Q on basic principles of weight and mass

[44HP]

Hi Newbie here so please go easy on me..!

To keep this within my ability to understand can we please live in the perfect world of the junior physics class i.e. uniform density, no friction g=10 etc - thanks

I understand that 'weight' is mass x g

So the weight of something is how hard the Earth is 'sucking' that's why things weight differently on Earth and the Moon

But the objects mass remains the same...

Now I'm arguing with somebody (he's an engineer and I'm a pen pusher so I'm guessing he'd be right) on another forum about things like energy and momentum so we are slinging p1v1=p2v2 and e=(mv^2)/2 at each other.

The other chap is telling me the weight of something needs to be converted from pounds weight to pounds force before we apply the equations...

the energy and momentum equations both list Mass as a factor. The gravitational constant is necessary to convert the weight of the object (pounds force) to Mass (pounds-mass). When you weigh an object on a scale that is a reading in pounds-force, but for our energy and momentum equations we need pounds-mass.

I don't get this, I thought for example 1 litre of water weighed 1kg and also has 1kg of mass.

This came about because I challenged his use of the gravitational constant in an equation to calculate the energy and momentum of an object. When I asked why it was there I got the answer above.

I'm perfectly prepared to admit I'm wrong but can some body shed some light?

And as an aside - does the term weight really get used in physics? Or is it a term used to allow regular folk get a grip on something you guys would call potential energy or something LOL

Thanks in advance...

 

[Doc Al]

I don't get this, I thought for example 1 litre of water weighed 1kg and also has 1kg of mass.

In standard physics usage, the term "weight" refers to force, not mass. Mass is measured in kg; weight = mg, and is measured in Newtons (in SI units).

In everyday usage it doesn't matter.

 

[44HP]

Thank you for the quick reply.

So, if I wanted to calculate the energy of a potato traveling at a certain speed...

If I weighed the potato and found it to be 500 grammes and it was doing 10 m/s

Would the equation be e=(0.5*10^2)/2 or e=((0.5*9.81^2)*10^2)/2

I though it was the first, but I'm getting confused because I'm reading things like this in Wikipedia

To convert between weight (force) and mass we use Newton's second law, F = ma (force = mass × acceleration). Here, F is the force due to gravity, m is the mass of the object in question, and a is the acceleration due to gravity, on Earth approximately 9.8 m/s² or 32.2 ft/s². In this context the same equation is often written as W = mg, with W standing for weight, and g for the acceleration due to gravity.

Thanks

 

[Doc Al]

So, if I wanted to calculate the energy of a potato traveling at a certain speed...

If I weighed the potato and found it to be 500 grammes and it was doing 10 m/s

Even though you "weighed" the potato, 500 grams is its mass (not its weight).

Would the equation be e=(0.5*10^2)/2 or e=((0.5*9.81^2)*10^2)/2

I though it was the first, but I'm getting confused because I'm reading things like this in Wikipedia

It's the first. Kinetic energy is defined as:
$$(1/2)m v^2$$
where m is the mass, not the weight. (The weight of the potato would given by mg = (.5 kg)*(10 m/s^2) = 5 Newtons.)

 

[44HP]

Sir, I thank you..!

That one line solves the flame war - the act of weighing it gave me it's mass (not it's weight LOL)

So I was right in what I was saying, but I was using the wrong terms.

So on Earth, weighing scale are calibrated to give the mass of the object placed on them.

And when I become a Moon tourist (after that lottery win) I would need a Moon calibrated set of scales to find out the mass of my potato on the moon.

Cheers

 

[Doc Al]

And when I become a Moon tourist (after that lottery win) I would need a Moon calibrated set of scales to find out the mass of my potato on the moon.

If you use a http://en.wikipedia.org/wiki/Weighing_scale#Balance", it will work just the same on the moon. (Since you are just comparing the weights of two masses, and weight and mass are proportional.)

 

[44HP]

:rofl::rofl::rofl:

That's too funny..!

Goes to prove when phrasing a question, there is no such thing as too much clarification