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Need help with Newton's Law of Physics!

  1. Nov 13, 2015 #1
    I've been stuck on part b for several hours now (feels like the whole day).

    1. The problem statement, all variables and given/known data:
    6gQ0sk5.png

    2. Relevant equations:
    2 k * x sin(θ') = 100/g * a
    sum(Fy) = F net
    Ma = 2kxsin(angle) - 100
    a = (2kxsin(angle) - 100) / (100 / 9.8)
    2*kxsin(θ')-100 / 10.2

    3. The attempt at a solution:

    What I've done so far...
    x component of d = d cos(θ) = 1.11 cos(31)= 0.95 (.9514557038)

    y component of d = d sin(θ) = 1.11 sin(31) = 0.57 (.5716922632)

    added 10 cm to y component = 0.67 (.6716922632)

    θ' =tan-1( 0.67 / 0.95)

    θ' = 35.19° (35.22066645)

    d' = sqrt((x component of d)^2 + (New y component of d) ^2 )

    d' = sqrt(0.95^2 + 0.67^2) (1.164662377)

    d’ = 1.164662377

    x= d' - Unstretched length of spring

    x = 1.164662377 - 1.02

    x = 0.1446623771 m

    exact values (for those wondering):
    < = 35.22069645

    d' = 1.164662377

    mg = 100N

    m = 100/g

    I used 2 k * x sin(θ') = 100/g * a

    And plugged in 2*1060*0.1446623771*sin(35.22069645)*9.8/100 and got 17.33356261 m/s^2 ;and it's still incorrect...

    I also tried another equation
    Ma = 2kxsin(angle) - 100
    a = (2kxsin(angle) - 100) / (100 / 9.8)
    a = 2*1060*0.1446623771*sin(35.22069645) - 100 / 100/9.8
    a = 7.533575457 m/s^2
    ^(still incorrect)

    currently at 71/100 for attempts.....

    So far, I've slumped 3 tutors on Cheggs, math majors, physics major, you name it. And yes I have tried contacting the teacher for the past few days, no replies back yet. This is starting to be very frustrating.

    god bless

    p.s.; this is how I got part a:
    2*1060*sin(31) = 1091.880719
    100/1091.880719 = .0915850956
    1.11-.0915850956 = 1.018414904
    I rounded it to 1.02 and got it right on webassign (not sure if it made a difference)
     
    Last edited: Nov 14, 2015
  2. jcsd
  3. Nov 13, 2015 #2

    haruspex

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    Your first two attempts above are clearly wrong in that you forgot to subtract mg. The last two are much better, but I still get a slightly higher number than the last, by about 4%. You need to be particularly accurate when taking a difference of two numbers that are fairly close together, such as happens when you find the new extension. Try keeping 5 digits throughout.

    Usually it's better to keep everything algebraic until the very end, only plugging in numbers then. Sometimes this circumvents those 'small difference' steps, other times it suggests an approximation method that's better. But in the present case, the final algebraic expression is pretty messy
     
  4. Nov 13, 2015 #3
    The numbers for the x and y components or the numbers for x in the equation?
     
  5. Nov 13, 2015 #4

    SteamKing

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    I agree with your calculations to this point.

    As the acceleration on the weight is produced by the net force, you should draw a free body diagram of the weight and the forces acting on it.
    It's not clear what value of 'angle' is being used here.
    You're getting close. Don't give up!
     
  6. Nov 13, 2015 #5

    Student100

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    Your work is kind difficult to follow, the linear restoring force of the spring system is $$\vec{F}=-k\vec{x}$$

    In your problem we're concerned about the displacement from equilibrium. Obviously to find the instantaneous acceleration we're going to want to look at $$\sum\vec{F}=m\vec{a}$$
    Are you looking at the sum of the forces in the direction of acceleration? What're the forces in the y direction upon release?
     
  7. Nov 13, 2015 #6
    I have the 2 tension forces upwards and gravity downwards
     
  8. Nov 13, 2015 #7

    haruspex

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    In respect of keeping at least 5 digits? For everything on the path to finding the new extension (so includes finding the relaxed length).
    If you keep five, you would normally have four reliable at the end. But the extensions are only about 10% of the relaxed length, so when you take one of those length differences you lose a digit of precision.
     
  9. Nov 13, 2015 #8
    also what equation did you use for it to be 4% higher?
     
  10. Nov 13, 2015 #9

    ehild

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    Keep 4-5 decimals during the calculation steps, including the unstretched length. Calculate x with the more precise value of it. As the lengths are close, you loose one significant digit during subtraction.
     
  11. Nov 13, 2015 #10
    I have updated all the numbers and kept all of the decimals. Can someone tell me what I'm doing wrong or if there's another equation that I can use?
     
  12. Nov 13, 2015 #11

    ehild

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    We do not see your calculations.
     
  13. Nov 13, 2015 #12

    Student100

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    Looks as though he posted it in the original post:

     
  14. Nov 14, 2015 #13

    ehild

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    The original length is not 1.02 m with 4 digit accuracy.
     
  15. Nov 14, 2015 #14

    ehild

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    Did you updated the original length too?
     
  16. Nov 14, 2015 #15
    2*1060*sin(31) = 1091.880719
    100/1091.880719 = .0915850956
    1.11-.0915850956=
    1.018414904
     
  17. Nov 14, 2015 #16
    WOW i guess it did made a difference
     
  18. Nov 14, 2015 #17

    ehild

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    You see at last? Never round down too much during calculations.
     
  19. Nov 14, 2015 #18
    Solve it, thanks to ehild! You had to use all the numbers in part a to solve it correctly (instead of just 1.02)
    Then plug it back into a = (2kxsin(angle) - 100) / (100 / 9.8)
    which you will get 7.7235 m/s^2
    Thank you to everyone that tried to help!

    Again big shout out to ehild!
     
  20. Nov 14, 2015 #19

    ehild

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    That is exactly the same result I got.
    You will remember the importance of significant digits forever I hope, :smile: But you need to round off to the desired accuracy at the end. It is 3 digits in this problem, so you have to give the acceleration as 7.72 m/s2.
     
  21. Nov 14, 2015 #20

    haruspex

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    I guess you did not read this properly:
     
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