Work-Energy Conservation Question

[Arman777]

Homework Statement

A $10.0kg$ block falls $30.0m$ onto a vertical spring whose lower end is fixed to a platform.When the spring reaches its maximum comprassion of $0.200m$,it is locked in place.The block is then removed and spring apparatus is transported to the Moon,where the gravitational acceleration is $\frac g 6$.
A $50.0kg$ astronaut then sits on top of the spring and the spring is unlocked so that it propels the astronaut upward.How high above the initial point does the astronaut rise ?

Homework Equations

[/B] $Δ(ME)=0$
$U_s=\frac 1 2kx^2$

The Attempt at a Solution

$\frac 1 2kx^2-mg(30.2)=0$
then we take the system to the moon.
Let's write another $Δ(ME)=0$
$mgh-\frac 1 2kx^2=0$ In the highest point the kinetic energy will be zero.So the change in kinetic energy is zero.

In both equations $\frac 1 2kx^2$ are same and If we put new mass and new $g$ we get same answer which its $30.2m$ but answer says its $36m$.

Probably I forget some energy term somewhere or I took wrong height Idk

Thanks

 

[TomHart]

I think you just have a math error in your work somewhere.

 

[Arman777]

I think you just have a math error in your work somewhere.

In which equation ?

 

[TomHart]

In which equation ?

Your work wasn't shown, so I don't know where it is. I thought I understood your logic behind the problem - that the potential energy on the Earth gets converted to potential energy on the moon. That's what I thought you worked.

 

[TomHart]

In other words, because I think your method is right, I conclude that there must be a math error somewhere.

 

[Arman777]

Your work wasn't shown, so I don't know where it is. I thought I understood your logic behind the problem - that the potential energy on the Earth gets converted to potential energy on the moon. That's what I thought you worked.

so both have same spring potantial energy
so in Earth $U_{g_e}=mg(30.2m)$

İn moon $U_{g_m}=6m\frac g 6(h)$

since they must be equal so h=30.2m

 

[QuantumQuest]

Homework Statement

A 10.0kg10.0kg10.0kg block falls 30.0m30.0m30.0m onto a vertical spring whose lower end is fixed to a platform.When the spring reaches its maximum comprassion of 0.200m0.200m0.200m,it is locked in place.The block is then removed and spring apparatus is transported to the Moon,where the gravitational acceleration is g6g6\frac g 6.
A 50.0kg50.0kg50.0kg astronaut then sits on top of the spring and the spring is unlocked so that it propels the astronaut upward.How high above the initial point does the astronaut rise ?

Homework Equations

Δ(ME)=0Δ(ME)=0Δ(ME)=0
Us=12kx2Us=12kx2 U_s=\frac 1 2kx^2

The Attempt at a Solution

12kx2−mg(30.2)=012kx2−mg(30.2)=0\frac 1 2kx^2-mg(30.2)=0
then we take the system to the moon.
Lets write another Δ(ME)=0Δ(ME)=0Δ(ME)=0
mgh−12kx2=0mgh−12kx2=0 mgh-\frac 1 2kx^2=0 In the highest point the kinetic energy will be zero.So the change in kinetic energy is zero.

In both equations 12kx212kx2\frac 1 2kx^2 are same and If we put new mass and new ggg we get same answer which its 30.2m30.2m30.2m but answer says its 36m36m36m.

Probably I forget some energy term somewhere or I took wrong height Idk

Thanks

When you take the system to the moon, how much energy is stored in the spring? You can find it from the first part. Then, when the astronaut is on the higher point after his propelling, what does the previous mentioned energy has transformed to? How many unknowns have you there?As an extra hint, remember to account for the decompressed length of the spring.

 

[Arman777]

When you take the system to the moon, how much energy is stored in the spring? You can find it from the first part. Then, when the astronaut is on the higher point after his propelling, what does the previous mentioned energy has transformed to? How many unknowns have you there?As an extra hint, remember to account for the decompressed length of the spring.

Could you tell me in which equation ı made wrong..cause I found spring potantial energy,and in moon it turns to potantial energy.Theres just one unknown.If we add kinetic energy I need two equations I guess but again since its asking max height it will be zero ? .I included decompressed length didnt I ?

 

[QuantumQuest]

Could you tell me in which equation ı made wrong..cause I found spring potantial energy,and in moon it turns to potantial energy.Theres just one unknown.If we add kinetic energy I need two equations I guess but again since its asking max height it will be zero ? .I included decompressed length didnt I ?

Can you post the final equation (with numbers) you have in order to find the height above the initial point that the astronaut rises?

 

[Arman777]

$\frac 1 2kx^2=mg(30.2m)=2959.6J$ in earth
so in moon

$2959.6J=Ma_gH=6m\frac g 6 H=mgH$

$H=30.2m$

 

[QuantumQuest]

2959.6J=MagH=6mg6H=mgH2959.6J=MagH=6mg6H=mgH2959.6J=Ma_gH=6m\frac g 6 H=mgH

H=30.2mH=30.2mH=30.2m

H is not 30.2 m. Taking $g = 9.81 \frac{m}{sec^2}$ you have $2962.62 J$ for the stored energy and you have to equate this with the potential energy at max height. So, substitute for the mass of astronaut, the value of g on the moon $\frac{g_{earth}}{6}$ and the unknown height plus the decompression in the right side and you have the answer that is approx. $36 m$ as you state in your OP.

 

[Arman777]

I see.I am taking total mass $(m_{box}+m_{a})$ I should just take $m_a$ Idk why did I such a silly mistake thanks

 

[Arman777]

In other words, because I think your method is right, I conclude that there must be a math error somewhere.

yep there was :)