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Problem with f=ma and E=0.5mv^2 
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#1
Mar2412, 04:13 PM

P: 4

EDIT: The question has been answered, and I don't need help anymore. Thanks a lot everybody!
Hi. This is something that has been puzzling me for a while, probably just due to my own stupidity. It seems to me that this should be really simple, but I don't know what I'm doing wrong though, so now I'm putting it out here. Imagine an object with a mass of 1 kg being accelerated at a rate of 1 m/s^2 over 10 seconds. We can then calculate the force being applied to the object F=1*1=1 N We can also calculate how far the object has moved over these 10 seconds According to Wikipedia, the work (W) is the product of a force times the distance through which is acts, measured in Joules, so W=F*d=m*a*d=55 J The velocity v at which the object moves is t*a=10 m/s Then we can calculate the kinetic energy E E=0.5*m*v^2=50 J But shouldn't E and W be the same? For example, if we take the object, now with the kinetic energy E and make it go up a hill on a planet where the gravitational acceleration is 1 1m/s^2 (which is =a from before, of course) we will convert the kinetic energy to potential energy, and we can calculate how far up the hill the object will go, as E(kin)+E(pot) is constant 0.5*m*v^2=m*a*h 0.5*1*10^2=1*1*h h=50 m But shouldn't h be 55 metres, as that was the distance the object moved when we accelerated it? I hope you understood what I meant, I'm absolutely horrible at explaining my thoughts. Thanks in advance! :) 


#2
Mar2412, 04:29 PM

P: 1,506

The distance travelled is not 55m.
Check your calculation 


#3
Mar2412, 04:48 PM

P: 4

after 1s: 1 m (velocity 1 m/s) 2s: 3 m (velocity 2 m/s) 3s: 6 m (velocity 3 m/s) 4s: 10 m (velocity 4 m/s) 5s: 15 m (velocity 5 m/s) 6s: 21 m (velocity 6 m/s) 7s: 28 m (velocity 7 m/s) 8s: 36 m (velocity 8 m/s) 9s: 45 m (velocity 9 m/s) 10s: 55 m (velocity 10 m/s) 


#4
Mar2412, 04:57 PM

Mentor
P: 15,070

Problem with f=ma and E=0.5mv^2
Your calculations are incorrect. Given a constant acceleration a, distance d as a function of time is given by [itex]d=\frac 1 2 a t^2[/itex].



#5
Mar2412, 05:13 PM

P: 4

Ah, well that obviously solves it. Thanks a lot.



#6
Mar2412, 06:10 PM

P: 963

[QUOTE=Nal;3832030]
We can also calculate how far the object has moved over these 10 seconds [/QUOTE Can you elaborate more on the method you use in finding the distance? 


#7
Mar2412, 06:41 PM

P: 4

[QUOTE=azizlwl;3832175]



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