**Question:** For the ball with acceleration up and mg down we have ma=-mg, so you now say acceleration is negative, and so it acts down?

**Answer:** This is not really correct language. When you look at Newton's Second Law, F = ma, what that is really saying is that the sum of the forces on an object *produces* or *implies* the acceleration. Forces act, and are causes. Accelerations react to forces, and are effects.

**Question:** How can this be if i give the ball acceleration in positive (up) direction?

**Answer:** All of this depends on when you "start the clock", so to speak. If you were to give the ball an initial known upward velocity and position, and then tell me to predict what happens next, I would start the clock at the instant the ball left your hand. That's t = 0, and at that point exactly, your hand is no longer a force on the ball. The only force on the ball is gravity, with the usual -mg formula. So you've accelerated the ball initially with the force of throwing, but once the ball leaves your hand, gravity takes over and is the only force on the ball (if you neglect air resistance). Does that make sense?

**Question:** I can see deccelartion will occur but how do i see from a=-mg there is an upward acceleration?

**Answer:** From the initial velocity.

**Question:** Or is your remark acceleration is negative wrong and i use the first paragraph to say as mg acts down it is negative to -(mg) is positive?

**Answer:** No, m and g are always positive, by convention. You assign a sign to the quantity based on your choice of coordinate system. Gravity always acts down (to a good approximation) while on Earth, so if y is positive upward, the gravitational force is approximated by -mg. If, on the other hand, you define y positive downward, which is a good choice in some problems, then the gravitational force is +mg. Make sense?

**Question:** If I wanted to accelerate the spring in the positive direction (to the right) how would I write F+ma for this case?

**Answer:** I'm assuming you meant to write F = ma in that question. In this situation, if I wanted to accelerate the spring in the positive direction, to the right, and if the spring was already to the right of its equilibrium position, then you would have to exert an outside force on the system to make that happen. If you were then to let the system go (in other words, remove your outside force), and you were interested in the motion, then I would start the clock at the instant you removed the outside force, use the new initial position and velocity, and go from there. It would be simpler, I think.

**Question:** As -kx is spring force would i not end up with ma=-kx >?

**Answer:** Could you please re-phrase this question? If you mean "will the spring force not end up being in the positive x-direction at some point", I would say, "Yes, any time the masses on the spring go farther left than the equilibrium position of the spring, the spring force is going to change sign (essentially, x will then be negative, thus canceling out the negative sign in front of -kx)." Is that what you meant? The spring force is always opposed, in direction, to the position of the spring relative to its equilibrium point, at least if you're in the region governed by Hooke's Law. You can exceed that if you wish, probably damaging the spring in the process.