Why is the Velocity equal to joules per kilogram?

[Chestermiller]

except for the 1/2. yeah. f=ma m*a*d = mv^2

I see how kinetic energy appears to be half a joule. it throws me off that the only way to get that 1/2 is by the integration. it doesn't make sense if you try to derive it by manipulating formulas.

edit: so far it doesn't make sense

$$F = m\frac{dv}{dt}$$
Multiplying both sides by velocity:
$$Fv = mv\frac{dv}{dt}$$or equivalently:$$F\frac{ds}{dt} = mv\frac{dv}{dt}$$where s is the distance over which the force is applied. Multiplying both sides by dt yields:
$$Fds = mvdv$$Integrating yields:
$$Fs=\frac{1}{2}mv^2$$
Notice that the left hand side of this equation is the force times the displacement over which it is applied. This in expressed in Joules. Notice that the right hand side of the equation is the increase in kinetic energy, equal to the number of Joules of work done by the force. So kinetic energy Joules are the same as force times distance Joules.

 

[kuruman]

To add my two bits at this point, there is another dimension for confusion concerning Joules. If you break down the dimensions of energy units, you get$$[J]=\frac{[M][L]^2}{[T]^2} $$ but so do you when you break down dimensions of torque. Two completely different beasts, same dimensions.

 

[JR Jonsson]

The area of a rektangel is for ex.
2 m • 4 m = 8 m^2

The area of a triangel is for ex.
2 m • 4 m /2 = 4 m^2

Make a diagram for velosity and time in the case of free fall and you get a triangel.
The area of this triangel is for the first second
1 s • 10/2 m/s = 5 m
( gravity 10 m/s^2, or 10 m/s every second of free fall, friction neglected)

The kinetik energy for 10 m/s and potential energy for 5 m

For two seconds we get;
2 s • 20/2 m/s = 20 m
And speed 2 s • 10 m/s^2 = 20 m/s

If you can follow this you see that the speed has doubled while the fall is four times longer.

That is why the 2 is needed between kinetik and potential energy, for bodies as well as for fluids.

Abaut the torque;
It really helps if you think of torque as Nm/rad

If you turn a wheel then you get a work done;
Nm/rad • rad = Nm = J

For power;
Nm/rad • rad/s = W

PS. An angel is not considered to have a dimension by the expertise, but it helps me to think of radius as a special case of length; m/rad DS

 

[Ricardo Reyes]

I just want to add my logic for the units.

J= Kg*m2*s2, so J/Kg is the same as Kg*m2*s2/kg Got it? In this case, the kilogram unit cancels out, and we're left with m2/s2 and if we take the square root of that we have the velocity units (m/s).
Hope that this helps anyone

 

[kuruman]

I just want to add my logic for the units.

J= Kg*m2*s2, so J/Kg is the same as Kg*m2*s2/kg Got it? In this case, the kilogram unit cancels out, and we're left with m2/s2 and if we take the square root of that we have the velocity units (m/s).
Hope that this helps anyone

That's good, but this thread is more than 4 years old.