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hello478
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Poster reminded to type their questions into PF instead of posting images
- Homework Statement
- a general question asked below
- Relevant Equations
- a= 9.81 m/s^2
You're free to choose either sign, as long as other quantities are consistent. For example, if a ball is thrown upward with a velocity of 20 m/sec, the acceleration due to gravity will be in the opposite direction, so would be -9.81 m/sec^2.hello478 said:Homework Statement: a general question asked below
Relevant Equations: a= 9.81 m/s^2
View attachment 342392
so like ifMark44 said:You're free to choose either sign, as long as other quantities are consistent. For example, if a ball is thrown upward with a velocity of 20 m/sec, the acceleration due to gravity will be in the opposite direction, so would be -9.81 m/sec^2.
Yes, but keep in mind that we're talking about a particular moment. It's probably better to consider "up" as the positive direction, though.hello478 said:object moving up :
i would take down as + and up as -
so object moving up would have velocity -20
and acceleration 9.81
If we're talking about the same scenario as above, wouldn't the velocity be 20 (m/sec) at the moment in question?hello478 said:and for object moving down the acceleration would still be 9.81 and the velocity would be 9.81
my bad, was an errorMark44 said:Yes, but keep in mind that we're talking about a particular moment. It's probably better to consider "up" as the positive direction, though.
If we're talking about the same scenario as above, wouldn't the velocity be 20 (m/sec) at the moment in question?
ok thank youMark44 said:Although I said you could go either way, I recommend that you treat upward velocities as positive so that g is negative. It's probably less confusing to do things this way.
My preference is otherwise. For me, ##g## is the named constant for the magnitude of the acceleration of gravity at the Earth's surface. Because it is a magnitude, it is always positive.Mark44 said:Although I said you could go either way, I recommend that you treat upward velocities as positive so that g is negative. It's probably less confusing to do things this way.
For me it's more than a preference, it's a MUST to avoid confusion when free body diagrams (FBD) and their translation into an equation through Newton's second law is involved.jbriggs444 said:My preference is otherwise. For me, ##g## is the named constant for the magnitude of the acceleration of gravity at the Earth's surface. Because it is a magnitude, it is always positive.
kuruman said:For me it's more than a preference, it's a MUST to avoid confusion when free body diagrams (FBD) and their translation into an equation through Newton's second law is involved.
Your logic eludes me.hello478 said:what about when we have gravitational potential energy and work?
then the acceleration cannot be negative so acceleration is always to be positive?
Why can the acceleration not be negative? When you say "acceleration" all by itself, the default interpretation is the acceleration vector which can be positive or negative depending on how you choose your axes as I indicated in post #10. It looks like you are still confusing the basic idea of acceleration with the specific number g = 9.8 m/s2.hello478 said:what about when we have gravitational potential energy and work?
then the acceleration cannot be negative so acceleration is always to be positive?
if an object is raised and then droppedkuruman said:Why can the acceleration not be negative? When you say "acceleration" all by itself, the default interpretation is the acceleration vector which can be positive or negative depending on how you choose your axes as I indicated in post #10. It looks like you are still confusing the basic idea of acceleration with the specific number g = 9.8 m/s2.
Please illustrate what you mean with an example.
Absolutely. Acceleration has magnitude (size) and direction. Near the surface of the Earth the acceleration of a falling object has direction towards the center of the Earth and magnitude 9.81 m/s2 (don't forget the units). The magnitude of the acceleration is just a number with units, 9.81 m/s2, and has no direction. We use the symbol "g" instead of that number so that we don't have to write 9.81 m/s2 all the time.hello478 said:so are g=9.81 and acceleration different things?
What exactly cannot be negative? Fill in your reasoning, please.hello478 said:it cant be negative as energy is scalar
See if this helps.hello478 said:if an object is raised and then dropped
so the gravitational potential energy would be mgh
but what if i choose g to be negative
it cant be negative as energy is scalar
so are g=9.81 and acceleration different things?
You mean so that the (vertical component of the) acceleration is negative. For example, if we take the upward direction to be the positive ##y##-direction, then ##a_y=-g \approx -9.8 \ \mathrm{m/s^2}##. Because ##g \approx +9.8 \ \mathrm{m/s^2}##.Mark44 said:I recommend that you treat upward velocities as positive so that g is negative.
gravitational potential energyjbriggs444 said:What exactly cannot be negative? Fill in your reasoning, please.
The direction of motion tells you nothing about the sign of the acceleration. You can be moving in the positive direction and your acceleration can be either positive or negative. Likewise, you can be moving in the negative direction and your acceleration can be either positive or negative.hello478 said:
yes it did help a lotSteve4Physics said:See if this helps.
We can take any convenient position to define where gravitational potential energy (GPE) is zero. We often take GPE=0 at the earth’s surface but that’s not essential.
E.g. the position where GPE=0 could be taken as the floor of the physics lab’ in the basement. Sometimes we don’t even need to bother choosing where GPE=0.
That’s because we are really only interested in changes in GPE. ##\Delta (GPE) = mg\Delta h##.
This equation assumes ##g## is a positive constant (as in previous posts) and that ##h## increases going upwards, i.e. that we are using the ‘upwards-is-positive’ sign-convention.
E.g. when you drop a ball ##\Delta h## is negative. The kinetic energy gained equals the GPE lost (assuming no air resistance). In this case ##\Delta (GPE)## is a negative quantity because GPE is reduced.
Warnnig: if you choose a sign-convention where upwards is negative, this reverses the sign of ##\Delta h##. So maybe to avoid confusion, if using ##\Delta (GPE)##, stick to upwards-is-positive. Or just stick to upwards-is-positive always.
Side-note: things are a bit more complicated if ##g## is not constant (e.g. for a rocket travelling from earth to a satellite’s orbit). So you can’t use ##\Delta (GPE) = mg\Delta h## in these situations.
Scalars can be negative; magnitudes cannot be negative.hello478 said:it cant be negative as energy is scalar
Which can be negative, as explained at least twice.hello478 said:gravitational potential energy