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anonymous

  • one year ago

PHYSICS!!! YAY

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  1. anonymous
    • one year ago
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    there is friction between the 5.00 kg block and tabletop. The coefficient of kinetic friction is 0.25.

  2. anonymous
    • one year ago
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  3. anonymous
    • one year ago
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    So this is a pulley problem. Identifying the acceleration of the whole system given the kinetic friction force amounting to 0.25;)

  4. anonymous
    • one year ago
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    So far what I've got is the force acting due to gravity on 5.0kg object which is (5kg)(9.8N/kg)=49N

  5. anonymous
    • one year ago
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    And force acting on the 6.0kg object which is (6.0kg)(9.8m/s^2)=58.8N

  6. anonymous
    • one year ago
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    The thing is force acting on the 5.0kg is horizontal hence not overly responsible for the Fmg had by 5.0kg itself.

  7. anonymous
    • one year ago
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    Given the kinetic energy of 0.25 I believe that 25% of force once it starts to move is lost to thermal energy.

  8. anonymous
    • one year ago
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    Therefore my reasoning is that 5.0kg object is pulled to the right horizontally by a force of 58.8N-58.8*0.25

  9. anonymous
    • one year ago
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    58.8N-14.7N=44.1N

  10. anonymous
    • one year ago
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    Given the force acting to the right of 5.00 kg object, using the equation F=ma 44.1N=(5.00kg)a a=8.82m/s^2

  11. anonymous
    • one year ago
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    What do you think @IrishBoy123

  12. IrishBoy123
    • one year ago
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    you need to draw these first, free body diagrams, then you can solve in a jiffy :p

  13. IrishBoy123
    • one year ago
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    |dw:1439464554560:dw|

  14. anonymous
    • one year ago
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    So in this question I need to find the friction force, tension force, as well as the acceleration of the whole system.

  15. IrishBoy123
    • one year ago
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    \(T - 5 \mu g = 5 a\) [A] \(6g - T = 6 a\) [B] [A] + [B] => \(g(6 - 5 \mu) = 11 a\) \(\large a = \frac{g(6 - 5 \mu)}{11} = \frac{9.8(6 - 5 * 0.25)}{11} m/s^2\)

  16. IrishBoy123
    • one year ago
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    that is for acceleration the friction force is the \(5 \mu g\) i mention, ie \(F = \mu R = \mu mg\) and for tension, from [B]: \(T = 6(g - a) \) not trying to do your coursework but you see the value is setting it out nicely. especially with a drawing. because pretty quickly these get complicated and you will need to be organised. the fact that it is the Tesnion T that connects the bodies is crucial. it is assumed that T is constant throughout the rope, and also that the pulley is frictionless and massless/inertialess.

  17. anonymous
    • one year ago
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    Oh I see now I am starting to see it right now. It's quite complicated that I need organization rather than just diving into the problem.

  18. IrishBoy123
    • one year ago
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    yeah, you're clearly very smart, maybe just slow down a bit :-)

  19. IrishBoy123
    • one year ago
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    "free body diagrams" more jargon, but everyone uses it! https://en.wikipedia.org/wiki/Free_body_diagram

  20. anonymous
    • one year ago
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    lol physicists have their own way to convince the public of their esotericism.

  21. anonymous
    • one year ago
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    The above solution only seems to concern static friction which is the force required to move the object... This question concerns with kinetic friction which is the force that once moved will sustain the proportional friction depending on the force applied.

  22. anonymous
    • one year ago
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    T-5um=5a In other words Tension force subtracted by the weight times the friction force is equal to the force required to move 5.0kg object whose acceleration is yet unknown.

  23. anonymous
    • one year ago
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    oh the equation assumes a sliding body yeah i think it rings my bell.

  24. IrishBoy123
    • one year ago
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    you have 2 \(\mu\)'s static is when the body is .... drum roll .....static! then once it moves, \(\mu\) decreases. push a book across a table from rest and it is "harder" to get it going than to keep it going or further accelerate it.. if this problem concerned static friction, neither body would be moving and there would be no motion. so you would set a to zero in the equations and you could still solve the equations. i think you would need to have glued the 5kg mass to the surface though, for that to make any practical sense !

  25. anonymous
    • one year ago
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    Oh I see now. The prospect of having 5a and 6a presupposes the movement

  26. anonymous
    • one year ago
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    Tension force being subtracted by the proportional kinetic friction as in the mug seems to make more sense. I see as the mass increases the kinetic friction subjected increases hence still holding proportionality...

  27. anonymous
    • one year ago
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    At first it seems as if the T force is taken away its constituent force by fixed amount but upon closer examination this equation reveals proportionality between the mass and kinetic friction.. Ha! @IrishBoy123 You rock

  28. anonymous
    • one year ago
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    Obviously the force that remains is equivalent to the Force which causes acceleration of the mass with given weight.

  29. anonymous
    • one year ago
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    So both equations T-5ug=5a and 6g-T=6a hold true.

  30. IrishBoy123
    • one year ago
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    for static \(\mu\), imagine a mass on an inclined plane at angle \(\theta\), you raise the plane until the mass just moves |dw:1439466158420:dw| \(mg \ cos \theta = \mu mg \ cos \theta\) \(\mu = tan \theta \) so you can spend all day working out \(\mu\) for everyday household items !! what fun. yeah, the tension throughout the rope is assumed to be the same. if it is different, the the rope experiences a force imbalance and you have to apply F = ma to that piece of rope. of course it's not the same because the rope is accelerating to, but we also assume that it is massless. so it's really just a connection between the 2 masses.

  31. anonymous
    • one year ago
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    mg cosθ(normal force)=μmg cosθ(u being equal to or less than 1)

  32. IrishBoy123
    • one year ago
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    i think you're going at this the right way, BTW you have to get a feel for what is happening or else you will forever be stuck with trying to fiddle about with equations without ever really knowing what is going on. so imagine you are the rope, if that helps. imagine you are one or other of the blocks.

  33. anonymous
    • one year ago
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    .

  34. anonymous
    • one year ago
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    incredible how physicist can use the equations to speak out.

  35. anonymous
    • one year ago
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    In stead of words.

  36. IrishBoy123
    • one year ago
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    yes, if you get to \(\mu = 1\) then the inclined plane is at \(90^o\) and the block still hasn't slid, you are in superglue territory if \(\mu > 1\) then you have broken gravity and get a Nobel prize !!

  37. anonymous
    • one year ago
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    Without inter interlocking mechanism that is.

  38. anonymous
    • one year ago
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    Friction measurement assumes two objects in contact are not interlocked nor glued but yet electrons repelling each other on a supposedly rugged or smooth surface?

  39. IrishBoy123
    • one year ago
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    yes, the rope is the interlocking mechanism, perfect!! thing to remember about friction is that it opposes motion. but yes it is the same idea, a way of interlocking the block and the surface. key is: friction opposes motion. so if in doubt, model first without friction, and that will tell you which way friction vector is pointing -- in other direction!

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