Physics 101 refresher question -- Confused about the definition of a Joule

Click For Summary

Discussion Overview

The discussion centers around the definition and understanding of a Joule, particularly in relation to kinetic energy (KE) and its calculation. Participants explore the relationship between mass, velocity, and energy, as well as the implications of different physical equations in the context of physics concepts.

Discussion Character

  • Conceptual clarification
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • Some participants express confusion about the relationship between kinetic energy and the definition of a Joule, particularly regarding the equation for KE (1/2mv^2) and its units.
  • One participant clarifies that a Joule is defined as the work done by a force of 1N over a distance of 1m, not directly related to the kinetic energy of a mass moving at a certain speed.
  • Another participant points out that the calculation of KE involves squaring the velocity, which is essential for deriving the units of Joules (kg*(m^2/s^2)).
  • There are repeated assertions that some participants may be conflating momentum with energy, particularly in their interpretations of mass and velocity.
  • Several participants suggest that working through practical examples could help clarify the relationship between force, distance, and kinetic energy.
  • One participant mentions the concept of power and its relation to changes in kinetic energy, referencing the work-energy theorem.
  • Discussions also touch on the practical applications of these concepts in racing and performance optimization, including the use of HP-seconds and kilowatt-hours.

Areas of Agreement / Disagreement

Participants express varying levels of understanding and confusion regarding the definitions and relationships between kinetic energy, Joules, and momentum. There is no consensus on the best way to articulate these concepts, and multiple viewpoints are presented without resolution.

Contextual Notes

Some participants indicate a need for further clarification on the definitions and relationships involved, suggesting that there may be missing assumptions or misunderstandings about the units and equations used in physics.

zanick
Messages
383
Reaction score
25
we all know the equation for KE
1/2mv^2

you do this simple math and you get an number that is in KE units , Joules.
so, why is the KE unit equation, 1 Joule = Kg x (m^2/s^2) ??
1Kg traveling at 1m/s would have a KE of 1Joule, but using the KE equation, it would be 1/2 a Joule.

why is that. I can't explain it to someone asking me. (forgot the basis! ;)
 
Physics news on Phys.org
The definition of Joule is not as the energy of 1 kg moving with a speed of 1m/s.
The Joule is defined as the work done by a force of 1N over a distance of 1 m.
 
  • Like
Likes   Reactions: CWatters and Redbelly98
so, the KE of 2Kg moving with a speed of 1m/s is 1 Joule. Maybe you can clarify some more of the KE units which are Joules and the KE unit equation which also are in Joules so I can articulated it best.. so , a watt-second is a joule too, its one watt applied for one second.
 
I think you might have momentum mixed up with energy.

>1Kg traveling at 1m/s would have a KE of 1Joule

It looks like you're taking mass*velocity and concluding this is the energy. Actually, this is the momentum (mass x velocity), which has units of kg*m/s. (not 1 kg*m^2/s^2= 1 J energy)

To calculate the KE of the object, you have to use KE=1/2*m*v^2. It is by squaring the velocity that you get Joules = kg*(m^2/s^2).

Edit: typo
 
Last edited:
It all starts with F = ma. $$F=m\frac{dv}{dt}$$ If we multiply both sides of this equation by v = dx/dt, we get $$F\frac{dx}{dt}=mv\frac{dv}{dt}=\frac{d(m\frac{v^2}{2})}{dt}$$If we integrate this equation with respect to time, we get:$$\int{Fdx}=m\frac{v^2}{2}$$This is where the 2 comes from.
 
  • Like
Likes   Reactions: CJ Howard
CJ Howard said:
I think you might have momentum mixed up with energy.

>1Kg traveling at 1m/s would have a KE of 1Joule

It looks like you're taking mass*velocity and concluding this is the energy. Actually, this is the momentum (mass x velocity), which has units of kg*m/s. (not 1 kg*m^2/s^2= 1 J energy)

To calculate the KE of the object, you have to use KE=1/2*m*v^2. It is by squaring the velocity that you get Joules = kg*(m^2/s^2).

Edit: typo
I understand the KE of an object, that's why i referenced 1 joule being 2Kg traveling at 1m/s. but the KE Unit has no "1/2" , (its like momentum with a squared velocity factor) and then i was reminded of the definition of the Joule, being the energy required for applying force of 1N over 1m. so the KE unit is the part I am having a hard time explaining (or remembering ;) )
 
zanick said:
I understand the KE of an object, that's why i referenced 1 joule being 2Kg traveling at 1m/s. but the KE Unit has no "1/2" , (its like momentum with a squared velocity factor) and then i was reminded of the definition of the Joule, being the energy required for applying force of 1N over 1m. so the KE unit is the part I am having a hard time explaining (or remembering ;) )
You are aware that there is a difference between a quantity and its units, right? Is the speed of light 1 m/s, or is it 3 x 108 m/s?
 
zanick said:
...we all know the equation for KE = 1/2mv^2

...you do this simple math and you get a number that is in KE units, joules.

The formula only expresses a relationship between physical quantities.
It doesn't say anything about units of measure.
 
zanick said:
I understand the KE of an object, that's why i referenced 1 joule being 2Kg traveling at 1m/s. but the KE Unit has no "1/2" , (its like momentum with a squared velocity factor) and then i was reminded of the definition of the Joule, being the energy required for applying force of 1N over 1m. so the KE unit is the part I am having a hard time explaining (or remembering ;) )
Yes, I understand... I guess I just need to work through an example... like applying the 1N to a 1Kg object for a few meters and then see If the KE (in J) agrees with the unit measure vs the KE equation.
 
  • #10
zanick said:
Yes, I understand... I guess I just need to work through an example... like applying the 1N to a 1Kg object for a few meters and then see If the KE (in J) agrees with the unit measure vs the KE equation.
It should agree with both. The KE should equal to force times distance and the KE should also be equal to 1/2 mv2. See Chet's post #5 above for the demonstration of that.
 
  • #11
zanick said:
Yes, I understand... I guess I just need to work through an example... like applying the 1N to a 1Kg object for a few meters and then see If the KE (in J) agrees with the unit measure vs the KE equation.

Check out khan's video on the work-energy theorem. Here
 
  • #12
jbriggs444 said:
It should agree with both. The KE should equal to force times distance and the KE should also be equal to 1/2 mv2. See Chet's post #5 above for the demonstration of that.
that's a good point. thanks (chets post makes sense)
But, I always like to see how and why things work out. It gives me a better visual of why things are happening.
Kind of like thinking about equations like with Inertial of rolling mass.. MR^2 vs 1/2MR^2... there is a lot that shows the derivation of a ring vs a disc.. in the end, thinking about it... 1/2MR^2 is just like a ring, or point mass, but totally thick ring, so the center of mass for R would be 1/2 that of the center of mass of the ring.
 
Last edited:
  • #13
EddiePhys said:
Check out khan's video on the work-energy theorem. Here
[]
good video...Net work! (change in KE)... the rate of change of KE would be power...HP-seconds... unit measure of work, correct ? :)
 
  • #14
zanick said:
HP-seconds... unit measure of work, correct ?
Correct, although I don't think I've ever seen seen anyone actually use hp-seconds.

Kilowatt-hours (kWh), on the other hand... I bet you've seen those on your electric bill.
 
  • #15
jtbell said:
Correct, although I don't think I've ever seen seen anyone actually use hp-seconds.

Kilowatt-hours (kWh), on the other hand... I bet you've seen those on your electric bill.
As a racer, its all about power. Finding the desired rate of change of KE over a speed range, can point to how many HP seconds we need to make that change. divide by seconds, and we get the HP we need. :) Then, its back to the shop to try and find it. other uses is finding reasons that brakes might be overheating. again, KEstart vs KEend (at braking point speed and release speed point) and we get total KE dissipated in a time then dividing by the time to slow in seconds to get average Power required for the slow down. this is most useful when comparing to others that might not have an issue based on their slow down rate and mass.
By using HP-seconds, it allows us to find the right RPM based on the HP curves to optimize acceleration over a lap. (vs using engine torque and having to produce thrust graphs for each gear)
 

Similar threads

High School Understanding Newtons vs Joules - Beginner
  • · Replies 14 ·
Replies
14
Views
4K
Undergrad Same old question, I still don't get it: E=1/2mv^2 - more E with V?
  • · Replies 41 ·
2
Replies
41
Views
4K
Undergrad Confusion about conservation of energy vs. momentum
  • · Replies 3 ·
Replies
3
Views
2K
Undergrad Is it more efficient to eject burnt fuel as reaction mass in rockets?
  • · Replies 2 ·
Replies
2
Views
1K
High School I have a few questions about the joule/Newton
  • · Replies 2 ·
Replies
2
Views
2K
Undergrad Confused about unit conversion involving natural units
  • · Replies 2 ·
Replies
2
Views
2K
High School Understanding Watt & Newton: Questions for Physics Forum
  • · Replies 9 ·
Replies
9
Views
2K
High School Questions about Feynman's contrasting Definitions and Laws in physics
  • · Replies 15 ·
Replies
15
Views
2K
Undergrad Confusion about work performed by a horizontal push or pull
  • · Replies 5 ·
Replies
5
Views
3K
High School What are the differences between relict and relic?
  • · Replies 28 ·
Replies
28
Views
4K