A problem involving Newton's second law.

  • #26
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actually disregard post 24 its nonsense.
 
  • #27
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ok so (delta)L = 2SA +SB

(delata)L = 0.5m

so Sb=0.5-2(0.5) = -0.5m = xa

subbing this into (1/2)kxa^(2)+(1/2)*kxb^(2) = (1/2)MbVb^(2) -(1/4)MaVb^(2)
and solving for vb i get 6.66m/s
 
  • #28
gneill
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Nearly there.

Keep in mind that kinetic energy doesn't have a direction. All kinetic energies for a system are positive and sum to the total KE.

For the PE added to the spring when the displacement occurs, take the difference between the PE stored in the spring initially and the PE at its full extension. That means finding the initial (equilibrium) stretch (xi) and the final stretch (xf) and taking the difference in the PE's at those positions of the spring:
$$\Delta PE = \frac{1}{2} k x_f^2 - \frac{1}{2} k x_i^2$$
The initial (equilibrium) stretch ##x_i## can be found by balancing the forces at the top of the spring when the system is in equilibrium -- the spring's force and the weight of Ma balances the upward force applied by Mb through the pulley system.
 
  • #29
gneill
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Be sure to take a close look at the dynamics of the pulley system motions. If there's a mechanical advantage involved, there will always be a similar ratio of motions involved. You've already identified that the force applied to mass A through string tension is doubled due to the two string segments holding up the Mass A pulley.

Mass A moves half the distance that mass B moves. Similarly, mass A's velocity and acceleration rates are half that of mass B.
 
  • #30
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ok so my attempt at finding xf and xi is as follows:
finding xi,
kxi-Ma*g+2T = 0

we know T = Mb*g =98.1
so subbing into equation and solving i get -0.05

now finding xf
-kxf-Ma*g+2T = 0 at the instant the string as gone 0.5m.
i solved for T previously and now can directly solve for xf
doing so i get 0.51m

the (delta)P = (1/2)(800)(0.22)^(2)-(1/2)(800)(0.05)^(2) = 18.36
 
Last edited:
  • #31
gneill
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Those values don't look right.

When the system is static at equilibrium, mass Mb pulls on the string with weight Mb*g. So that's the tension in the string, as you've written.

Twice this force (2T) is balanced by the weight of Ma and the force due to the spring. So write

2T = Ma*g + k*xi

Find xi.

You don't need any force or tension equations to find the amount by which the spring length changes when Mb is displaced. The displacement of Mb is coupled to the displacement of Ma and the spring via the pulley system. How much does Ma move for a given movement of Mb?
 
  • #32
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2T=Ma*g +k*xi,
subbing in values xi = 0.27m

now this is the part where i'm confused- if the string is inextensible then isn't the displacement of Ma and the spring the same value as the displacement of Mb?
 
  • #33
gneill
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2T=Ma*g +k*xi,
subbing in values xi = 0.27m <--- Looks a bit high. Check the calculation.

now this is the part where i'm confused- if the string is inextensible then isn't the displacement of Ma and the spring the same value as the displacement of Mb?

No. There is a mechanical advantage involved, since there's two segments holding up the Ma pulley and only one holding the Mb mass. Check the operation closely. Draw a detailed figure with accurate lengths and measure them if required (choose your own scale and initial lengths).
 
  • #34
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2SA=SB, that is to say that the vertical motion of B is twice the vertical motion of A

for the calculation i got

(2(98.1)-2(-9.81))/800 = 0.27m
 
  • #35
gneill
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2SA=SB, that is to say that the vertical motion of B is twice the vertical motion of A

for the calculation i got

(2(98.1)-2(-9.81))/800 = 0.27m

Why the "-" on the 9.81? At the top of the spring, the 2T from Mb is pulling upwards, while Ma and the spring pull downwards. Considering force magnitudes, Ma*g should sum with k*xi.
 
  • #36
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oh yes, sloppy on my part. so the new calculation comes out as 0.221m. so am i correct in saying 2SA=SB?
 
  • #37
gneill
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oh yes, sloppy on my part. so the new calculation comes out as 0.221m. so am i correct in saying 2SA=SB?

Yes.
 
  • #38
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Ok just to double check my working here is my final attempt:

(1/2)kxf^(2)-(1/2)kxi^(2)=(1/2)Ma*va^(2)+(1/2)Mb*vb^(2)

now i have xi = 0.221 and thus xf must equal double that so that is 0.442

now i have Va=-Vb/2

subbing this into the equation and solving i get
5.25Vb^(2)-58.6092=0, so vb comes out at 3.3412m/s
 
  • #39
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it also seems odd to me that i never used the fact that block B was pulled down 0.5 m from the rest position in this problem.
 
  • #40
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gneil can i just get you to confirm my working please as i have no answer key to this question.
 
  • #41
gneill
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Ok just to double check my working here is my final attempt:

(1/2)kxf^(2)-(1/2)kxi^(2)=(1/2)Ma*va^(2)+(1/2)Mb*vb^(2)

now i have xi = 0.221 and thus xf must equal double that so that is 0.442
xf is not double xi. It's xi plus however much the spring is stretched due to the movement of Mb.
xi represents the initial, equilibrium stretch of the spring. When Mb is moved away from equilibrium, the spring is stretched an additional amount that depends upon how far Mb is moved. That's where your ratio SA/SB comes in.
now i have Va=-Vb/2
Yes, the velocity relationship is okay.
subbing this into the equation and solving i get
5.25Vb^(2)-58.6092=0, so vb comes out at 3.3412m/s
 
  • #42
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so 2SA=0.5, then SA = 0.25 so Xf = 0.221+0.25=0.471
 
  • #43
gneill
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so 2SA=0.5, then SA = 0.25 so Xf = 0.221+0.25=0.471

That looks better.
 
  • #44
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so (1/2)kxf^(2)-(1/2)kxi^(2)=(1/2)Ma*va^(2)+(1/2)Mb*vb^(2)
then subbing in real values, (1/2)(800)(0.471)^(2)-(1/2)(800)(0.221)^(2) = (1/2)(2)(vb^(2)/4)+(1/2)(10)(vb^(2))

simplifying i get 5.25Vb^(2)-69.2=0

solving this however i get imaginary numbers.
 
  • #45
gneill
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so (1/2)kxf^(2)-(1/2)kxi^(2)=(1/2)Ma*va^(2)+(1/2)Mb*vb^(2)
then subbing in real values, (1/2)(800)(0.471)^(2)-(1/2)(800)(0.221)^(2) = (1/2)(2)(vb^(2)/4)+(1/2)(10)(vb^(2))

simplifying i get 5.25Vb^(2)-69.2=0

solving this however i get imaginary numbers.

Really? I see real roots. Try again.
 
  • #48
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Thanks for the guidance gneil!
 

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