A problem involving Newton's second law.

  • #1
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Homework Statement


see this link http://i45.tinypic.com/10gwhw2.jpg


Homework Equations



F=kx, F=ma

The Attempt at a Solution



ok first I find the total length of the pulley system,
2Sa+Sb=L, differentiating with respect to time i get,
2Va + VB=0 and thus 2Aa+Ab =0
So Aa =-Ab/2

next I first analyze block A and do a force balance to get,
-Ma*g - k*x + 2T = Ma*Aa

following this i analyze Block B doing a force balance again,

T-Mb*g = Mb*Ab

so to eliminate one of the variables I input Aa=-Ab/2 into the equation describing block A.

I get -Ma*g-k*x+2T+Ma(-Ab/2)

then i get -2(Ma*g-k*X+2T)=Ab

so now i substitute this in Block B with real values

T-98.1=10(19.62-400+2T)
so T = 195.04
then Ab = 19.62-400+2(195.04) = 9.7 m/s^2

So now I have (delta)y = 0.5 m A=9.7
using the following equation i can solve for t, Δy=(1/2)at^(2)
doing so i get 0.32s
thus v = at = 9.7(0.32) =3.104 m/s

However i do not have a solution for this exercise and unsure if my attempt is correct or not. Any help is greatly appreciated thanks!
 

Answers and Replies

  • #2
gneill
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Since the length of the spring changes with time, the force will vary with time, too. That means the accelerations will not be constant, so you can't just apply the constant acceleration kinematic equations. Instead, you'll end up with a differential equation to solve.

Have you considered using a conservation of energy approach? Muuuuch simpler.
 
  • #3
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Ok so I assume my working is correct up to the point with which i calculated T.

so my new attempt is as follows:

W=(1/2)*m*vf^(2) - (1/2)*m*vi^(2)
F*D= (1/2)*m*vf^(2) - (1/2)*m*vi^(2)

so F*D = (-Ma*g-k*x+2T)*Δy = (1/2)*mvf^(2)

subbing in values and solving for vf i get 0.98m/s^2

I'm guessing however this is incorrect because the force that the spring exerts is not constant?
 
  • #4
gneill
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Ok so I assume my working is correct up to the point with which i calculated T.

so my new attempt is as follows:

W=(1/2)*m*vf^(2) - (1/2)*m*vi^(2)
F*D= (1/2)*m*vf^(2) - (1/2)*m*vi^(2)

so F*D = (-Ma*g-k*x+2T)*Δy = (1/2)*mvf^(2)

subbing in values and solving for vf i get 0.98m/s^2

I'm guessing however this is incorrect because the force that the spring exerts is not constant?
Yup. This approach must yield a differential equation. The spring force is not constant, so the accelerations are not constant and you cannot apply the constant acceleration kinematic formulae.

Try an energy conservation approach.
 
  • #5
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Ok my next attempt is as follows

(1/2)ky^(2) = Mb*vf^(2)

so subbing in values to solve for vf i get 4.472m/s.
 
  • #6
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actually maybe this is the equation i need?

(1/2)ky^(2) +Ma*g*y=(1/2)Mb*Vf^(2)
and solving i get 4.686 m/s
 
  • #7
gneill
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Suppose you set the individual gravitational PE's of the two masses to zero when the system is at equilibrium. Also set the spring's PE to zero at equilibrium. (You can do this because PE is a relative measure).

Now, whenever the system is at the equilibrium position, those PE's will be zero.

How much additional energy will be stored in the spring when mass B is displaced by 0.5m and held there? What's the total KE at that time? (Remember, the masses are stationary). So what's the total energy PE+KE at that time?
 
  • #8
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PEa = Ma*Aa*y and KEa = 0
PEb= Mb*Ab*y and KEb =0

I can't use g as acceleration right? because the forces acting on the objects interfere with its normal functioning? or am i way off the mark :(
 
Last edited:
  • #9
gneill
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PEa = Ma*Aa*y and KEa = 0
PEb= Mb*Ab*y and KEb =0
Okay, not what I asked. What's the change in PE stored in the spring when mass MB is displaced by 0.5m?
I can't use g as acceleration right? because the forces acting on the objects interfere with its normal functioning? or am i way off the mark :(
You won't need ANY accelerations or forces using the conservation of energy method.
 
  • #11
gneill
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ok so that would be -kx

No, What's the formula for the energy stored in a spring?
 
  • #13
gneill
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So... What's the additional energy stored in the spring when mass B is displaced?
 
  • #15
gneill
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No. There's no motion at this point. Mass B is displaced and held stationary at its new position. But the spring has been stretched. What energy does SPRING gain?
 
  • #17
gneill
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Make an effort. Calculate the value!
 
  • #18
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Ok I'm thinking, (1/2)k(xf^(2)+xi^(2))=1/2(Mb)vf^(2)

subbing in values i get

6.32 m/s?
 
  • #19
gneill
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Ok I'm thinking, (1/2)k(xf^(2)+xi^(2))=1/2(Mb)vf^(2)

subbing in values i get

6.32 m/s?

How do you justify the equation above? (What does it represent?)

What about mass A? Won't it be moving too?

The reason I ask you to calculate the potential energy added to the spring when the displacement is carried out is that it represents the energy added to the overall system, and will stay with the system when it returns to the equilibrium position. When the system returns to the equilibrium position all the gravitational potential energy changes due to displacement disappear -- everything is back in its original position. What's left is the new energy that was stored in the spring, which has then been released as KE since the spring, too, is back in its original position.

So why not calculate that spring energy? How much is the spring extended when MB is displaced by 0.5m? How much energy does that put into the spring, in Joules?
 
  • #20
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" How much is the spring extended when MB is displaced by 0.5m?"
the spring is extended 0.5m.

"How much energy does that put into the spring, in Joules?"
(1/2)k(y)^(2) = (1/2)*800*(0.5)^2 = 100 J
 
  • #21
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ok new attempt total potential energy = (1/2)kx^(2)+(1/2)*kx^(2) = (1/2)MbVb^(2) -(1/2)MaVa^(2)

we know Va = -Vb/2
so
(1/2)kx^(2)+(1/2)*kx^(2) = (1/2)MbVb^(2) -(1/4)MaVb^(2)

then we get 2kx^(2)=(Mb-(1/2)Ma)Vb^(2)
and i need to solve this equation?
 
  • #22
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my reasoning is as follows:
block A
(1/2)kx^(2)-(1/2)MaVa^(2)= 0
and
block b
-(1/2)(kx^2)+(1/2)MbVb^(2)=0
 
  • #23
gneill
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Confirm how far the spring is stretched when the mass Mb moves by a given amount. The pulley arrangement provides a mechanical advantage...
 
  • #24
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ok so 2SA+SB=L

then SB=-2SA

so SB/-2 = SA then SA=-0.25m=xa
so
(1/2)kxa^(2)+(1/2)*kxb^(2) = (1/2)MbVb^(2) -(1/4)MaVb^(2)
subbing in values and simpilfying i get 4.5Vb^(2)-125=0
solving i get 5.27 m/s
 
  • #25
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actually tho won't the spring get stretched by the same amount as the displacement at B since the cord is inextensible?
 

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