Rubber strings pulled and realeased -- find the max height

[Helly123]

Homework Statement

Homework Equations

The Attempt at a Solution

I tried to find constant k
f = k $\Delta$x
1/2 * 9,8 = k * 0,1
k = 49 N/m

is rubber string the same as spring?

Ep + EK at 0,1 meter = Ek + EP at 0,3 m
1/2 k x^2 + 0 = 1/2 k x^2 + 1/2 mv^2
v^2 = 2*49(0,08)
v = 7*0,4 = 2,8 m/s

max when v = 0?
can I use vertical motion equation?

maybe I can find T
and the max is when 1/2 T
but still wrong.

can you give me clue? thank you.

 

[mfb]

is rubber string the same as spring?

Only in tension. It doesn't resist compression.

Ep + EK at 0,1 meter = Ek + EP at 0,3 m

Which two things do you compare here? What is x? Between figure 2 and 3 the mass gains energy.

What is the energy in tension of the string in figure 3?

 

[Helly123]

Only in tension. It doesn't resist compression.Which two things do you compare here? What is x? Between figure 2 and 3 the mass gains energy.

What is the energy in tension of the string in figure 3?

figure 2, the x is 0,1 meter and it has new equilibrium position.
if we set x = 0,1 as reference point
Ep = 0. Also, Ek = 0 (equilibrium)

figure 3,
x = 0,2 meter (from reference point)
Ep = 1/2 kx^2 = 0,98 J
Ek = max = 1/2mv^2

if mechanical energy is conserved then
I can say :
EM at fig 2 = EM at fig 3
0 = 0,98 J + 1/2mv^2

It is just what I assume.. I might be wrong. Please help

 

[mfb]

Also, Ek = 0 (equilibrium)

Only if it actually rests at this point. It does so in figure 2, but it won't be at rest there after you release it at the table.
And where is the energy in the string?

x = 0,2 meter (from reference point)
Ep = 1/2 kx^2 = 0,98 J

That is a problematic approach. Not impossible, but it can easily lead to mistakes. Better consider the total energy in the string.

EM at fig 2 = EM at fig 3

No. We pulled the mass downwards with our hand in between.

x = 0,2 meter (from reference point)
Ep = 1/2 kx^2 = 0,98 J
Ek = max = 1/2mv^2

It is at rest at the table in figure 3.

 

[Helly123]

Only if it actually rests at this point. It does so in figure 2, but it won't be at rest there after you release it at the table.
And where is the energy in the string?That is a problematic approach. Not impossible, but it can easily lead to mistakes. Better consider the total energy in the string.No. We pulled the mass downwards with our hand in between.It is at rest at the table in figure 3.

Should i take into account the mgh ?

My logic is
In fig 3 there is a lot of Ep no Ek. When it released, all Ep will be transformed into Ek when it reached the new equilibrium position.

The mass will keep moving forward compress the string (it reaches max height), but no have restoring force. (Since it rubber string)

Is it right?

 

[haruspex]

Should i take into account the mgh ?

My logic is
In fig 3 there is a lot of Ep no Ek. When it released, all Ep will be transformed into Ek when it reached the new equilibrium position.

The mass will keep moving forward compress the string (it reaches max height), but no have restoring force. (Since it rubber string)

Is it right?

Yes, except "Ep will be transformed into Ek " is not the whole story.

 

[Helly123]

Yes, except "Ep will be transformed into Ek " is not the whole story.

Not a whole story?
When going from floor to equilibrium point
Ep floor - mgh = Ek (equilibrium)

When going from equilibrium to max point
Ek - mgh = Ep (max point)

Is it right?

 

[haruspex]

- mgh

That's the bit that was missing.

 

[Helly123]

That's the bit that was missing.

But, why i get wrong answer? This question is hard for me.

 

[Helly123]

When going from floor to equilibrium point
Ep floor - mgh = Ek (equilibrium)
(.5)(49)(0.04) - (0.5)(9.8)(.2) = Ekmax

When going from equilibrium to max point
Ek - mgh = Ep (max point)
(.5)(49)(0.04) - (0.5)(9.8)(.2) - (.5)(9.8)(x+0.2) = (.5)(49)(x^2)

 

[haruspex]

When going from floor to equilibrium point

Is that when the string becomes slack?

 

[Helly123]

I find constant k
Kx = mg
K = 49N/m

I find acceleration because of restoring force
Restoring force - mg = ma (fig 3)
49*(0.3) -(.5)(9.8) = (.5)a
a =

 

[Helly123]

I find constant k
Kx = mg
K = 49N/m

Find V when max after all Ep transformed into Ek
Ep in fig 3 - mgh = Ek max (x = 0)
1/2.49.(0.3)^2 - 1/2.9.8.0 = 1/2.1/2.v^2
V^2 = 49.0.09 . 2

Find the max distance when the v become zero when acceleration gravity slow it down by 9.8
Vt^2 = vo^2 -2as
0 = 49.0.09.2 - 2*9.8*S
S = 98.0.09 / 2*9.8
S = 0.45 (above the initial position)

S total is 30 cm + 45cm

Is it right?
Is the max height always surpass the initial point? No matter what mass and what k?

 

[tnich]

This part is correct:

I find constant k
Kx = mg
K = 49N/m

Here it is not clear what you are trying to do because your notation is so difficult to decipher:

Find V when max after all Ep transformed into Ek
Ep in fig 3 - mgh = Ek max (x = 0)
1/2.49.(0.3)^2 - 1/2.9.8.0 = 1/2.1/2.v^2
V^2 = 49.0.09 . 2

It would be much easier to help you if used LaTeX.
I suggest that you start by writing the equations for the total energy at the floor (Figure 3) and the total energy when the rubber string contracts back to its original length. Then use conservation of energy to solve for the kinetic energy at the original length. An important equation that I have not seen in your posts so far is the equation for the potential energy of the rubber string. You are going to need that to write the total energy equations.

 

[haruspex]

1/2.49.(0.3)^2 - 1/2.9.8.0

The mgh term is not zero. The mass rises from its lowest position to the relaxed string position..

 

[Helly123]

energy total in figure 3
only Ep spring

energy total when go to original length
Ek + mgh

energy total in fig 3 and at original length is the same
Ep spring = Ek + mgh
$\frac{1}{2}k*x^2$ = $\frac{1}{2}m*v^2$ + mgh
$\frac{1}{2}49*(0.09)$ = $\frac{1}{2}\frac{1}{2}*v^2$ + $\frac{1}{2}*9.8*0,3$
1/2 v^2 = 49*(0.09) - 49(0,06)
v^2 = 49*(0.06)
max height is when v = 0
vt^2 = vo^2 - 2as
0 = 49*(0.06) - 2*9.8*s
s = $\frac{49*(0.06) }{2*9.8}$
s = $\frac{0.3}{2}$
s = 0.15 meter

s total from the floor
0.15 m + 0.3 m = 0.45 m = 0.45 cm

is it right??

 

[haruspex]

Looks right, but there is a quicker way.
You know k(0.1)=mg, and from work conservation you can write ½k(0.3)²=mgh, so h=0.45.

 

[Helly123]

Looks right, but there is a quicker way.
You know k(0.1)=mg, and from work conservation you can write ½k(0.3)²=mgh, so h=0.45.

thank you ^^
but why
½k(0.3)²=mgh ?
when max height Ep spring not zero right?

 

[Helly123]

oh, the rubber string doesn't resist compression?
If it were a spring then the Ep spring not zero?

 

[haruspex]

oh, the rubber string doesn't resist compression?
If it were a spring then the Ep spring not zero?

Right.