Simple physics question with friction/acceleration/force.?

Question

Please answer it using only SI units, none of that primitive crap they use in the USA. Please show all work along with any formulae used, preferrably in steps, so that I can learn how to do this on my own as I'm rather stuck at the moment with this question. The question is as follows:

A sled of mass 50kg is pulled along snow-covered, flat ground. The static friction coefficient is 0.30, and the sliding friction coefficient is 0.10.

d)Once moving, what total force must be applied to the sled to accelerate it 3.0m/s^2?

The help would be most appreciated! I need this damn physics course to begin my career as an electrician.

Answer 1

First the static frictional force must be overcome before it starts to move.
Ff static = usFn = 0.30(50kg)(9.8 m/s^2) = 147 N

Fnet = Fa - Ff(kinetic friction) = ma
Fa - uk(mg) = ma
Fa - 0.10(50 kg)(9.8 m/s^2) = 50 kg(3.0 m/s^2)
Fa = 50kg(3.0ms^2) + 0.10(50 kg)(9.8 m/s^2)
Fa = 199 N

All in all a total force of 147 N + 199 N = 346 N is required
from the moment the static frictional force is overcome to moment it is given an acceleration of 3.0 m/s^2.

teddy boy

Answer 2

I've tried to avoid questions in this topic, but I'm impressed that there's finally a question asking about the science. You are quite right about greenhouses, oddly enough, they are mostly warmed by their lack of convection, and not by the greenhouse effect, so in that sense the greenhouse effect is misnamed. However, because the effect is misnamed doesn't mean the science is wrong, and while you always make some simplifications in moving from an actual physical phenomenon to a scientific model, the key is to understand the magnitude of the approximations that you're making, and I think people have a pretty good understanding of the physics. First you have to understand that there are several methods of energy transfer involved--one is convection of both sensible and latent heat, and the other is radiative transfer. The lowest part of the atmosphere is the troposphere (I think you're a bit confused on the meaning of that), and the troposphere is characterized by both radiation and convection being important, that's why atmospheric scientists will refer to "radiative-convective equilibrium" as setting the lapse rate for the troposphere. These forms of energy transfer take place at vastly different rates, convection occurs at meters per second, or tens of meters per second for the strongest thunderstorms, while radiation occurs at the speed of light. So when you say something "couldn't heat the ground air" because it would counteract convection, well it does so constantly. Now carbon dioxide is fairly well mixed throughout the troposphere, so it occurs in the upper levels also. Perhaps you think that any absorption of radiation from the ground will be trapped near the surface, but this is false, because just as CO2 absorbs radiation it also radiates it, both upwards and downwards, so some of that radiation gets transferred to the upper levels, and some will still escape into space. It's just that adding more CO2 will mean that a bit more of it will be re-radiated back in the downward direction and shift the equilibrium temperature of the Earth to be slightly higher. When people say "it's simple physics," I think they do so because it's a very simple calculation to consider the Earth with no atmosphere and calculate an equilibrium temperature, then add an absorbing layer and re-calculate it and find that the equilibrium temperature goes up. Of course, to be a realistic model you need to get a lot more sophisticated, but you can make the layers arbitrarily thin and add more of them and keep re-doing the calculations until they're sufficiently accurate for your purposes. So, it's simple to understand that adding CO2 will increase the equilibrium temperature, but to understand how much requires much more involved modelling. Also, while as you say the atmosphere is in a constant state of flux, which complicates things, if you were to average over a sufficiently long time period the mean equilibrium temperature of Earth MUST be set by the radiative equilibrium--because there is no convection or conduction to space, only radiative transfer. That's why if you decrease the amount of IR radiated to space, the temperature of the Earth must go up. You can shuffle it between the ocean, atmosphere (and to a smaller extent the ground), but the energy content will go up. [As an aside, the temperature in the troposphere typically falls with height (although certainly not always) because the upper atmosphere is farther from the heat source (the ground). The sun's radiation is not very effective at heating the atmosphere because air is transparent to visible light. The sun does heat up the ground, which warms the air by conduction, convection and radiation.]

Answer 3

Rick got it right. The force to overcome the static friction goes away as soon as the sled starts moving and is replaced by µ(k)mg. (= 49 N) This added to the force to accelerate (ma = 50*3 = 150) → 199 N, the correct answer.......

Answer 4

Well, this may be a trick question.

To me, "total force" means "net force". If that's what is meant you can use this equation:

Net force = m × a

Since they give you "m" and "a" already, just multiply. The friction is irrelevant.

On the other hand, it may be that by "total force" they mean, "how hard you have to pull on the sled (with friction pulling in the opposite direction)." I think this is a misuse of the term "total force," but perhaps this is the interpretation they have in mind. In that case, let F_p be the force with which you're pulling on the sled, and let F_f be the force of friction. In that case,

F_net = (F_p − F_f) = ma

So:
F_p = F_f + ma
= (weight)(μ_k) + ma
= mgμ_k + ma

They give you m, μ_k and "a", and g is of course 9.8m/s². So just plug in the numbers