# Mechanics question with variable force

1. Dec 22, 2006

### Sink41

Question copied word for word

1. A particle of mass m moves in a straight line under the action of force F where F = ms, s being the displacement of the particle from O, a fixed point on the line.
When s = -a the velocity of the particle is u. Find the velocity of the particle when s = 0.

2. F = ma or F= -ma (i think...)

a = v(dv/ds)

3. If ms = -ma, then s is always equal to -a, so the velocity would always be u, which doesnt look right... Or if I intergate:

s = -a

s = -v(dv/ds)

s^2 = -(v^2) + c

replace s with a and v with u.

a^2 + u^2 = c

s^2 + v^2 = a^2 + u^2

when s = 0, a = 0

v = u

Where have i gone wrong?

2. Dec 22, 2006

### neutrino

F = ms? Force = MassxLength?

I think something is amiss.

3. Dec 22, 2006

Doesn't have to be a miss, the given force is a function of the displacement.

4. Dec 22, 2006

### neutrino

Okay, fine. I suppose the constant in some set of units take the value of 1. Sorry about that.

5. Dec 22, 2006

### HallsofIvy

Staff Emeritus
F= ma. Don't guess- look it up.

Why would the velocity always be u? If s is not 0 then -a is not 0: with non-zero acceleration, the velocity changes.

You aren't really saying anything here. What is v if not u?

You are correct that $v^2+ s^2= C$, a constant.

Is "a" here some given constant? In that case, you have $C= u^2+ a^2$ and so the general formula is $v^2+ s^2= u^2+ a^2$ where u and a are given constants. When s= 0, $v^2= u^2+ a^2$ and so $v= \pm \sqrt{u^2+ a^2}$.

6. Dec 22, 2006

### Sink41

If F = ma then the particle would speed off in one direction, unless it ended up or started at O with a velocity of zero. If F = -ma then you would get harmonic motion... I thought it might be plus or minus because in the question it didnt say whether the force was towards or away from O.

I was thinking that a was acceleration, which was why i got confused... The velocity can't always be u. In the other questions the teacher used k for any generic constant and used the word acceleration instead of a symbol. I'll just guess that he meant to write k and made a mistake.

Last edited: Dec 22, 2006