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

  • Thread starter Thread starter richard9678
  • Start date Start date
  • Tags Tags
    Energy Units
AI Thread Summary
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.
richard9678
Messages
93
Reaction score
7

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?
 
Last edited:
Physics news on Phys.org
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.
 
  • Like
Likes 1 person
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.
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
Thread 'Voltmeter readings for this circuit with switches'
TL;DR Summary: I would like to know the voltmeter readings on the two resistors separately in the picture in the following cases , When one of the keys is closed When both of them are opened (Knowing that the battery has negligible internal resistance) My thoughts for the first case , one of them must be 12 volt while the other is 0 The second case we'll I think both voltmeter readings should be 12 volt since they are both parallel to the battery and they involve the key within what the...
Thread 'Trying to understand the logic behind adding vectors with an angle between them'
My initial calculation was to subtract V1 from V2 to show that from the perspective of the second aircraft the first one is -300km/h. So i checked with ChatGPT and it said I cant just subtract them because I have an angle between them. So I dont understand the reasoning of it. Like why should a velocity be dependent on an angle? I was thinking about how it would look like if the planes where parallel to each other, and then how it look like if one is turning away and I dont see it. Since...
Back
Top