Function Posted October 28, 2013 Share Posted October 28, 2013 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 Link to comment Share on other sites More sharing options...
Strange Posted October 28, 2013 Share Posted October 28, 2013 (edited) 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, 2013 by Strange Link to comment Share on other sites More sharing options...
Function Posted October 28, 2013 Author Share Posted October 28, 2013 Ah yes, well.. I forgot to put in the value of g.. My bad! Link to comment Share on other sites More sharing options...
swansont Posted October 28, 2013 Share Posted October 28, 2013 If you are doing unit analysis, you drop the 2, since it is unitless. Link to comment Share on other sites More sharing options...
Sensei Posted October 29, 2013 Share Posted October 29, 2013 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 Link to comment Share on other sites More sharing options...
Function Posted October 30, 2013 Author Share Posted October 30, 2013 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? Link to comment Share on other sites More sharing options...
imatfaal Posted October 30, 2013 Share Posted October 30, 2013 Elastic potential Energy is U=1/2kx^2 Link to comment Share on other sites More sharing options...
Function Posted October 30, 2013 Author Share Posted October 30, 2013 Now I'm embarrassed. Link to comment Share on other sites More sharing options...
swansont Posted October 30, 2013 Share Posted October 30, 2013 That's the lesson of unit analysis. If the units are wrong you know you made a mistake somewhere. Link to comment Share on other sites More sharing options...
imatfaal Posted October 31, 2013 Share Posted October 31, 2013 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. 1 Link to comment Share on other sites More sharing options...
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