# Proof of F=ma from second law of newton

1. Feb 16, 2010

### hasanhabibul

from the second law of newton we know... applied force is proportional to the change of momentum...that means F= k( mv-mu) where k is constant .....from here now proof F=ma

2. Feb 16, 2010

### Stonebridge

The second law doesn't say that. Look at it again.
You need to start with the correct equation before you can prove F=ma

3. Feb 16, 2010

### saunderson

I think i can help you, but with my own labelling ;)

$$F \sim m \cdot \frac{v^\prime - v}{\delta t}$$​

$$v^\prime$$ is the speed after a short time interval $$\delta t$$ related to the instant of time $$t$$. So the equation above can be written in a more descriptive way

$$\delta t \, F \sim m \cdot \Bigl( v(t+\delta t) - v(t) \Bigr)$$​

if the time interval $$\delta t$$ is very small, we can approximate $$v(t+\delta t)$$ according to Taylor (keyword Taylor expansion).

$$v^\prime = v(t + \delta t) \approx v(t) + \dot v(t) \, \cdot \, \delta t + \mathcal{O}\bigl((\delta t)^2 \bigr)$$​

putting it in the initial equation that yields

$$\delta t \, F \sim m \cdot \Bigl( ~ \bigl( v(t) + \dot v(t) \, \cdot \, \delta t \bigr) ~ - v(t) \bigr) = m \, \cdot \, \dot v(t) \, \delta t$$​

Here you see, that the time interval $$\delta t$$ must be very small, otherwise the first term of the Taylor series isn't a good approximation any longer.