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Homework Help: Inclined slope

  1. Jan 6, 2010 #1
    1. A 5kg block rests on a smooth plane inclined at an angle of 20degrees to the horizontal. The block is now pulled up the plane by a rope with a tension of 300N parallel to the plane for 10s.
    a) What is the acceleration during this period?
    b) What is its speed at the end of this period?
    c) How far does it travel during this period?
    d) The rope snaps - how much further will the block travel up the slope?
    e) What is the time interval between the rope snapping and the block returning to this point?




    2. Relevant equations
    F=ma
    v=u+at
    s=0.5(u+v)t


    3.
    a) F=ma. The force is given but do I need to calculate mgsin(theta). I have a force but do not have velocity so cannot use v=u+at to find acceleration for the time period. Using F=ma I get F/m=a = 300N/5kg= 60ms^-2.
    b) v=u+at 0+60ms^-2x10s =600ms^-1
    c) distance=speed x time = 600ms^-1 x 10s = 6000m

    Am I on the right track?
     
  2. jcsd
  3. Jan 6, 2010 #2

    tiny-tim

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    Hi lemon! :smile:

    (try using the X2 tag just above the Reply box :wink:)
    Yes, F = ma is correct.

    (F = ma is physics, v = u + at (etc) is just geometry. This part of the question is physics, the rest is geometry.)

    But you need Ftotal = ma, and you've left out gravity. :wink:
    Correct method. :smile:
    No, distance=speed x time is only for constant speed.

    You need one of the standard constant acceleration equations … which one?
     
  4. Jan 6, 2010 #3
    Hello tiny-tim:
    a) so to include gravity into the calculation I calculate Fsin(theta)
    300Nsin(20)=103N
    F-Fsin(theta)=ma
    therefore, 300N-103N=5kg x a - acceleration= 197/5 =39.4ms-[/SUP]

    b) v=u+at 0+39.4ms-[/SUP] x 10s = 394ms-[/1]

    c) C is asking for distance, so s=1/2(u+v)t should do it, or s=ut+1/2at[t][/SUP]
    first one - 1/2(0+394)10=1970m

    any better?

    not getting on with the html thingys very well sorry
     
  5. Jan 6, 2010 #4

    ideasrule

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    It's mgsin(theta), not Fsin(theta).
    Both methods are right, but your answer to a) is wrong.
     
  6. Jan 6, 2010 #5

    tiny-tim

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    whatever's up …

    Hello lemon! :smile:

    Your tags should look like [noparse] and [/noparse] around whatever's up. :wink:
     
  7. Jan 6, 2010 #6
    a) so to include gravity into the calculation I calculate mgsin(theta)
    5x9.81(20)=16.8N
    F-F1=ma
    therefore, 300N-16.8N=5kg x a - acceleration= 283.2/5 =56.6ms-2

    b) v=u+at 0+56.6ms-1 x 10s = 566ms-1

    c) C is asking for distance, so s=1/2(u+v)t should do it, or s=ut+1/2at[t][/sup]
    first one - 1/2(0+566)10=2830m

    Thanx tt
     
  8. Jan 6, 2010 #7
    d) The rope snaps - how much further will the block travel up the slope?
    e) What is the time interval between the rope snapping and the block returning to this point?

    d) v2=u2+2as
    s=(v2)-u2/2a
    s=(0ms-1-566ms-1)2/2(56.6ms-1)=2830m

    e) v=u+at
    t=v-u/a
    = (56.6ma-1-0)/56.6ms-2=10s
    and back to the original point = 2 x 10s = 20s
     
  9. Jan 6, 2010 #8

    tiny-tim

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    d) and e) have the correct formulas, but you must use a different acceleration now (also, either a or s must be negative :wink:)
     
  10. Jan 6, 2010 #9
    ahh! Of course. If the rope has snapped then the original tension is no longer acting upon the object. Only the gravitational force remains.

    d) mgsin(theta)=ma
    new acceleration = -16.8N/5kg = -3.36ms-2

    s=(0ms-1-566ms-1)2/2(-3.36ms-2)=-47672.0N
    47700N 3s.f.
    This puppy's being yanked really hard for a 5kg block!

    e) t=2(566/47700N)=0.024s 2s.f.

    I think I may have gone wrong here somewhere. The force seems too big and the time too short.
     
    Last edited: Jan 6, 2010
  11. Jan 6, 2010 #10

    tiny-tim

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    erm :redface: … you've used the wrong formula for e) :rolleyes:

    btw, there's no need to keep writing the units (N, s, etc). …

    if you make sure that you start with everything in SI units, then the result will automatically be in SI units also (that's the whole point of a system of units). :wink:
     
  12. Jan 6, 2010 #11
    Indeed. I put in the -44.7Km distance instead of acceleration. Is this figure right for distance though? It seems awfully big.

    e) t=2(566/-3.36) = -336.9s
    -5min 37s
     
  13. Jan 6, 2010 #12

    tiny-tim

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    Well, it was moving at half a kilometre a second …

    that is pretty fast. :smile:
     
  14. Jan 6, 2010 #13
    Thanks for everything today Tiny-Tim
     
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