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

1. May 2, 2016

### zanick

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 cant explain it to someone asking me. (forgot the basis! ;)

2. May 2, 2016

### nasu

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.

3. May 2, 2016

### zanick

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.

4. May 2, 2016

### CJ Howard

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: May 2, 2016
5. May 3, 2016

### Staff: Mentor

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.

6. May 3, 2016

### zanick

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 im having a hard time explaining (or remembering ;) )

7. May 3, 2016

### Staff: Mentor

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?

8. May 3, 2016

### David Lewis

The formula only expresses a relationship between physical quantities.
It doesn't say anything about units of measure.

9. May 4, 2016

### zanick

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. May 4, 2016

### jbriggs444

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. May 4, 2016

### EddiePhys

Check out khan's video on the work-energy theorem. Here

12. May 4, 2016

### zanick

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: May 4, 2016
13. May 4, 2016

### zanick

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good video...Net work! (change in KE)...... the rate of change of KE would be power.......HP-seconds.... unit measure of work, correct ? :)

14. May 4, 2016

### Staff: Mentor

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. May 4, 2016

### zanick

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)