# Homework Help: Snow shoveling, how much work

1. Oct 7, 2005

### Erik_at_DTU

This may be a unfamiliar problem for those of you from warmer countries, but here in the north we sometimes have to shovel snow. This task is about snow shoveling:

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It has fallen 10 cm of snow. Estimate the amount of work needed to shovel a sidewalk free of it. The sidewalk is 10 m long and 1.5 m wide. A liter snow weighs almost the same as 100 ml water. State all your assumptions.

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I feel like I have to make at least 5-6 assumptions, is that correct or is it an easier way to do it? To estimate the amount of work, I have to make assumptions about length and acceleration. But also a lot more, in order to make it realistic.

Though I could of course say that I shovelled it 10 meters away, using a horisontal force which gave an average acceleration of 2 m/s^2. 'Giving me the expression: W = F*s = m*a*s = (150kg)*(2 m/s^2)*(10m) = 3000 J

I will appreciate all the help I can get!

2. Oct 7, 2005

### Integral

Staff Emeritus
You should be able to do this from energy considerations. You have a mass of snow on the ground, you must raise it above the ground some distance to throw it off the path. Make a reasonable guess to how high you must raise it, now compute the work done.

3. Oct 7, 2005

### mezarashi

I'm 90% sure that this question if given as homework is a kinetics or energy question. The equation you should be working with would be the potential function: mgh

Use the dimensions to find the total mass of snow, and make an assumption on how high you're going to need to lift it up in order to dump it somewhere else. Assume you let gravity do the dumping, so no force or energy needed on your side to let the snow on the shovel go on the way down. Sound less complicated? :D

4. Oct 8, 2005

### Erik_at_DTU

hmm...

I think I understand, but what about the horisontal motion?

Thanks for the help so far

5. Oct 8, 2005

### mezarashi

From an energy point of view. When you are walking horizontally with snow in your shovel, you are doing no work. That is because the force you are exerting on the snow is in the vertical direction to keep it up against gravity. You are however moving in the horizontal direction, which is perpendicular to the direction of force. If you remember, W is the dot product of F and d.

6. Oct 8, 2005

### Erik_at_DTU

I sort of understand

I think I understand, even though I don't :tongue:

I am confused, because:

1. The potential energy you give the mass = the work you have to do, so W=mgh

But I also have to move the snow in the horisontal direction, so:

2. W = F*d*cos(x)

Which should I use, or should I combine them? Right now I am thinking maybe it is enough to just explain the different assumptions I make (one is: to move the snow I have to lift it 0.5 meters up), and say that the total work I have to do is W=mgh=150*9.8*0.5

But on the other hand it seems too simple...

7. Oct 8, 2005

### mezarashi

W = mgh is a special case of W = F*d*cos(x).
You apply force in the y direction (F = mg), you move it up d or h meters in the y direction. cos 0 = 1. Now in the horizontal direction, you are lifting the snow with force F in y, and are moving it along x, so cos 90 = 0. You do no work while you move the snow sideways.

And it is that simple! ^^

8. Oct 8, 2005

### Erik_at_DTU

wow

Thanks, you have just made my day!