Solve Energy Homework: Understand (2) & (3) of 1 J = 1 N m

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Expressions (2) and (3) illustrate how the Joule (J) can be expressed in terms of the base SI units of mass, length, and time. Specifically, expression (2) shows that 1 Joule equals 1 Newton meter, where 1 Newton is defined as 1 kg·m/s². Expression (3) further simplifies this to kg·m²/s², emphasizing the relationship between force, mass, and acceleration. The discussion clarifies that the work done on a mass can be calculated using these units, regardless of the gravitational context. Ultimately, the key takeaway is that all expressions are consistent representations of energy in the SI unit system.
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Homework Statement



I cannot understand what is written in my textbook.

Homework Equations



i.e:

1 J = 1 N m (I get that) ........(1)

= 1 Kg m s-2 x m ......(2)

= 1 Kg m2 s-2...(3)

The Attempt at a Solution



If I push a 1 Kg mass (in space) at an acceleration of 9.8 m s-2) then the Force will be 1 N. And if I push the mass 1 m, that is 1 N m of work done.

What are expressions (2) and (3) saying?
 
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richard9678 said:
1 J = 1 N m (I get that) ........(1)

= 1 Kg m s-2 x m ......(2)

= 1 Kg m2 s-2...(3)If I push a 1 Kg mass at an acceleration of 9.8 m s-2 then the Force will be 1 N.

A force of 1 N will accelerate a 1 kg mass at 1 m/s2.

And if I push the mass 1 m, that is 1 N m of work done.

What are expressions (2) and (3) saying?

They are just showing how the Joule can be expressed in terms of the three base SI units: kg, m, s.
 
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richard9678 said:

Homework Statement



I cannot understand what is written in my textbook.

Homework Equations



i.e:

1 J = 1 N m (I get that) ........(1)

= 1 Kg m s-2 x m ......(2)

= 1 Kg m2 s-2...(3)

The Attempt at a Solution



If I push a 1 Kg mass at an acceleration of 9.8 m s-2) then the Force will be 1 N.
it will be 9.8 kg.m/sec^2, which is called 9.8 N , since a Newton is defined as a kg.m/sec^2.
And if I push the mass 1 m, that is 1 N m of work done.
9.8 N.m of work done, or 9.8 kg.m^2/sec^2 in longhand form.
What are expressions (2) and (3) saying?
Same thing, in SI basic units of mass, length, and time. Force is a derived quantity.
 
Aah, I was making a mistake with the definition of the Newton. I was (for some reason) thinking that the definition of the Newton had something to do with Earth's gravity. Of course, gravity exerts a force of 9.8 Newtons on a 1 Kg mass, not 1 Newton.

(1) is talking about a unit called Newton metres.

(2) is talking about a unit called?

(3) is talking about a unit called?
 
(2) and (3) are combinations of base SI units. You can call (3) a "kilogram meter-squared per second-squared" :smile:
 
Having thought about this, these are my thoughts:

The definition of the Newton is the force it takes to accelerate a 1Kg mass (in space, outside of a gravitational field) at 1 m/s/s.

So, really, whatever work done that it takes to move that 1 Kg, can be expressed in terms of Kg per metre of movement - of a body in space. If the mass is 2 Kg, then work done doubles. Move a 1 Kg mass at an acceleration of 1 m/s/s, 1 metre and work done is 1 J.

You can also express the work done in terms of Kg per second of the movement of a body in space. Double the time, and work done doubles. Move a 1 Kg mass at an acceleration of 1 m/s/s for 1 second and work done is 1 joule.

Of course, here were talking about a mass hanging in space. And referring to mass, time, distance, acceleration.

If we get away from mass in space, and units associated with that situation, we may talk about Newton metres.

A force of 1 Newton moving anything, whatever it's mass, no matter how long it takes, 1 metre, equals 1 J of work done. There will be no acceleration if the force remains constant. And it is this Newton metre measure, that is really an equivalent to the other measures.

My 2 cents. I think I solved my problem.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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