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Inclined Planes

Discussion in 'MCAT Study Question Q&A' started by inaccensa, Aug 17, 2011.

  1. inaccensa

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    Absent all other forces, a block sliding down an inclined plane experiences an acceleration =gsin(theta), ie fraction of g.
    -->Is it safe to say that the block is solely moving under the influence of gravitational force, if so what about the frictional force? Is the force due to gravity > frictional force?
    -->If we assume the inclined plane to be infinitely long, at what point ( will it only stop at the bottom of the incline and due to what condition does the block come to a stop? Ignoring air resistance
    -->Work done moving it up directly at height "h" is equal to the work done pulling the block up the distance "d." If there is indeed friction, will the work done pulling the block up d < work done pulling the block at height h? What frictional force is present when we pull the block directly up?
    --> Lastly, ignoring friction,if the work done is same, then KE is equal and the velocity or rather change in velocity will be sam. If I look according to v=(2gh)^1/2; v'=(2g(h/sintheta)^1/2; v'=velocity moving down the incline is less, which makes sense since the acceleration is less, but doesnt this seem contradicting?
     
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  3. costales

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    If the problem says "absent all other forces", then yes, gravity is what gives the block that acceleration a = gsin(theta). If friction enters into the picture, then in order for the block to move, the weight's component along the surface mgsin(theta) must exceed the frictional force.
    If the incline is infinitely long and there is no air resistance and there are no other forces, then the block will keep gaining speed at v^2 = 2(gsin(theta))(d) until d is infinite and v is infinite which is nonsense - this is where relativity takes over because nothing can go faster than light. The only way to stop this motion is to dissipate the kinetic energy through heat, friction, etc.
    If you define d as distance along the incline, h = dsin(theta), the work done pulling the block straight up is mgh, the work done sliding it up the incline without friction is mgdsin(theta), and greater if you have to work against friction.
    There is no contradiction. v^2 = 2(gsin(theta))(d) = 2gh because h = dsin(theta).
     
    #2 costales, Aug 17, 2011
    Last edited: Aug 17, 2011
  4. inaccensa

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    If you define d as distance along the incline, h = dsin(theta), the work done pulling the block straight up is mgh, the work done sliding it up the incline without friction is mgdsin(theta), and greater if you have to work against friction

    Can you please elaborate. I understand that the work is same in absence of friction, but with friction, will the work done by pulling an object (mgh >mgsin(theta)d)
     

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