quickquestion Posted April 24, 2017 Author Share Posted April 24, 2017 (edited) It is not clear where you get the numbers for the KE equation, but they are clearly wrong. Official KE equation is 1/2mv2 I wrote 0.5*m*sqr(v) imatfaal's equations seem to imply that both m and sqr(v) should be multiplied by 0.5. this is the descrepancy. 1/2m looks to me like .5m. But apparently it means that it applies to the whole thing, not just m. Edited April 24, 2017 by quickquestion Link to comment Share on other sites More sharing options...
imatfaal Posted April 24, 2017 Share Posted April 24, 2017 Official KE equation is 1/2mv2 I wrote 0.5*m*sqr(v) imatfaal's equations seem to imply that both m and sqr(v) should be multiplied by 0.5. this is the descrepancy. 1/2m looks to me like .5m. But apparently it means that it applies to the whole thing, not just m. OK you need to learn very basic maths. You also need to learn to use notation in the way the rest of the world does 1. multiplication - it doesn't matter if you multiply the m by 1/2 or multiply the m by v^2 and then by 1/2; they are both exactly the same. This is basic junior school maths 2. sqr(v) does not normally mean v2 and it is very close to be confused with sqrt(v) which stands for the square-root When you are getting confused DO THE ACTUAL MATHS - don't guess! Link to comment Share on other sites More sharing options...
quickquestion Posted April 24, 2017 Author Share Posted April 24, 2017 OK you need to learn very basic maths. You also need to learn to use notation in the way the rest of the world does 1. multiplication - it doesn't matter if you multiply the m by 1/2 or multiply the m by v^2 and then by 1/2; they are both exactly the same. This is basic junior school maths 2. sqr(v) does not normally mean v2 and it is very close to be confused with sqrt(v) which stands for the square-root When you are getting confused DO THE ACTUAL MATHS - don't guess! 2. I thought sqr(v) is the same as v2. What else could it mean? one. So if 1 is true, then the PE is false because of what I said earlier. 4*9.8*9.8!=2*9.8*9.8 And yeah I am sleepy. So I made a mistake in #26. So we can agree on point 1. But i still think PE is wrong because. 4*9.8*9.8!=2*9.8*9.8 Link to comment Share on other sites More sharing options...
Bender Posted April 24, 2017 Share Posted April 24, 2017 Why do you assume that v=9.8 m/s? It isn't. v=13.9 m/s (sqrt(192) from Imatfaals post) Link to comment Share on other sites More sharing options...
Strange Posted April 24, 2017 Share Posted April 24, 2017 Official KE equation is 1/2mv2 I wrote 0.5*m*sqr(v) But you have the wrong value for v. How did you calculate it? Link to comment Share on other sites More sharing options...
quickquestion Posted April 24, 2017 Author Share Posted April 24, 2017 But you have the wrong value for v. How did you calculate it? Why do you assume that v=9.8 m/s? It isn't. v=13.9 m/s (sqrt(192) from Imatfaals post) Ok that is the thing i'm not getting. if acceleration is 9.8 m/s, then shouldn't v be 9.8 m/s? it takes 1 second to move 9.8 m/s. therefore velocity is 9.8 m/s. Link to comment Share on other sites More sharing options...
Mordred Posted April 24, 2017 Share Posted April 24, 2017 (edited) Start with your height 9.8 metres calculate the time for an object to fall 9.8 metres. It will not be 1 second. [latex] t=\sqrt{\frac{2h}{g}}[/latex] [latex]v=\sqrt{2gh}[/latex] Use the first formula to calculate time of fall the second formula the final velocity. There is a difference between acceleration and velocity that you are overlooking. Ok that is the thing i'm not getting. if acceleration is 9.8 m/s, then shouldn't v be 9.8 m/s? it takes 1 second to move 9.8 m/s. therefore velocity is 9.8 m/s. acceleration is 9.8 m/s^2 not 9.8 metres per second. The mistake was assuming the object will drop 9.8 metres in 1 second. If you use the formulas above you will find it takes longer than 1 second. Think about the meaning of acceleration. Then consider your objects starts at zero velocity. If it helps recall that just like your car you don't instantly go from 0 to 9.8 metres per second. It takes a falling object 1 second to accelerate to a velocity of 9.8 metres/sec. it does not mean the falling object travelled 9.8 metres in the first second. Edited April 25, 2017 by Mordred Link to comment Share on other sites More sharing options...
swansont Posted April 25, 2017 Share Posted April 25, 2017 Ok, so what about this though. Let's make it even simpler and give a situation with zero air resistance. On earth an object will take 1 second to fall 9.8 meters. PE equation is mgh. Lets say mass is 4 kg, gravity is 9.8 m/s and height is 9.8 m. So it will tell me PE: 4*9.8*.9.8 KE=.5*m*sqr(v) KE=2*9.8*9.8 These do not match. So what am I doing wrong? Or is the equation itself, wrong? Since bullet experiments in water proved KE is correct equation, then I say PE is the wrong equation. It's already been pointed out to you that your equation for KE is wrong. Also, an object dropping from rest for 1 second will fall 4.9 meters, not 9.8. (s = 1/2 gt^2) At that time it will have a speed of 9.8 m/s KE = 0.5(4kg) (9.8m/s)^2 = 192 J PE = (4 kg)(9.8 m/s^2)(4.9 m) = 192 J Link to comment Share on other sites More sharing options...
quickquestion Posted April 25, 2017 Author Share Posted April 25, 2017 It's already been pointed out to you that your equation for KE is wrong. Also, an object dropping from rest for 1 second will fall 4.9 meters, not 9.8. (s = 1/2 gt^2) At that time it will have a speed of 9.8 m/s KE = 0.5(4kg) (9.8m/s)^2 = 192 J PE = (4 kg)(9.8 m/s^2)(4.9 m) = 192 J my equation for KE is not wrong, it is the same as yours. But I fear you are right about the acceleration. I must ponder these equations and how a linear equation can return the same result as an exponential equation. These are truly equations of magical proportions. As physicists say..."the equation is beautiful." Link to comment Share on other sites More sharing options...
imatfaal Posted April 25, 2017 Share Posted April 25, 2017 my equation for KE is not wrong, it is the same as yours. Yes it is - you are using the wrong value for v. But I fear you are right about the acceleration. As he has a further degree in physics and this is first year school physics I reckon he is right too. I must ponder these equations and how a linear equation can return the same result as an exponential equation. There is no exponential equation there. x^2 is not exponential. 5^x is exponential Learn some basic maths and basic physics - it is well worth it. Many people here will help - but fewer will help if you pretend great knowledge and then demonstrate great ignorance 1 Link to comment Share on other sites More sharing options...
quickquestion Posted April 25, 2017 Author Share Posted April 25, 2017 (edited) Yes it is - you are using the wrong value for v. As he has a further degree in physics and this is first year school physics I reckon he is right too. There is no exponential equation there. x^2 is not exponential. 5^x is exponential Learn some basic maths and basic physics - it is well worth it. Many people here will help - but fewer will help if you pretend great knowledge and then demonstrate great ignorance ^2 is an exponent. Therefore it is exponential. If I draw a graph of x^2 it will be non-linear. Also, my equation was not wrong, my equation input was wrong. An equation does not care what input you put into it, it is a nonliving entity. Edited April 25, 2017 by quickquestion Link to comment Share on other sites More sharing options...
Strange Posted April 25, 2017 Share Posted April 25, 2017 ^2 is an exponent. Therefore it is exponential. Why do you always insist you are right, even when you are talking about things you don't understand. It is very frustrating. https://en.wiktionary.org/wiki/exponential_equation Link to comment Share on other sites More sharing options...
quickquestion Posted April 25, 2017 Author Share Posted April 25, 2017 Why do you always insist you are right, even when you are talking about things you don't understand. It is very frustrating. https://en.wiktionary.org/wiki/exponential_equation My apologies. I shall reformulate my question as: How does an equation with a variable with exponent, return the same results as a linear equation. I accept it does this, but I do not understand why. Link to comment Share on other sites More sharing options...
Strange Posted April 25, 2017 Share Posted April 25, 2017 My apologies. I shall reformulate my question as: How does an equation with a variable with exponent, return the same results as a linear equation. I accept it does this, but I do not understand why. I don't know why you think there should be a problem with this. If you take the equation for velocity that Mordred provided: [latex]v=\sqrt{ 2 g h}[/latex] And substitute for v in the equation for kinetic energy: [latex]e = \frac 1 2 m v^2[/latex] [latex]e = \frac 1 2 m {(\sqrt{2 g h})}^2[/latex] [latex]e = \frac 1 2 m 2 g h[/latex] [latex]e = m g h[/latex] In other words, the equation for potential energy. 1 Link to comment Share on other sites More sharing options...
quickquestion Posted April 25, 2017 Author Share Posted April 25, 2017 I don't know why you think there should be a problem with this. If you take the equation for velocity that Mordred provided: [latex]v=\sqrt{ 2 g h}[/latex] And substitute for v in the equation for kinetic energy: [latex]e = \frac 1 2 m v^2[/latex] [latex]e = \frac 1 2 m {(\sqrt{2 g h})}^2[/latex] [latex]e = \frac 1 2 m 2 g h[/latex] [latex]e = m g h[/latex] In other words, the equation for potential energy. Ah ok, so that's what I was forgetting. That inside the equation, was bundled another equation which had a sqrt function inside of it. Link to comment Share on other sites More sharing options...
Sriman Dutta Posted April 26, 2017 Share Posted April 26, 2017 I must point out here that you may get confused between 'sqrt' and '^2' . 1 Link to comment Share on other sites More sharing options...
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