# Homework Help: Difference between newton's first and second law

1. Nov 19, 2008

### alpha372

1. The problem statement, all variables and given/known data
I'd just like some verification really: see step three

2. Relevant equations

net force = 0 --> equilibrium; net force = ma

3. The attempt at a solution
I've come to the conclusion that the difference between newton's first law and second law is acceleration:

Newton's first law:
absence of acceleration

Newton's second law:
presence of acceleration

I was wondering if it would be safe to say:

"A particle not accelerating in an inertial frame of reference implies that the net force acting on the particle is zero"

(after all, if it is not accelerating, wouldn't that automatically imply that the particle is a rest or moving at a constant velocity?)

Instead of what the book more or less says:
"A particle at rest or moving at a constant velocity in an inertial frame of reference implies that the sum of the forces acting on the particle is zero"

2. Nov 19, 2008

### PhanthomJay

That's right, both statements are correct. Newton's first law is just a special case of his 2nd, when a=0. A particle at rest or moving with constant velocity, will remain at rest or moving with constant velocity, unless acted on by a net unbalanced force (Newton 1). If a net unbalnced force acts on a particle,it will accelerate in the direction of the unbalanced force (Newton2: Net Force = rate of change of momentum, or f=ma for constant mass).

3. Nov 20, 2008

### alpha372

Thank you. I like how you pointed out that Net Force = rate of change of momentum, or f=ma for constant mass, I didn't know about the "Net Force = rate of change of momentum" equation.

4. Nov 20, 2008

### alpha372

oh, it has been awhile since I've been in a calc class.

Does, "rate of change of momentum" mean the derivative of momentum?

5. Nov 20, 2008

### PhanthomJay

It's been awhile for me, too! Yes, it's the first derivative of the momentum with respect to time. Newton 2 may be written as $$F_{net} = d(mv)/dt$$. When mass is constant, this boils down to $$F_{net} = m(dv/dt)$$, and since dv/dt =a, then $$F_{net} = ma$$. When mass is not constant (like in rocket propulsion problems where the rocket is burning off fuel), you've got to use the more general equation.