Work-Energy Conservation Question

1. Feb 1, 2017

Arman777

1. The problem statement, all variables and given/known data
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 ?

2. Relevant equations
$Δ(ME)=0$
$U_s=\frac 1 2kx^2$

3. The attempt at a solution
$\frac 1 2kx^2-mg(30.2)=0$
then we take the system to the moon.
Lets 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

2. Feb 1, 2017

TomHart

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

3. Feb 1, 2017

Arman777

In which equation ?

4. Feb 1, 2017

TomHart

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.

5. Feb 1, 2017

TomHart

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

6. Feb 1, 2017

Arman777

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

7. Feb 1, 2017

QuantumQuest

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.

8. Feb 1, 2017

Arman777

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 ?

9. Feb 1, 2017

QuantumQuest

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

10. Feb 1, 2017

Arman777

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

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

$H=30.2m$

11. Feb 1, 2017

QuantumQuest

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.

12. Feb 1, 2017

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

13. Feb 1, 2017

Arman777

yep there was :)