Work-Energy Conservation Question

In summary, the problem involves a block of mass 10.0kg falling 30.0m onto a vertical spring, compressing it to 0.200m before being locked in place. The spring apparatus is then taken to the moon where the gravitational acceleration is 1/6 of that on Earth. A 50.0kg astronaut sits on top of the spring and is propelled upward when the spring is unlocked. The question asks for the height above the initial point that the astronaut will rise. Using the equations Δ(ME)=0 and Us=1/2kx2, the spring potential energy is found to be 2959.6J on Earth and equated to the potential energy on the moon, giving
  • #1
Arman777
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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
 
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  • #2
I think you just have a math error in your work somewhere.
 
  • #3
TomHart said:
I think you just have a math error in your work somewhere.
In which equation ?
 
  • #4
Arman777 said:
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.
 
  • #5
In other words, because I think your method is right, I conclude that there must be a math error somewhere.
 
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  • #6
TomHart said:
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
 
  • #7
Arman777 said:

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.
 
  • #8
QuantumQuest said:
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 ?
 
  • #9
Arman777 said:
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?
 
  • #10
##\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
Arman777 said:
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.
 
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  • #12
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
TomHart said:
In other words, because I think your method is right, I conclude that there must be a math error somewhere.

yep there was :)
 
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Related to Work-Energy Conservation Question

1. What is the work-energy conservation principle?

The work-energy conservation principle is a fundamental law of physics that states that the total work done on an object is equal to the change in its kinetic energy. This means that the work done by all forces acting on an object, including non-conservative forces like friction, must result in a change in the object's kinetic energy.

2. How is the work-energy conservation principle applied in real-world situations?

The work-energy conservation principle is applied in various real-world situations, such as calculating the speed of a roller coaster at different points along the track or determining the amount of force needed to launch a rocket into space. It is also used in engineering and design processes to optimize the efficiency and performance of machines and systems.

3. Can the work-energy conservation principle be violated?

No, the work-energy conservation principle is a fundamental law of nature and cannot be violated. In an isolated system, the total amount of energy, including both kinetic and potential energy, remains constant regardless of any changes in the system. This is known as the law of conservation of energy.

4. What is the difference between conservative and non-conservative forces in relation to the work-energy conservation principle?

Conservative forces, such as gravity and elastic forces, do not dissipate energy and therefore do not change the total mechanical energy of a system. On the other hand, non-conservative forces, like friction and air resistance, dissipate energy and result in a decrease in the total mechanical energy of a system.

5. How is the work-energy conservation principle related to the law of conservation of momentum?

The law of conservation of momentum states that the total momentum of a system remains constant in the absence of external forces. The work-energy conservation principle is related to this law, as the change in an object's kinetic energy is equal to the net work done on it, which is also equal to the impulse, or change in momentum, of the object. Therefore, the work-energy conservation principle can be seen as an extension of the law of conservation of momentum to include changes in kinetic energy.

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