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Work and Friction

Discussion in 'MCAT Study Question Q&A' started by tinylilron, Dec 7, 2008.

  1. tinylilron

    tinylilron Member
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    I can't seem to understand what the deal is with work and friction. I understand that friction is going to create heat and thus internal energy of the objects so therefore its not perfect. But I don't seem to understand everything that EK is trying to say in their explanation and I can't find an explanation in my Columbia review text books. Could anyone explain the concept to me and the equation that EK gives us?
     
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  3. isaacmn

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    Based on Kaplan perspective, you right that friction does create some kind of internal energy especially when two bodies are rubbed together. But basically friction is a non conservative force. The total enery and initial are not equal in some sense. it acts opposite of the object motion. the work done by friction on an object being acted upon by gravity is just - force of that gravitational force. Workdone by friction is -force of gravityX h ....weight or gravity =mg...-mgh.................or in the case of incline object . the work done by friction is -mgsintheta
     
  4. Airplay355

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    work= (force parallel to the distance) x the distance

    so for friction, you'd take the frictional force (which is the friction coefficient x the normal force) and multiply that by the distance.

    since you are multiplying a force (N) x a distance travelled or displacement (m) you end up with Nm as a unit which is equal to a joule (J)

    I'm pretty sure that's right...it should be right, I just took a physics final this morning!! :scared:

    I hope that helps.
     
  5. dougkaye

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    EK does a great job of butchering this topic. My physics textbook explained it much better. Use this to find the work performed on an object by non-conservative forces (on the MCAT this will generally be just friction or air resistance):

    Wnc = dE = dK + dU

    A more intuitive way of thinking about this is to consider someone rappelling down a cliff. They want to slow down and so use friction to dissipate (reduce) mechanical energy, which in this case is transformed into heat via friction. This creates a reduction in mechanical energy (dE) which equals the amount of work done by friction. So energy is NOT conserved but is dissipated.

    This makes the somewhat cryptic EK formula actually make sense:

    Fkd (cos theta) = dK + dU = dE

    NOTE: if only conservative forces are involved, then dE = zero. This formula works out because there IS a dE due to non-conservative forces (in this case, friction) doing work on the system.
     
    #4 dougkaye, Jan 15, 2011
    Last edited: Jan 16, 2011
  6. TwoPaddles

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    I saw EK's explanation of work and friction and found it not very useful as well. What I recommend you do is more passages (and more questions from the EK 1001 book related to work and friction). FYI, there are numerous videos on youtube explaining work and friction in a very simple manner with problems.
    Try to think of many equations when you are doing a friction problem.
    For example,
    W = F*d; this is the formula for work where F is the parallel force (not perpendicular!) and d is the distance.
    also, W = -f*d; where f is the friction force opposing the parallel force F. It's negative because it is opposing the parallel force.
    Well, lets see... what else do we know about Work formulas?
    We know from work-kinetic energy theorem that
    W = ΔKE = 0.5mvf^2 - 0.5mvi^2
    where vf = final velocity and vi = initial velocity
    What else do we know? We know that f = µN
    Look what we have so far:
    W = F*d = ΔKE = µN
    What else? we know that on a horizontal surface, the normal force equals the weight (mg)
    so the equation becomes:
    W = F*d = ΔKE = µmg

    Couple examples may be:
    1) Let's say you are lifting a weight, holding it still above your head. You're just holding it; not moving it anywhere. What's the "work" done? ZERO. Because you are not moving it anywhere therefore d = 0; notice that even if d = 0, the whole thing (W = F*d = ΔKE = µmg) becomes zero.

    2) You apply a force of 20N horizontally on a box and move it 10 meters. What's the work done? Well, simply 20*10 = 200.

    3) What's the work done by the normal force on the previous problem? Zero.. because F in the W=F*d formula is the parallel force.

    About heat...
    Let's say you start at a height and slide down the hill and come to a stop. What happens is, at the top you have some potential energy (mgh, because you are at a height).. As you slide down, that potential energy is turning into kinetic energy. You come to a stop due to friction. Heating occurs because some of the energy is lost as heat as you come to a stop.

    So this is mostly how I think of it. You may have already know all these or not.

    Hope it helps
     

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