Potential/kinetic/mechanical energy

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The discussion centers on a physics problem involving a toy rocket that weighs 10 N and has an initial kinetic energy of 40 J. At the top of its trajectory, the total mechanical energy is stated to be 140 J, leading to confusion about the correct height the rocket reaches. One participant calculates the height as 10 meters based on gravitational potential energy, while another argues it should be 14 meters, asserting that at the top, all energy is potential due to zero kinetic energy. The conversation reveals discrepancies in answer keys found online, suggesting that the problem and its solutions may have been inaccurately propagated. Ultimately, the consensus leans towards 14 meters as the correct height, highlighting the importance of verifying sources in physics problems.
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Homework Statement


A toy rocket that weighs 10 N blasts straight up from ground level with an initial kinetic energy of 40J. At the exact top of its trajectory, its total mechanical energy is 140J. To what vertical height above the ground does it rise, assuming no air resistance?

2a. Relevant equations
gravitational potential energy = mgh
potential + kinetic = mechanical energy

3a. The attempt at a solution
if the rocket has 40J of energy on the ground, it has gained 100J at the top of its trajectory*.
GPE = 100J
GPE = mgh
100 = (10)h
h = 10 meters

*I think this is true, but I can't explain why. It's not like it lost 40J of energy to reach the top, but if it's at the top, it isn't moving anymore, so the final kinetic energy is zero, yes?

also, is there a way to use kinematic equations to solve this problem? here's what I tried:
2b. Relevant equations
gravitational potential energy = mgh
potential + kinetic = mechanical energy
kinetic energy = 1/2mv^2
weight = mass * gravity
v(final)^2 = v(initial)^2 + 2ad

3b. The attempt at a solution
10N = m(10); m = 1kg
initial kinetic energy = 40J
40 = 1/2 m v^2; initial v^2 = 80
final velocity = 0
0 = 80 + 2(-10)d
d = 4 meters

I feel like I'm missing something fairly obvious. Thanks for any light you can shed.
 
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Yes, you are missing something obvious. It is a toy rocket, toy engines are working for a while ...
Let me know if you need more hints
 
J827 said:

Homework Statement


A toy rocket that weighs 10 N blasts straight up from ground level with an initial kinetic energy of 40J. At the exact top of its trajectory, its total mechanical energy is 140J. To what vertical height above the ground does it rise, assuming no air resistance?

2a. Relevant equations
gravitational potential energy = mgh
potential + kinetic = mechanical energy

3a. The attempt at a solution
if the rocket has 40J of energy on the ground, it has gained 100J at the top of its trajectory*.
GPE = 100J
GPE = mgh
100 = (10)h
h = 10 meters

*I think this is true, but I can't explain why. It's not like it lost 40J of energy to reach the top, but if it's at the top, it isn't moving anymore, so the final kinetic energy is zero, yes?

also, is there a way to use kinematic equations to solve this problem? here's what I tried:
2b. Relevant equations
gravitational potential energy = mgh
potential + kinetic = mechanical energy
kinetic energy = 1/2mv^2
weight = mass * gravity
v(final)^2 = v(initial)^2 + 2ad

3b. The attempt at a solution
10N = m(10); m = 1kg
initial kinetic energy = 40J
40 = 1/2 m v^2; initial v^2 = 80
final velocity = 0
0 = 80 + 2(-10)d
d = 4 meters

I feel like I'm missing something fairly obvious. Thanks for any light you can shed.
Yes, you are missing something.

The rocket goes straight up. At the top of its trajectory, what is its speed? Is it zero? If so, what is its kinetic energy at that instant?
 
Henryk said:
Yes, you are missing something obvious. It is a toy rocket, toy engines are working for a while ...
Let me know if you need more hints

I'm sorry...I don't know what you're trying to hint at here. Are you trying to say that the kinetic energy is still 40 at the top because of the engine? That doesn't make any sense, and is in direct conflict with this:

SammyS said:
The rocket goes straight up. At the top of its trajectory, what is its speed? Is it zero? If so, what is its kinetic energy at that instant?

I said that before. The speed at the top is zero, so kinetic energy should be zero. That would make the 140J all potential, resulting in an answer of 14 meters.

I have been told that the answer to the problem is 10 meters, but I don't understand how the GPE is only 100J. energy isn't a vector quantity, so can't be a negative value...
 
The rocket is launched straight up. That means no horizontal velocity and we can assume that it moves straight up.
At the top, the velocity is zero, therefore, kinetic energy is zero and all energy is potential.
Therefore, the height is 140 J/10 N = 14 m.
 
J827 said:
I have been told that the answer to the problem is 10 meters
By a usually reliable source?
Everyone here seems to agree it's 14m.
 
haruspex said:
By a usually reliable source?
Everyone here seems to agree it's 14m.

yeah...the answer key. however, just because it's usually reliable doesn't mean it's infallible. however, this one seems to have been perpetuated for a long time... through Google, I found this problem on several websites, and the answer given almost every time was 10 m. The only site where I found a handwritten solution gave the answer as 14 m. Gets to the point where it's really better to talk to real people, you know? Thanks for the help.
 
J827 said:
yeah...the answer key. however, just because it's usually reliable doesn't mean it's infallible. however, this one seems to have been perpetuated for a long time... through Google, I found this problem on several websites, and the answer given almost every time was 10 m. The only site where I found a handwritten solution gave the answer as 14 m. Gets to the point where it's really better to talk to real people, you know? Thanks for the help.
Looks like the question and answer have been blindly copied around from some common erroneous source.
 
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