Question

12) Can the kinetic energy of an object be negative? Explain your answer.

14) Two identical objects move w/ speeds of 5.0 m/s and 25.0 m/s. What is the ratio of their kinetic energies?

Answer 1

Hello @Lessis, that was fast! xD

Answer 2

Was browsing questions. xP

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How is the kinetic energy of an object defined?

Answer 3

No idea. hold on

Answer 4

\[KE=\frac{1}{2}mv^2\]\[W_net=\Delta KE\]

Answer 5

So.
\[k = \frac{ 1 }{ 2 }mv^{2}\]

In regards to 12)
The kinetic energy of an object is defined by its mass, and the magnitude of its velocity squared. Both of this are positive, right?

Answer 6

By the way, my book is 2007 Copyright Holt

Answer 7

Multiplying two positive numbers always gives you a positive number. Therefore, K is positive.
For 14) I'll assume "identical" means they have the same mass.

Therefore, the ratio of their kinetic energies is:

\[\frac{ k_{1} }{ k{2} }= \frac{ \left( \frac{ 1 }{ 2 } mv_{1}^{2}\right)}{ \left( \frac{ 1 }{ 2 } mv_{2}^{2}\right) }\]
\[= \frac{ v_{1}^{2} }{ v_{2}^{2} }\]

Answer 8

Wait, so #12 is NO... but why? o_o

Answer 9

Funfact; if Tachyons exist (Particles that can move faster than the speed of light), you can have negative kinetic energy. However, this particles have only been postulated and have never been found to actually exist. (That's also the reason as to why time travel to the past is impossible, at least in theory)

Answer 10

The kinetic energy of an object is defined as a product of its mass and its velocity squared.
Every object has positive mass, and the square of any number (In this case, the objects velocity) is also positive. Therefore, if you multiply this two numbers, you'll get a positive number.

Answer 11

fun facts ftw

Answer 12

Actually let me re-read that -- @AllTehMaffs that took a long time for you to come online -- and I'll get back to you on that.

Answer 13

I never check the email that's linked to this account ^_^ Surprises whenever I log in!

Answer 14

@Lessis would the negative K be because of "negative mass" or because the velocity is imaginary (i^2 = -1?)

Answer 15

You can't have negative kinetic energy. D:

At least, not in what is our current understanding of the universe.

I've never really understood what negative mass, imaginary time, etc meant. ._.
Also, Isn't velocity defined as a real number? Like when you define a wave as the real part of an exponential with an imaginary power.

Answer 16

Thanks @Lessis for explaining #12, I got it. [:

Answer 17

De nada. ;D

Answer 18

@Lessis I think that the "realness" of velocity has to do with it being an observable in QM - I don't know though :/

@kittiwitti1 How's about #14?

Answer 19

Well. Velocity is a vector in a metric space. And V^2 is obviously referring to the magnitude of velocity squared. And the magnitude of any vector in a metric space is defined as positive. So I'd say its magnitude has to be real, because of where it "lives".
(You also can't have imaginary positions, and iirc, time can only be imaginary for gauge bosons or something like that, so any "normal" particle can't have an imaginary time, therefore, considering velocity as dx/dt, it must be real too)

Answer 20

I need help on #11 and 13 because the stupid answer key doesn't tell me the answers.

11) A person drops a ball from the top of a building while another person on the ground observes the ball's motion. Each observer chooses his/her own location as the level for 0 potential energy. Will they calculate the same values for:
a) the potential energy associated w/ the ball?
b) the change in potential energy associated w/ the ball?
c) the ball's kinetic energy?

13) Can the gravitational potential energy associated with an object be negative? Explain your answer.

Answer 21

My answer for #11 was No, Yes and Yes
Reasons:
a) No, because they are viewing the ball's motion with different levels for zero potential energy (or something like that)
b) Yes, because no matter where the level for zero potential energy is the change is still the same
c) NOT sure.

Answer 22

Assuming the observers are static, they will calculate the same kinetic energy. This is because they both are seeing that the ball has the same mass and moves with the same velocity.

Answer 23

Also, isn't gravitational potential energy defined as negative? o.o

Answer 24

Okay, but are the other two parts correct?

Answer 25

I think so.

Answer 26

Alrighty, thanks.
How's #13 solved, though? c:

Answer 27

http://hyperphysics.phy-astr.gsu.edu/hbase/gpot.html

"The gravitational potential energy near a planet is then negative, since gravity does positive work as the mass approaches. This negative potential is indicative of a "bound state"; once a mass is near a large body, it is trapped until something can provide enough energy to allow it to escape. "

Answer 28

Although, now that I think about it, I'm not sure about what you're defining as gravitational potential energy. D:

Answer 29

Wait, whut o.o
That sounds like #15 though

Answer 30

Which is in this other question I closed b/c it was answered...
http://openstudy.com/updates/528bf363e4b09a30e7655cfa

Answer 31

Gravitational potential energy is negative because it's the energy with which an object (Usually, the earth) "pushes" you to it due to its mass. In that case, it can never be positive (Unless, you know, negative masses and all that crazy stuff).

Answer 32

Let's ignore the "crazy stuff" and just focus on the most logical, alright? xD

Answer 33

- But the crazy stuff is the reason why I study physics. :< Crazy is pretty -

Ignoring the crazy stuff. It depends on how you define gravitational potential energy.
Is it U=mgh?

Answer 34

LOL.
Not that I know of, let me check.

Answer 35

:>

Answer 36

the equation I found is:\[PE_g=mgh\]

Answer 37

Gravitational potential energy = mass x free-fall acceleration x height?

Answer 38

Technically, yes. At least if you're close to the earth. Although, now I'm curious: @AllTehMaffs
Question: Gravitational Potential Energy is usually defined as negative. But why is it possitive when using it in the simpler form U=mgh? O:

Answer 39

He's busy with two other problems lolol @Lessis

Answer 40

And now I just read about it. D:

"Since the zero of gravitational potential energy can be chosen at any point (like the choice of the zero of a coordinate system), the potential energy at a height h above that point is equal to the work which would be required to lift the object to that height with no net change in kinetic energy. "

So, U = mgh.
M is possitive. g is possitive.
What about h? h depends on your reference point (Your zero). And can be above or bellow said point. If it's above, it's a positive energy (Positive*Positive*Positive = Positive), and it's the energy required to get to that point.
If it's below, it's negative (Positive*Positive*Negative = Negative), and that means the system (In this case, the Earth) pushes you towards it.

Answer 41

I'm watching this thread too....

Answer 42

@AllTehMaffs does @Lessis mean that it can't be negative?

Answer 43

@Lessis I think it's positive on the smaller scale because the potential energy is defined in relation to work done
\[W=Fd \quad \quad ; \quad \quad W = -\Delta U\]
\[ F=mg \quad d=h\]
And work done against gravity is negative; so the potential energy of raising an object up is positive.

@kittiwitti1 I'm not 100% sure - it mostly just depends on where you say your zero is... but physically talking, having a negative gravitational potential doesn't make much sense. It would mean that gravity would want to push the object away....?

Answer 44

@Lessis (I know you typed that work stuff like half an hour ago, but I had written it out and forgot to post it and so... I lose :P)

Answer 45

Lessis isn't here, @AllTehMaffs

Answer 46

I KNOWZ

Answer 47

^^

Answer 48

So by your estimate, Lessis is saying that gravitational potential energy can't be negative?

OH WAIT. mgh: if m and h are positive, then whether g is positive or negative doesn't affect the sign of the end result; then gravitational potential energy is positive! :D

@AllTehMaffs does my explanation make sense?

Answer 49

unfortunately no :/

+++ = +
++- = - <------ this is your scenario if g = -
+-- = +
--- = -


And it's not whether or not g is positive or negative, it has to do with h. We're looking at U, not g.

Answer 50

But h can't be negative?

Answer 51

That's the kicker. If you say h=0 is on top of a mountain, then everything below that mountain is negative h. It's super confusing :P

Answer 52

Wait, so mgh; if g is positive then h can be either negative or positive?
I understand that mass must stay positive for obvious reasons.

Answer 53

yah

Answer 54

isnt gravitational potential enegry always negative?

Answer 55

by defitnition itself?

Answer 56

For stellar distances, yes - convention says it's generally positive near earth, though

Answer 57

Then my explanation would be "Since g and h can both have negative/positive values, the equation can produce an end value of either negative OR positive gravitational potential energy"?

Answer 58

leave the g out of it - it's dependent on h.

Answer 59

but if the value of gpe is 0 at infinity, it will always be less than zero for a finite distance

Answer 60

Oh, good point. :p

@AllTehMaffs, I'll message Lessis and tell him to give you a medal since you're continuing his work :3

Answer 61

No, Maffs is right @xxyyxyyx, the gravitational potential energy CAN be negative AND positive.

Answer 62

@xxyyxyyx is getting more technical - the actual definition of gravitational potential says (like he said) that U=0 @ infinity, so anything closer than infinity is negative.

Answer 63

@xxyyxyyx http://en.wikipedia.org/wiki/Potential_energy#General_formula

but it's based on convention

Answer 64

Well, yes, but I'm not learning about that atm

Answer 65

exactly

Answer 66

Maffs, how do I write the explanation? @ - @
Mine goes like "Yes - because the h can be negative..."
But if g is negative then it'd make a positive, so I HAVE to include the g in the explanation, right?

Answer 67

Don't get caught up on g - for this purpose it'll always be positive

Answer 68

Oh, because the projectile is always being thrown downwards?

Answer 69

@xxyyxyyx she hasn't done gravitation yet

Answer 70

What? c:

Answer 71

@kittiwitti1 gravity for all intents and purposes right now never changes direction; whether something is moving up or moving down it has the same value - but in potential energy, its value will always be positive

Answer 72

Intents?
But in forces we learned that deceleration -9.8 m/s/s was when a projectile went upwards...

Answer 73

you've never seen
\[F_g = G\frac{M m}{r^2}\] right?

Answer 74

No... I haven't... what is that ? o.o

Answer 75

Gravitation

Answer 76

Ahh, okay. Gotcha

Answer 77

k, on to #14!

Answer 78

for projectile motion, you deal with vectors. ***Energetics doesn't deal with vectors*** Gravitational acceleration is 9.8m/s^2 in the negative y direction - that's where the negative comes from.

Answer 79

what was the explanation for the question?

Answer 80

Gotcha. So which formula do I use for #14?
It's okay @xxyyxyyx, I got the answer

Answer 81

lessis has already answered 14

Answer 82

14) Two identical objects move w/ speeds of 5.0 m/s and 25.0 m/s. What is the ratio of their kinetic energies?

Kinetic energy ratio? o_o

Answer 83

yah, like the 7th post in this thread Lessis did it

Answer 84

Multiplying two positive numbers always gives you a positive number. Therefore, K is positive.
For 14) I'll assume "identical" means they have the same mass.

Therefore, the ratio of their kinetic energies is:

k1k2=(12mv21)(12mv22)

=v21v22

Answer 85

wat

Answer 86

square of the first speed/ square of the second speed

Answer 87

OH. Hold up while I go read that~

Answer 88

sry was copy pasting lessis answer. didnt turn out excatly

Answer 89

@xxyyxyyx sorry that I'm grouchy - super tired and on a sugar crash :P

Answer 90

Sugar crash? xD

Answer 91

@xxyyxyyx this one?
I thought it'd make things easier.

Answer 92

yah, that one

Answer 93

yes. sry i have a lousy internet

Answer 94

So starting from the last part, \[\frac{v_1^2}{v_2^2}=\frac{5.0^2}{25.0^2}=\frac{25}{625}=0.04\]

Answer 95

The ratio is 0.04?