Hello everybody

 

I have a question concerning the formulas of energy and its unit

 

[math]E=\frac{m\cdot v^2}{2}[/math]

 

[math]1 J=\frac{1 kg\cdot 1 m^2}{2\cdot 1 s^2}[/math]

 

and

 

[math]E=m\cdot g\cdot h[/math]

 

[math]1 J=1 kg\cdot\frac{1 m}{1 s^2}\cdot 1 m=\frac{1 kg\cdot 1 m^2}{1 s^2}[/math]

 

And because [math]E_k=E_p[/math] (e.g. a mass falling from a height its potential energy will be converted to kinetic energy when on its lowest point):

 

[math]J=\frac{kg\cdot m^2}{s^2}=\frac{kg\cdot m^2}{2\cdot s^2}[/math]

 

[math]\Leftrightarrow \frac{1}{2}=1[/math]

 

What is wrong with my reasoning?

 

Thanks!

 

Function

What will the velocity of the mass be after falling from a height of 1m?

Clue: it won't be 1m/s. (In fact, it will be a vlaue that nicely cancels things out.)

 

(And what value of g are you using?)

Edited October 28, 201312 yr by Strange

Ah yes, well.. I forgot to put in the value of g..

 

My bad!

If you are doing unit analysis, you drop the 2, since it is unitless.

Ep = m*a*h

 

Distances traveled by object with acceleration a is:

 

h=1/2*a*t^2

 

so

 

m * a * 1/2*a*t^2 =

1/2*m*a^2*t^2

 

kg * m^2/s^4 * s^2 = kg * m^2/s^2 = kg * (m/s)^2

 

And we are receiving other equation back.

 

Ek= 1/2 *m * v^2

 

Thanks, Sensei. Your proof is very clear

Another problem: elastic potential energy:

 

[math]E_{pe}=\frac{k\cdot\Delta l}{2}[/math]

 

[math]J=\frac{\frac{N}{m}\cdot m}{2}=\frac{N}{2}[/math]

 

[math]E_{pe}=\frac{F}{2}[/math]

 

[math]E_{pe}=\frac{m\cdot a}{2}[/math]

 

[math]J=\frac{1}{2}\cdot kg\cdot \frac{m}{s^2}[/math]

 

[math]J=\frac{kg\cdot m}{2s^2}\neq \frac{kg\cdot m^2}{2s^2}[/math]

 

Where did the extra m go?

Elastic potential Energy is U=1/2kx^2

Now I'm embarrassed.

That's the lesson of unit analysis. If the units are wrong you know you made a mistake somewhere.

Now I'm embarrassed.

No need for that

 

Think - Work is force and displacement.

 

Simplistically - Mass * Accel due to Gravity is the force to move something higher - multiply that by the height you displace the object and you get the Work required; the Gravitational Potential Energy through lifting an object on earth.

 

For a spring we know from Hookes law that the Force is equal to the Spring Constant * displacement - integrate that over distance (cos in this case the force varies with distance) and you get U_e=1/2 k x^2

 

You cannot remember everything - but if you remember some basics and how to get from basics to more advanced then you are sorted.