i.e. 0 ≤ θ < π/2)"

If the sled goes down,

m.g.sinø>m.g.µ.cosø → tanø>µ → arctan(µ) < ø ≤ π/2

m.g.sinø>m.g.µ.cosø → tanø>µ → arctan(µ) < ø ≤ π/2

No doubt about that, it is what I mentioned in another post that it should hold:

sinθ - μcosθ > 0 => tanθ > μ.

This is the condition so that the sled comes back down the slope. (One may argue that it is in fact tanθ > μ', where μ' > μ is the static friction coefficient.)

As for the condition 0 < θ < π/2 (the "logical" θ I mentioned) is for stupidly ensuring:

cosθ>0 => sinθ+μcosθ ≥ sinθ - μcosθ (equality holds if μ=0) => β≥α and consequently t_up≤ t_tot/2.

Anyway, it would be absurd if θ lied outside these limits. Certainly the forces applied to the sled wouldn't be the same in such cases. Therefore, the condition 0 < θ < π/2 is already implied in the solution.

Whatever, the point I wanted to make is that t_up≤ t_tot/2 with the equality being valid for μ=0.

tkot,

You are very close to the answer ...

Abe

The time for the sled to reach the top of the trajectory is simply a function of the vertical acceleration of gravity which is equal to the vertical component of the force. So just take the sin of theta times the force and use that result for the force/acceleration equation.

What about the friction?

Friction is just a part of the deceleration equation for the cycle.

I'm all but lost on this one, but isn't he asking the time it will take to reach the TOP of the trajectory? This doesn't even touch on the descent. Unless friction is Zero and the slope is great enough to only deal in gravity, then rate of descent could be applied. I've pushed sleds up a hill and they ascend quickly, but with a slight slope may take a while to even change direction and accelerate downward.

I look forward to this solution! Just trying to put my 2 cents worth in.

Yes, the challenge is asking how long it will take to the top from when you release it to start moving under its momentum.
It then tells you the time for the return trip back to the release point. Based on that, the angle of the slope and the coefficient of friction, it asks you to calculate how long the first leg of the trip (bottom to top) will take.

tkot assumes a distance 's' up the slope that the sled travels, and from that calculates the up and the down times (based on knowing the angle of the slope and the coefficient of friction). As the ratio of the up and the down times turns out to be independent of the distance 's', he can use this to calculate the up time in terms of the round trip time.

I get:

t_up = √(sinΘ - 2μcosΘ)

Anybody confirm?

Obviously a lot missing - it bears no relation to the total time.

Component of acc along the snow(incline plane) = g * sin(w), where w = angle

Force due to sled mass normal to snow = m * g * cos(w)

Friction on sled = u * m * g cos(w), where c = coeff of friction

Total acc acting on sled along snow = g sin(w) - cmg cos(w)

Using eqn's of motion v=u+at, v^2 = u^2 +2as, s = ut + .5at^2

v = 0 = end speed, u = start speed, a = acc, t = time

t = u / (g sin(w) - cmg cos(w))

ffeJ

It will reach it's trajectory about 5 seconds after I fall and bust my butt for the third time and I fling the dang thing as far as I can throw it.

How come nobody has touched on the fact that the coefficient of friction between the sled and the snow changes as the velocity approaches zero?

Just trying to stir it up...

Clearly, constant coefficient of friction is a simplifying assumption to allow thought experiments.
If the sled is running on snow then the kinetic energy transferred to the displaced snow will depend on the sled's velocity. Running water-over ice is also a complex situation.

I'd just go with the (ice?) flow.

Hi, Tkot!

The snow's being compacted by the sled's runners on the way up will reduce the co-of-f for the ride down directly in the old tracks. If the sled continues to sink into those tracks on the way down, the f forces due to meeting new snow that has to be crushed into the track will be greater than the f forces between what one could call non-reactive (just a track, no new sinking) snow & runners. These are two considerations that add to the idea that the reason the co-of-f for the snow was a given constant is that it isn't meant to be a timing consideration in the difference between the rides up and down in this ...er ...challenge...um...once we...uh ...puzzle it out.

Yeeee Haawwww!! Ouch! Youch! Wheee! Ooops! YOW! (shouldn't 'a had that hot chocolate & cigarette before I started this run. Shoulda listened to the wife...just like the last thousand times she was right!) Groooaaaan! BLURP!

Mark

Hi, Physicist? !

I think we only have to consider Heisenberg for very teeny tiny sleds that have trouble staying in their tracks.

And perhaps we don't have to quite go so far as to worry about 'thermal vibration' (which variety were you referring to?) either, since the inertia-breaking impetus for the sled comes from a push whose duration lasts for d₁, or some such, or is just accepted as the time taken for the state of the sled to gain velocity and momentum at which instant the pusher lets go and it begins to lose both on the uphill grade.

Hopefully, the sleighride is being repeated so that the runners of the sled don't have to create new tracks.

But more succinctly, I think the snow postulated by the challenge is lovely, flat, non-tracking supporting stuff that has a co-of-f of nothing of importance so that it won't present a problem in arriving at the real answer by being different on the way up and on the way down.

In fact, the very snow in this question is only a metaphor to place the object (sled) in this question at the base of a slope, impart a momentum to it (which we differ about re its origin's force, duration, and importance), and time its upward and downward movement.

What I've been purporting all along is that t_up = d₁+ (t_total - t_down). You can substirute various equations to describe t_down. d₁ is , although unknown, a given; but without it, t_up becomes mythical (instantaneous push) and leads to hypotheses that describe the trip to the top of the slope and back as being of equal duration. I think somebody even mentioned the good ol' trajectory of a thrown baseball as an analogy.

Without getting anal , one could merely (just throwing tkot's very fine analysis into this particular pot) conclude, for example, that another solution to the question might be something like

"If (μ=0) then t_up = [d₁ +] ( t_tot / (1 + [(tanθ + μ) / (tanθ - μ)]^1/2)) ."

Why? Because...

The initial question, "You give a sled a quick push up a snowy slope that makes an angle θ with the horizontal. In time t, the sled goes up and comes back down." seems to me to indicate that the push must be a part of the equation. It takes place at the beginning of the incline and thus occupies a place in the movement of the object up the slope; but the common thought expressed in this blogset depends upon measuring the termination of the "down" component at the same spot as the end of the push.

To me, unless the push had indicated

• a running start before the incline and
• ended with microscopic precision exactly as the object is fully engaged on the incline,

which it did not; the object definitely has

• farther to come down while expressing a normal rate of descent, and
• the time taken to go up will be differentiated by the duration of the push.

The question did NOT say that the push comes from a running start before the slope. It clearly states "push up ... the slope". The push impetus could even be imparted in the middle of an incline, and would still have to be taken into account, since the only given point of distance and time starts from the beginning of the point "quick push".

Mark

...oops! Didn't mean to let that show. (**emarassed blush**)

I think the friction was given as a constant so it could be ignored in the equation. The mass of the sled and the distance it travels are also the same in both directions. t_total is time both up and down. It'll be divided into t_up and t_down .

t_down is the square of time as applied to the velocity [9.8 m/s/s acceleration] over the common distance powered by the force of gravity acting on falling articles on the way down. The time on the way down can then be deducted from t_total and the remainder will be the time up. t_up = t_total - t_down.

Mark

If you could ignore friction, then time up would equal time down. That's the theory I advanced in my first post. However, you cannot ignore friction, because is shortens deceleration time (time_up), and increases acceleration time (time_down).

That t_up = t_total - t_down goes without saying, regardless of friction or lack thereof.

It's hard to find much wrong with tkot's explanation.

I guess the problem is a simple one if we're abstracting the pushing sequence during the sliding (up or down), the fundamental dynamics equation is: m[a]= µ[N]+m[g] - [] is for vector where [N] is the normal force resultant, meaning m.g.cos(θ) we can reduce the problem by projection on the tangent direction, making it: m.at = -µ.m.g.cos(θ) - m.g.sin(θ) it is then a simple integration of a linear diff equation: Vt= Vot - g(µ.cos(θ)+sin(θ)).t where Vot is the initial tangent speed (when you released the sled) the maximum height (and distance) is reached when Vt is 0, meaning Vot/(g.(µ.cos(θ)+sin(θ))) seconds after you let the sled go. you can see that if your slope is negative (going down), the sled will stop if the snow is too "sticky". Think I have it right... well, hope so, anyway :)

Hi, Guest! (BTW, joining this blogsite is free, and rewarding )

I had thought that the total sled trip was meant to include the initial push:

But I may have been incorrect, and the trip really was intended to be calculated starting from the point of the end of the push and ending there as well. Guess we'll find out sooner or later what the riddler meant.

Mark

Hi, Physicist? !

"as no data is given regarding the characteristics of the push, you can't say anything about the timing of the push"

No, you can't. But as the push is an unknown quantity that forms a part of the total unknown quantity of time of the total sleighride, about which --after all-- nothing concrete except formulae may be said, it's fair to include it. To declare it intantaneous by default is only --and perhaps justifiably (and perhaps not)-- avoiding a complication.

Mark

Hi, Physicist? !

Quite right about that curve, and apologies for the slap dash diagram. Here's a somewhat more considered version. I wasn't able to show the reduction of time down with a visually reduced t_down line on this portrayal of the ride, so I showed it in red and labelled it as 'incremental'. t_up is in blue and is meant to include the push curve:

Hopefully, the colours will show up when this is posted.

Mark

That is a bit close to what I feared. Once the push is finished, and the sled is moving uphill, the velocity will be reducing (not constant), and (given that we are told it does slip downhill again), the downhill velocity will increase continuously.

BTW, if we include the push time, t_up can be either less than or greater than t_down, depending on how hard you can push.

Hi Physicist? !

"the velocity will be reducing (not constant)"

Er...yup.... .

Consider the diagram so-altered.

Perhaps now an overall time derivation equation can be shown to cover the impetus, slope, t_up deceleration, and t_down acceleration back to the initial point of impetus followed by subraction of t_down to get t_up. I think all the variables are in. The diagram, although flawed, is still helpful.

Your go. In general, I prefer appreciating other people's math to writing it. Too long out of school, and too far away in scope of interest now to jump on it joyfully as I once did.

Mark

When I attempted to take a crack at this, I came up with the same answer, although via a different route. I also solved for the total time it takes the sled to stop at the top of its run, but did not used any integration. If T=0 after being released,

Velocity = (Starting Velocity at T=0) + (acceleration x time)

then T = (Starting Velocity) / (Acceleration), when velocity is zero at top of run.

The acceleration on the way up is

(frictional force + (gravitation force x cos(θ)))/mass, or

g(μ cos(θ) + sin(θ)), therefore

Time to top = starting velocity / (g(μ cos(θ) + sin(θ))

I also reached the same answer if solving for time for complete trip with sled reaching original position using same 'up the hill' acceleration for the complete trip, and then taking half that value. I know that's cheating a bit, and that the time it takes the sled goes up and down are different.

Friction is directed against the motion so it would also have a downward component on the way up. This can be solved using an incremental caclulation with small time steps. I'll try to recreate my golf ball trajectory with air resistance using a moving coordinate system in Mathcad and post it in a couple days.

I am stuck on something.

In Blink's original answer, he said that t_up = t_total / 2, which he now discounts because of tkot's answer. According to Blink's later comments, tkot's answer showed the effects of friction would make the times of ascent and descent assymetical, but tkot does state that if friction is zero, then t_up still equals t_total / 2.

The only way I can see t_up = t_total / 2 is if the upslope acceleration equals the acceleraton of gravity. Don't we have to apply a force stronger than gravity to start the sled moving uphill? Won't the force applied vary from person to persont? I know a weigthlifter could apply more force to the sled than I could. Wouldn't the applied force have to generate an acceleration of 2g for t_up = t_total / 2?

CR4 needs a head-scratching smilie!

"The only way I can see t_up = t_total / 2 is if the upslope acceleration equals the acceleraton of gravity."

It does not have to be the acceleration of gravity, but up slope acceleration must be the same as downslope acceleration, as in the idealized case of no friction.

"Don't we have to apply a force stronger than gravity to start the sled moving uphill? Won't the force applied vary from person to persont? I know a weigthlifter could apply more force to the sled than I could. Wouldn't the applied force have to generate an acceleration of 2g for t_up = t_total / 2?"

That's why I considered the time to start (T=0) at the release of the sled. There is no way of knowing for how long or how strong the sled was pushed. I know everyone dislikes assumptions, but they are a necessary evil at times.

Well... That's at least how I did it.

In Blink's original answer, he said that t_up = t_total / 2, which he now discounts because of tkot's answer. According to Blink's later comments, tkot's answer showed the effects of friction would make the times of ascent and descent assymetical, but tkot does state that if friction is zero, then t_up still equals t_total / 2.

My initial solution would be right in the case where mu = 0, but incorrect in all other cases. The assumption here is that t_total starts at the release point of the sled, and ends when the sled crosses that same point on the way down.

I think that's what the riddler had in mind, but I'm not sure. If it starts at the beginning of the push, then where does it end? If the sled is not stopped on the way down, then it could keep going for quite a while, depending upon terrain. I think the riddle is intended to catch unsuspecting fools like me who jump to the conclusion that the friction cancels out. That is does not cancel, should be more obvious than it was to me (and when I first read tkot's explanation, I thought he was over-complicating things, and I had to draw up little triangles and do some pondering to prove the error of my ways) -- maybe it was something I ate! (I wasn't drinking; I am almost a t_totaler!)

Wouldn't the applied force have to generate an acceleration of 2g for t_up = t_total / 2?

Given tkot's and my assumption that the time starts at release, then any positive acceleration prior to that will impart a velocity at the start point

BTW, I think if one assumes that the push is included, then the problem becomes extremely complicated: When does the release occur? Is the initial acceleration linear? When does t_total end?

A minor point; On a snowy slope the sled will not stop at the place it was pushed from, it will keep going down hill and accelerating due to gravity.

Hi, Guest! (same guest???)

Which is precisely why IF the push (impetus) were to have been factored in, the return trip to the starting point will have accelerated over the push distance, and take a shorter time than t_up.

While pushing a sled uphill in the first place is kinda strange, let's at least assume that the push starts from the bottom of the hill, and when the sled returns there it eventually comes to a point where its momentum won't carry it farther. Perhaps we can also assume that the snow changes --on the flat-- to the friction variety we used to get our footing when we gave it that instant motion.

Mark

t1 = t / (1+sqrt((cos(theta)+mu*sin(theta))/(cos(theta)-mu*sin(theta))))

I assumed an initial velocity then determined distance the sled traveled upslope to it's stopping point, eventually arriving at an inverse square relationship between acceleration i.e. force/mass and time elapsed.

Cheers, Sully

t = (time to drag sled up hill) - {(time taken to huff and puff and sweat)+(time taken to have a quick cigarette) + (time taken to explain to kid that this is far enough up hill)+( time taken to tell the misses that your going as fast as you can and that the Hot Chocolate and marshmallows in the chalet isn't going to run out any time soon).

t=t1- (t2+t3+t4).

Who cares how long it takes as long as you have fun doing it.

In case you get bored:-

http://linerider.com/play-line-rider-online

Going up

F = Fpp + Fr1 = - m.g.sinø – m.g.µ.cosø

a = F/m = - g . (sinø + µ.cosø)

V(t) = Vo – g.(sinø + µ.cosø).t

Tu : time going up

V(Tu) = 0 ; then

Vo = g.Tu.(sinø + µ.cosø)

X(t) = Vo.t + ½.a.t²

X(t) = g.Tu.t. (sinø + µ.cosø)-1/2.g. (sinø + µ.cosø).t²

X(Tu) = Xm : (X on top)

Xm = ½.g.Tu².(sinø + µ.cosø)

Going down

F = Fpp – Fr2 = - m.g.sinø + m.g.µ.cosø

a = - g.(sinø - µ.cosø)

X(t) = Xm – ½.(sinø - µ.cosø).g.t²

Td : time going down when the sled reaches it's orignal position

X(Td) = 0

½.g.Tu².(sinø + µ.cosø)-1/2.gTd².(sinø - µ.cosø) = 0

Tu² = Td².(sinø - µ.cosø)/ (sinø + µ.cosø)

K = (sinø - µ.cosø)/ (sinø + µ.cosø)

Tu² = Td².K

Working, bla, bla, bla …

if the sled goes down: arctan(µ) ≤ ø ≤ π/2 ; 0 < K < 1

T : total time

T = Tu + Td

Tu² = K.(T-Tu)²

(1 – K).Tu² + 2.K.T.Tu – K.T² = 0

Tu = [- 2.K.T ± (4.K².T² + 4.(1 – K).K.T²)^½]/(1 – K)

Remember:

K = (sinø - µ.cosø)/ (sinø + µ.cosø)

arctan(µ) ≤ ø ≤ π/2 ; 0 < K < 1

Just one option makes sense:

Tu = K^½.T.(1-K^½)/(1 – K)

From your expression:
Tu² = Td².K
we also use:
T_total = Tu+Td = Tu(1+1/k^1/2)
Tu = T_total/(1+1/K^1/2)

Same result as you give (which is itself the same as ktot), but a simpler route (and also a simpler final presentation). ktot (eventually) progressed it to a solution that used the parameters of the challenge...

Distance that sled travel up slope is same distance that it would travel under constant acceleration of (gsinθ + gµcosθ) for time = Tup with 0 initial velocity.

Although not obvious at first, indeed, we can "reverse" the upward motion and come to the formula
s=1/2g(sinθ+µcosθ)t_up² , in which case, there is no need to take into account the initial velocity! (In fact both formulas s = 1/2a_upt_up² and s = v_upt - 1/2a_upt_up² are equivalent, as v_up=a_upt)

Beauty lies in simplicity, so you get my vote!

The only I could add, is that if you convert the formula to the equivalent:

t_{up =} t_tot / (1+ [(tanθ+μ)/(tanθ-μ)]^1/2)

then you have less calculations and trigonometric look-ups to make.

Very nice. Logical, clear, concise. Welcome to CR4!

Agreed - I gave the first of the ratings.

My post #2 was a somewhat cynical response to a challenge that is open to interpretation, though I have found that others, with regard to other challenges, have been able to pin point the essence of what is necessary to answer the challenge?

I have not been able to post an answer, because, to just consider the point of release and return as the same point in space, I don't find convincing. The force 'g' that de-accelerates the sledge is the same force that accelerates the sledge after it comes to rest. So the velocity of the sledge on the way down passing the point of release would be the same as the velocity that was imparted to it on the way up. The challenge does not state that the sledge stopped at this point only that an overall distance was covered in time 't', that distance is unclear? In my mind I see a ratio between the time to accelerate and the time to cover the same distance accelerated by 'g', and this I cannot resolve?

Regards JD.

Maybe the problem is that you are reading secondary interpretations rather than the original posting?

The post says: "In time t, the sled goes up and comes back down". To be pedantic, it could say "The sled goes up the slope and returns to the release point in total time t", but there is nothing about "an overall distance".
Other than that, I think that (like many other respondents) you are ignoring a (probably necessary) pair of conventions in the way challenge questions are set:
i) if it is possible to interpret the question in a way that the question contains all the information needed to solve it, that is the intended interpretation (note that the previous challenge, "night fishing", did not allow this); and
ii) a void interpretation of the challenge is not valid (in this case that would be t being the time for the sled to go in each direction separately). {I.e. it is not a childrens' riddle}

Now to some more detail: other than friction, the post says nothing about any constraint on the sled on the way down - so we assume that it just goes up and comes back under the action of the forces specified in the challenge - gravity and friction. That would cause it to be moving with some velocity as it passes the release point. Clearly, the direction of the frictional force changes when the sled changes direction - so in the presence of friction the times up and down have to be different. [This supports the interpretation of t as being the time for the return journey].

I hope that accepting the above conventions and readings now makes a "natural" interpretation clear; but I may have missed something. If so, let me know and I'll try again.

Regards

Fyz

i) if it is possible to interpret the question in a way that the question contains all the information needed to solve it, that is the intended interpretation.

Brilliant: general purpose "rule" for these challenges.

Guys and Gals, like the math so far, but there's a couple of challenges that might be hidden in the question. (Look up "Hysteresis" for details of energy loops with inbuilt losses and also "damped oscilations".)

We don't get much snow here is Aus, but I understand the way a sled works is to slide on a microscopic film of "water" between the rails and the solid snow/ice.

Friction in a "non lubricated" condition is normally assumed to be independant of velocity and that seems the way that everyone is going so far.

If the question was posed by someone from a fluid mechanics background, friction in a "lubricated" condition is different. Not my field, but I suspect in lubricated sliding the friction is dependant on velocity (or even velocity squared) as it would be dealing with the shear forces in the "lubricant".

Everyone is also assuming the sled is not "pushing" through soft snow on the way up and then returning in the previously compacted track and that seems in keeping with the information in the question.