# Problem with f=ma and E=0.5mv^2

1. Mar 24, 2012

### Nal

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! :)

Last edited: Mar 24, 2012
2. Mar 24, 2012

### technician

The distance travelled is not 55m.

3. Mar 24, 2012

### Nal

Distance travelled over time

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. Mar 24, 2012

### D H

Staff Emeritus
Your calculations are incorrect. Given a constant acceleration a, distance d as a function of time is given by $d=\frac 1 2 a t^2$.

5. Mar 24, 2012

### Nal

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

6. Mar 24, 2012

### azizlwl

7. Mar 24, 2012