Question about acceleration of free fall

[hello478]

Homework Statement

a general question asked below

Relevant Equations

a= 9.81 m/s^2

 

[Mark44]

Homework Statement: a general question asked below
Relevant Equations: a= 9.81 m/s^2

View attachment 342392

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]

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.

so like if
object moving up :
i would take down as + and up as -
so object moving up would have velocity -20
and acceleration 9.81

and for object moving down the acceleration would still be 9.81 and the velocity would be 20m/s

am i right?

 

[Mark44]

object moving up :
i would take down as + and up as -
so object moving up would have velocity -20
and acceleration 9.81

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.

and for object moving down the acceleration would still be 9.81 and the velocity would be 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]

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?

my bad, was an error
i meant 20m/s btw

 

[Mark44]

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.

 

[hello478]

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.

ok thank you
will do that from now onwards

 

[jbriggs444]

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.

When you use $g$ in a problem, you have to decide which way gravity points relative to your coordinate system. If your coordinate system is upward positive then the acceleration due to gravity will be $-g$. If your coordinate system is downward positive then the acceleration due to gravity will be $+g$.

So $g$ is always 9.8 meters/sec². But the free fall acceleration $\vec{a}$ from gravity will be either +9.8 m/sec² or -9.8 m/sec² depending on your choice of coordinate system.

 

[hello478]

thanks!

 

[kuruman]

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.

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.

Example

A 3 kg mass is hanging vertically from a string in which the tension is 50 N as shown in the figure on the right. Find the net force acting on the mass.
Solution
1. Draw the FBD as shown on the right. Gravity defines the "down" direction. Clearly the direction opposite to "down" is "up".
2. The net force is the vector sum of all the forces and you want to write an algebraic equation for it. Before you do that, you have to choose which direction is positive and which is negative then proceed with the solution.

Choice I "Up" is positive
The net force is the sum of what is "up" with a "+" sign up front and what is "down" with a "-" sign up front. F_net = +T + (-mg) = T - mg = 50 N - 2 (kg) * 10 (m/s²) = 50 N - 20 N = 30 N.
Since "up" is positive and the algebraic result is positive, the answer is
Answer: The net force is 30 N up.

Choice II "Down" is positive
The net force is the sum of what is "up" with a "-" sign up front and what is "down" with a "+" sign up front. F_net = (-T) + (mg) = -T + mg = -50 N + 2 (kg) * 10 (m/s²) = -50 N + 20 N = -30 N.
Since "down" is positive and the algebraic result is negative, the answer is
Answer: The net force is 30 N up.

It should be clear how either choice gives the same answer. It should also be clear that, when the time comes to replace g with a number, that number must be positive in either choice because g is the magnitude of a vector as noted by @jbriggs444. The direction of the acceleration of gravity has already been taken care of by the transcription of "up" and "down" into "+" or "-" when one casts the FBD in algebraic form.

 

[hello478]

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.

what about when we have gravitational potential energy and work?
then the acceleration cannot be negative so acceleration is always to be positive?

 

[jbriggs444]

what about when we have gravitational potential energy and work?
then the acceleration cannot be negative so acceleration is always to be positive?

Your logic eludes me.

Acceleration can be negative. Even when $g$ is always positive.
Acceleration can be negative. Even when contemplating gravitational potential energy and work.

In addition, work can be negative. Gravitational potential energy can be negative. The rate of change of gravitational potential energy with respect to vertical position can be negative if you have a downward-positive coordinate system.

 

[kuruman]

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.

Please illustrate what you mean with an example.

 

[Steve4Physics]

Hi @hello478. Another way to think about the direction of acceleration is this...

The direction of an object’s acceleration is always the same as the direction of the net force on the object.

That’s because $\vec F_{net} = m \vec a$, so that $\vec a = \frac {\vec F_{net}}m$.

For example, think about a ball thrown vertically upwards (with negligible air resistance). The only force on the ball is its weight, always acting vertically downwards. The direction of acceleration is therefore the same, always vertically downwards. This is true when the ball is rising, at the high-point or falling. The sign of the acceleration (+ or -) only depends on what sign-convention you have chosen for up/down.

Minor edit.

 

[hello478]

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.

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?

 

[kuruman]

so are g=9.81 and acceleration different things?

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.

 

[jbriggs444]

it cant be negative as energy is scalar

What exactly cannot be negative? Fill in your reasoning, please.

Scalars^(*) can be negative. Magnitudes (aka norms) are always positive or zero. You are correct that kinetic energy is always non-negative.

However, potential energy is always relative to an arbitrarily chosen reference. It can be negative. For instance, the potential energy of a weight in a hole is negative if we choose our reference to be at ground level.

(*) Here I take "scalar" in the sense of a vector space where we have vectors in a space and scalars from a field. You can multiply a vector by a small scalar to scale it down, decreasing its magnitude but leaving its direction unchanged. You can multiply a vector by a large scalar to scale it up, increasing its magnitude while leaving its direction unchanged. You can multiply a vector by a negative scalar. This will invert its direction in addition to possibly scaling it up or down.

 

[Steve4Physics]

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?

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.

 

[Herman Trivilino]

I recommend that you treat upward velocities as positive so that g is negative.

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 $g \approx +9.8 \ \mathrm{m/s^2}$.

 

[hello478]

What exactly cannot be negative? Fill in your reasoning, please.

gravitational potential energy

 

[Herman Trivilino]

View attachment 342392

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]

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.

yes it did help a lot
thank you

 

[haruspex]

it cant be negative as energy is scalar

Scalars can be negative; magnitudes cannot be negative.
A change in energy can be negative.

 

[jbriggs444]

gravitational potential energy

Which can be negative, as explained at least twice.