# My Mistake? Understanding Friction Force & Work Done on Snowy & Icy Surfaces

• songoku
In summary: I feel like I'm missing something. Can you explain why the tension is 200 N in part 1 if it's not accelerating?In summary, the problem is about determining which stage of the sledge's journey has the most work done on it. The options are being carried back 20 m at a constant speed, being lifted up 1 m, being pulled 10 m across ice, or being pulled 10 m across snow. The correct answer is (D) being pulled 10 m across snow. The mistake in the conversation is assuming that the sledge is not accelerating on the ice, when in fact it must accelerate due to the frictionless surface. The correct free body diagrams show that
songoku
Homework Statement
A sledge of mass 50 kg sits on a snowy surface. It is pulled horizontally for 10 m against a frictional force of 200 N, then it is pulled horizontally across ice for 10 m. There is no friction between the ice and the sledge. It is lifted up vertically by 1 m and finally carried back at a constant speed to where it started. During which stage of its journey is most work done on the sledge?
A being carried back 20 m at constant speed
B being lifted up 1 m
C being pulled 10 m across ice
D being pulled 10 m across snow
Relevant Equations
W = Fd

W = mgh
The answer is (D), but I don't understand why.

Option (A) is wrong because the work done = 0. Then, I divide the motion into 3 parts:

1) motion on snowy surface
Since the sledge is being pulled horizontally (let assume to the right), there will be tension force T to the right and friction force of 200 N to the left so the work done will be (T - 200) x 10

2) motion across ice
The sledge is still pulled horizontally so I assume the value of T is the same as previous motion so the work done is T x 10

3) lift up the sledge
Work done = GPE

I assume again during motion (1) the sledge is not decelerating so T is at least 200 N so the most work done is during motion (2) so my answer is (C).

I can not imagine a case where the work done in motion (1) is more than in motion (2).

Where is my mistake? Thanks

songoku said:
Homework Statement: A sledge of mass 50 kg sits on a snowy surface. It is pulled horizontally for 10 m against a frictional force of 200 N, then it is pulled horizontally across ice for 10 m. There is no friction between the ice and the sledge. It is lifted up vertically by 1 m and finally carried back at a constant speed to where it started. During which stage of its journey is most work done on the sledge?
A being carried back 20 m at constant speed
B being lifted up 1 m
C being pulled 10 m across ice
D being pulled 10 m across snow
Relevant Equations: W = Fd

W = mgh

The answer is (D), but I don't understand why.

Option (A) is wrong because the work done = 0. Then, I divide the motion into 3 parts:

1) motion on snowy surface
Since the sledge is being pulled horizontally (let assume to the right), there will be tension force T to the right and friction force of 200 N to the left so the work done will be (T - 200) x 10

2) motion across ice
The sledge is still pulled horizontally so I assume the value of T is the same as previous motion so the work done is T x 10

3) lift up the sledge
Work done = GPE

I assume again during motion (1) the sledge is not decelerating so T is at least 200 N so the most work done is during motion (2) so my answer is (C).

I can not imagine a case where the work done in motion (1) is more than in motion (2).

Where is my mistake? Thanks
Draw a free body diagram of the sled (not sledge) in part 1, where the sled is being pulled through the snow. Then draw another FBD of where its being pulled across the ice. If it's not accelerating ( based on the answer this is what I expect - which we shouldn't have to do to solve the problem IMO) in either case, which case has larger tension in the rope?

One issue I can see is they failed to state whether or not its accelerating as its being pulled across the ice.

songoku
erobz said:
One issue I can see is they failed to state whether or not its accelerating as its being pulled across the ice.
If one assumes that ice is frictionless, the sled must accelerate as it is being pulled across the ice.

I think this problem is unclear in that (like many of its kind) it does not specify the force that is doing the work.

Now, if we assume that it is the external force ##F## exerted by the person doing all the moving, we also have to consider at what happens at the boundaries. Let's say that the sled is stopped before it starts the next leg of its journey. Then one can use the work-energy theorem with ##~\Delta K =0~## to find the net work done by ##F## during each leg. Specifically, the work done by ##F## during the ice leg is zero because there is only one horizontal force acting on the sled. Similar reasoning can applied to the other legs. I will let @songoku do it.

songoku and nasu
kuruman said:
If one assumes that ice is frictionless, the sled must accelerate as it is being pulled across the ice.
Tension in the rope pulling the sled (presumably-in the idealization) can go to zero?

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EDIT - strikethrough portion I wish now I had not posted:

I believe you are all over-thinking this. The problem is obviously a didactic exercise to teach -
(1) horizontal motion without friction does no work on the sledge, and
(2) vertical motion even without friction does do work on the sledge.

PS: sledge
noun:
(1) a vehicle of various forms, mounted on runners and often drawn by draft animals, used for traveling or for conveying loads over snow, ice, rough ground, etc.
(2) a sled.
(3) British. a sleigh.

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songoku and erobz
erobz said:
Tension in the rope pulling the sled (presumably-in the idealization) can go to zero?
And what stops the sled before it is lifted up? What stops the sled from moving up and gets it moving backwards? If we are to compare work done over the different legs, I think it is important to account for the kinetic energy, if any, at the boundaries and the forces that change this kinetic energy.

There is no mention of rope in the statement of the problem so "person" is just as good an assumption.

songoku
erobz said:
Draw a free body diagram of the sled (not sledge) in part 1, where the sled is being pulled through the snow. Then draw another FBD of where its being pulled across the ice.
FBD of the sled in part 1 is tension to the right and friction to the left.

FBD in part 2 is tension to the right.

erobz said:
If it's not accelerating ( based on the answer this is what I expect - which we shouldn't have to do to solve the problem IMO) in either case, which case has larger tension in the rope?
If it's not accelerating in part 1, then the tension is 200 N.

How can we determine the tension in part 2?

erobz said:
Tension in the rope pulling the sled (presumably-in the idealization) can go to zero?
So you mean in part 1 there is tension and in part 2 the tension is zero?

But I won't interpret "then it is pulled horizontally across ice for 10 m" as the tension being zero because I think it is not possible to pull with 0 N.

kuruman said:
Now, if we assume that it is the external force ##F## exerted by the person doing all the moving, we also have to consider at what happens at the boundaries. Let's say that the sled is stopped before it starts the next leg of its journey. Then one can use the work-energy theorem with ##~\Delta K =0~## to find the net work done by ##F## during each leg. Specifically, the work done by ##F## during the ice leg is zero because there is only one horizontal force acting on the sled. Similar reasoning can applied to the other legs. I will let @songoku do it.
In this scenario, I understand why work done by ##F## during ice leg is zero because it means ##F## must accelerate and also decelerate the sled so the sled can start from rest and also end at rest. No change in KE means work done by external force = 0.

gmax137 said:
I believe you are all over-thinking this. The problem is obviously a didactic exercise to teach -
(1) horizontal motion without friction does no work on the sledge, and
Sorry I don't understand this part. If the sledge is being pulled across frictionless ice, surely there should be work done on the sledge by the pulling force?

Thanks

songoku said:
Sorry I don't understand this part. If the sledge is being pulled across frictionless ice, surely there should be work done on the sledge by the pulling force?
There are no retarding forces, the tension in the rope ( or the in the arms or tongue of the person ) can go to zero in the idealization across the ice if the sled is not accelerating.

songoku
erobz said:
There are no retarding forces, the tension in the rope ( or the in the arms or tongue of the person ) can go to zero in the idealization across the ice if the sled is not accelerating.
Then the word "pulled horizontally" has no meaning if the tension can go to zero?

As @kuruman often says, this is going back to mind-reading exercise (sigh)

songoku said:
Then the word "pulled horizontally" has no meaning if the tension can go to zero?

As @kuruman often says, this is going back to mind-reading exercise (sigh)
This is an idealization. Frictionless ice. I think common parlance often fails in describing these purely theoretical scenarios. Language was developed to talk about reality...mostly? I don't think there is much to be gained in being pedantic about it.

songoku
erobz said:
I don't think there is much to be gained in being pedantic about it.
I agree. It would help this thread to know the level of the class or book that provided this problem. The level of the answers should be commensurate. Physics (teaching) is rife with elementary explanations in intro courses that are refined in later more advanced classes.

songoku and erobz
kuruman said:
And what stops the sled before it is lifted up? What stops the sled from moving up and gets it moving backwards? If we are to compare work done over the different legs, I think it is important to account for the kinetic energy, if any, at the boundaries and the forces that change this kinetic energy.

There is no mention of rope in the statement of the problem so "person" is just as good an assumption.
To @gmax137 point in #11. I'm sure you could go on and on about all the nuances. I doubt that depth of analysis was the intention of the problem given the answer?

I was damned if I didn't say something about the possibility that it could be accelerating across the ice (or snow - clouding judgement), and now I'm damned that I did (because I didn't go far enough)...gotta love physics!

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songoku
gmax137 said:
I believe you are all over-thinking this. The problem is obviously a didactic exercise to teach -
(1) horizontal motion without friction does no work on the sledge, and
(2) vertical motion even without friction does do work on the sledge.

PS: sledge
noun:
(1) a vehicle of various forms, mounted on runners and often drawn by draft animals, used for traveling or for conveying loads over snow, ice, rough ground, etc.
(2) a sled.
(3) British. a sleigh.
The motion with or without friction does not do work. Work is done by forces on objects not by types of motions.
If you lift the sledge vertically with constant speed the net work on the sledge is zero. The work done by either gravity or the lifting agent is not zero. Same for the motion on the portion with friction. The net work on the sledge can be very well zero too, if the motion is not accelerated.
It is not a good idea to teach things like these (1 and 2 are not even wrong), better teach the actual physics.

songoku
It is a sledge, not a sled. No self-respecting sled could have a coefficient of friction as high as this one. Nor would anyone carry a sled down a hill. You ride sleds down hills for goodness sake!

songoku said:
I assume again during motion (1) the sledge is not decelerating so T is at least 200 N so the most work done is during motion (2) so my answer is (C).
I get (C) as well.

songoku said:
A being carried back 20 m at constant speed
B being lifted up 1 m
C being pulled 10 m across ice
D being pulled 10 m across snow
A - no. Constant speed = zero work.
B - no. Constant speed = zero work. Work done by lift is equal and opposite to work done by gravity. Net zero. Nobody told us to consider only one force doing the work. Nobody told us to account for gravity as a potential rather than as a force. Alternately, one kg-gee over one meter is less than 4 kg m/s^2 over 10 meters.
C - yes. It is accelerating = positive work. Nobody told us how hard the pull was, so I assume same as when sledge was on snow.
D - no. Likely constant speed = zero work. In any case, less acceleration than on the ice.

Sure, they meant "how much work was done by the propulsion mechanism" for each step. And sure, they meant for us to assume constant speed during every step, not just the return. But dammit, they need to write a decent question.

songoku and erobz
kuruman said:
Now, if we assume that it is the external force ##F## exerted by the person doing all the moving, we also have to consider at what happens at the boundaries. Let's say that the sled is stopped before it starts the next leg of its journey. Then one can use the work-energy theorem with ##~\Delta K =0~## to find the net work done by ##F## during each leg. Specifically, the work done by ##F## during the ice leg is zero because there is only one horizontal force acting on the sled. Similar reasoning can applied to the other legs.

The similar reasoning doesn't apply to the vertical lift "leg" though, does it? It seems to me, that portion of the journey (B: being lifted up one meter) is the most unambiguous case of work being done on the sledge. OK, with the additional assumption that this scenario takes place in a gravitational field. Or am I missing something?

nasu said:
If you lift the sledge vertically with constant speed the net work on the sledge is zero.

I have to admit I do not understand this. Are you discounting the increase in potential energy of the sledge?

gmax137 said:
I have to admit I do not understand this. Are you discounting the increase in potential energy of the sledge?
The force of gravity can do work. We are free to regard gravity as a force and to disregard the associated potential, looking only at the forces that exist and the kinetic energy that results.

We can write:$$W = (F_\text{lift} + F_\text{gravity}) \cdot \Delta S$$and$$W = \Delta KE$$.Or we can write:$$W = F_\text{lift} \cdot \Delta S$$and$$W = \Delta KE + \Delta PE$$

songoku
But the question asks "During which stage of its journey is most work done on the sledge?"

gmax137 said:
But the question asks "During which stage of its journey is most work done on the sledge?"
And the answer is "it's ambiguous". Make a set of disambiguating assumptions. Write them down. Determine the result. Turn the whole thing in. That is what my teachers always told us to do in the face of ambiguity.

OK, but if we assume the journey occurs here on the earth's surface, can we not say for B, the work done on the sledge is 50 kg * 9.8 m/s^2 * 1 m ~500 Nm? I take "being lifted up 1 meter" means exactly what it says, and no more -- the velocity at the beginning and end of the lift is zero.

gmax137 said:
OK, but if we assume the journey occurs here on the earth's surface, can we not say for B, the work done on the sledge is 50 kg * 9.8 m/s^2 * 1 m ~500 Nm? I take "being lifted up 1 meter" means exactly what it says, and no more -- the velocity at the beginning and end of the lift is zero.
It is ambiguous.

Yes, we could say that. We could regard gravity as a potential field counting toward energy gained rather than as a force counting toward "work absorbed".

Or we could choose to regard gravity as just another force doing (negative) work as the sledge is lifted.

songoku and gmax137
jbriggs444 said:
C - yes. It is accelerating = positive work. Nobody told us how hard the pull was, so I assume same as when sledge was on snow.
Why not zero work for C? Whatever kinetic energy the sledge acquires moving horizontally has to go to zero at the end of the 10 m leg before it is lifted up. That requires negative work to be added to the positive work. We cannot ignore the sledge's gain in kinetic energy if net positive work is done on it during a leg when we want to compare with the work done by whatever during the next leg.

This is not a clearly formulated problem especially since the force doing the work is not specified. I give this problem a D.

songoku, jbriggs444 and gmax137
kuruman said:
Why not zero work for C?
kuruman said:
Whatever kinetic energy the sledge acquires moving horizontally has to go to zero at the end of the 10 m leg before it is lifted up. That requires negative work to be added to the positive work.
That could be impulsive work done at the corner of the trajectory. I would resist regarding impulsive work at the corner as part of the more vanilla work done during the preceding slide across the ice.

kuruman said:
We cannot ignore the sledge's gain in kinetic energy if net positive work is done on it during a leg when we want to compare with the work done by whatever during the next leg.
We have four legs on the trajectory and three (maybe four) corners. The problem should have allowed for possibility of work being done or extracted at each of the three (or four) corners.

kuruman said:
This is not a clearly formulated problem especially since the force doing the work is not specified. I give this problem a D.
Yes, I agree.

songoku
Nothing is constant, and nothing that pulls the sled has a constant response over a varying load. If animal is pulling it at some constant speed and the friction were to drop it’s going to let up unless it’s being told otherwise. If I'm pulling the kids around (we don't have a hill in the yard...) and it gets easier, I can assure you I take a break! It’s hardwired into us to be efficient. The work over the ice leg can be zero, and the work at the very end of the ice leg the negative its kinetic energy. It’s not absurd to assume the tension goes slack the moment it hits the ice and it just cruises along at constant velocity either.

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songoku
erobz said:
It’s not absurd to assume the tension goes slack the moment it hits the ice and it just cruises along at constant velocity either.
Yes, it is not absurd. However, it is absurd though to say that at the end of the displacement on ice the kinetic energy goes to zero without the expenditure of work. Of course, one can always dream up a a quarter-circle frictionless ramp that changes the velocity from horizontal to vertical. However in that case, the person will work that is less than ##mgh## because the initial kinetic energy is traded for potential energy as the sledge rises.

Conclusion: Poorly formulated problems are lose-lose situations.

nasu, songoku and erobz
Thank you very much for the help and explanation erobz, kuruman, gmax137, nasu, jbriggs444

erobz
gmax137 said:
I have to admit I do not understand this. Are you discounting the increase in potential energy of the sledge?
I said "net work". The work done by gravity is negative and can be described by an increase in potential energy. But this not the net work. It must be another force acting against gravity. This second force does positive work on the sledge. The sum of the two is the net work. If the sledge accelerates upwards, the net work is positive. If it moves up with constant velocity the net work is zero. In both cases the work done by gravity is the same, for a given height. The work done by the other force depends on how do you do the lifting.

gmax137
"During which stage of its journey is most work done on the sledge?"
Since the work done on a mass is its increase in KE, the issues are:
• are we to take each stage as an indivisible whole or are we to consider the maximum work done over any portion of each stage?
• what is its speed at each stage transition?

Only the net work is equal to the change in kinetic energy. But the problem does not even tell which work is asking for.

nasu said:
Only the net work is equal to the change in kinetic energy. But the problem does not even tell which work is asking for.
It specifies work done on a body. If it does not identify a particular force then surely net work is what we must assume.

Sorry, if you mean net work then I agree with everything you say. Just the sentence used is not clear. And the question is already ambigous as it is. As you can see from other posts, the work done by whatever pushes the sledge may be also assumed to be the goal of the problem. From the fact that they don't specify what work to calculate we may assume a careless author. This is a valid assumption too.

## What is the relationship between friction force and work done on snowy and icy surfaces?

Friction force and work done are closely related on snowy and icy surfaces. Friction force is the resistive force that opposes motion, and work done is the energy required to overcome this force over a distance. On snowy and icy surfaces, the friction force is generally lower, meaning less work is needed to move an object, but it also means there is a higher risk of slipping.

## How does temperature affect the friction force on snowy and icy surfaces?

Temperature significantly affects the friction force on snowy and icy surfaces. As temperature increases, ice can start to melt, creating a thin layer of water that acts as a lubricant, reducing friction. Conversely, at very low temperatures, ice becomes harder and more brittle, which can increase friction slightly, but generally, icy surfaces remain slippery.

## What materials provide the best traction on snowy and icy surfaces?

Materials such as rubber with deep treads, chains, and specialized winter tires are designed to provide the best traction on snowy and icy surfaces. These materials are engineered to maximize grip by increasing surface area contact and sometimes incorporating materials that can bite into ice.

## Why is it harder to walk on icy surfaces compared to snowy surfaces?

It is harder to walk on icy surfaces compared to snowy surfaces because ice has a much lower coefficient of friction. Snow, especially if it is packed, can provide some level of grip due to its granular nature. Ice, being smooth and often covered with a thin layer of water, offers very little resistance to sliding, making it more challenging to maintain balance.

## How can one calculate the work done in moving an object on a snowy or icy surface?

To calculate the work done in moving an object on a snowy or icy surface, you need to know the friction force and the distance over which the object is moved. The formula is Work = Friction Force x Distance. Friction force can be determined by multiplying the normal force (weight of the object) by the coefficient of friction for the surface. Once you have the friction force, multiply it by the distance to find the work done.

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