Mechanical energy and frames of reference.

AI Thread Summary
The discussion revolves around calculating gravitational potential energy (PE) and kinetic energy (KE) of a water balloon dropped from a height, using two different frames of reference: one from the chancellor at the balcony and the other from a student on the ground. In the chancellor's frame, the balloon's height is considered negative as it falls, leading to a decrease in potential energy, while the student's frame sees the height as positive, resulting in an increase in potential energy. The participants clarify that total mechanical energy appears not to be conserved when viewed from different frames due to the varying reference points for potential energy, but the change in energy remains consistent across frames. They conclude that the conservation of energy holds true, as the change in potential energy equals the change in kinetic energy, regardless of the frame used. The discussion emphasizes the importance of the chosen reference point in energy calculations.
InertialRef
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Homework Statement



a)Suppose the chancellor of the university drops a 2.00 kg water balloon from the administration
building balcony 10.0 m above the ground. The chancellor takes the origin of his vertical axis
to be even with the balcony. A student standing on the ground below the chancellor decides
she would rather have the origin of her coordinate system be the ground at her feet.
b)Calculate the value of the gravitational potential energy of the balloon before it is dropped and
just as it hits the ground for each of the frames of reference.

chancellor frame:
PE bef= (2.00)(9.81)(0 m) = 0 J
PE aft= (2.00)(9.81)(10 m) = 196.2 J

student frame:
PE bef= (2.00)(9.81)(10 m) = 196.2 J
PE aft= (2.00)(9.81)(0) = 0 J

c)Calculate the value of the kinetic energy of the balloon before it is dropped and just as it hits the ground for each of the frames of reference.

chancellor frame:
KE bef= (0.5)(2.00)(0)^2 = 0
KE aft= (0.5)(2.00)(9.81*(√(20/9.81)))^2 = -196.2

I calculated for the final velocity using the principle kinematics equation.

student frame:
KE bef= 0
KE aft= -196.2

d)Calculate the total mechanical energy of the balloon before it is dropped and just as it hits the ground for each of the frames of reference.

chancellor frame:
TME bef=
TME aft=

student frame:
TME bef=
TME aft=


Homework Equations



KE = (0.5)m(v^2)
PE = mgh

The Attempt at a Solution



I've solved for most of it, since it was pretty simple, but I'm stuck at part d. Shouldn't total mechanical energy always be conserved? How is it that for the president, total mechanical energy isn't conserved? The total energy initially = 0, then it increases. Why does it do that?

I understand that when looked at from one frame of reference only, the total energy is conserved. But when looked at from two frames of references, total energy only appears to not be conserved, but it actually is. Is there some way to correct for this, or is the only way to see if mechanical energy is conserved is to observe such motion from one frame of reference?
 
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Firstly, how is the kinetic energy negative?

Secondly, in the frame of reference of the chancellor, the balloon's height is negative. Do you see why?

Does the potential energy increase or decrease when an object comes closer to the Earth's surface?
 
Sourabh N said:
Firstly, how is the kinetic energy negative?

Secondly, in the frame of reference of the chancellor, the balloon's height is negative. Do you see why?

Does the potential energy increase or decrease when an object comes closer to the Earth's surface?

Whoops, sorry about that. Mixed up PE with KE, my mistake.

Yes, I can see why the balloon's height is negative. From the frame of reference of the student, the height is positive, so from the frame of reference of the chancellor, the height is negative. Potential energy decreases as it comes closer to the Earth's surface, but that's only if you take the frame of reference of the student, isn't it?
 
InertialRef said:
Whoops, sorry about that. Mixed up PE with KE, my mistake.

Yes, I can see why the balloon's height is negative. From the frame of reference of the student, the height is positive, so from the frame of reference of the chancellor, the height is negative.

Following your argument, PE aft in the chancellor's frame is changed, as below.

chancellor frame:
PE bef= (2.00)(9.81)(0 m) = 0 J
PE aft= (2.00)(9.81)(-10 m) = -196.2 J

Potential energy decreases as it comes closer to the Earth's surface, but that's only if you take the frame of reference of the student, isn't it?
No, potential energy varies, but CHANGE in potential energy is the same for both, since the only difference between the two frames is the zero of the potential energy.
 
Sourabh N said:
Following your argument, PE aft in the chancellor's frame is changed, as below.

chancellor frame:
PE bef= (2.00)(9.81)(0 m) = 0 J
PE aft= (2.00)(9.81)(-10 m) = -196.2 J


No, potential energy varies, but CHANGE in potential energy is the same for both, since the only difference between the two frames is the zero of the potential energy.

So, would I be correct in saying that since the change in PE is equal to the change in KE, in both frames of reference, then this indicates that the energy is conserved and the only reason the two observers don't calculate the same values for mechanical energy is because the zero of the potential energy varies in each frame?
 
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