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I'm brushing up on my physics, and I've been struggling with this problem in my head. I hope you can help.

Let's assume that we have some object/particle that only moves along a single axis ($x$). We observe that this object moves along the $x$-axis, and we note its position at different times.

We also note that a certain amount of force or work is done in the opposite direction for the object to move in the opposite direction.

For example:

At $t_0$, the object is at position $2$ along the $x$-axis. At $t_1$ it's at $1$ and $t_2$ it's at $12$ and $t_3$ it's at $10$ etc.

I know we can calculate the speed and velocity of this object. Is it possible to figure out what the object's mass is?

Is there anything else that could be understood or derived from this motion, perhaps the force needed or applied at $t_1$ and $t_2$ for this object to change its direction? Anything about energy or momentum or inertia that could be interesting that could be understood from this?

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  • $\begingroup$ Suppose the position was known explicitly as a function of time $x(t)$, we could find both $v(t)$ and $a(t)$ by differentiation. But we cannot get any information about the forces or masses. Applying Newton's law $F=ma$ you see there are two unknowns: force and mass. If you specify one then the other may be calculated $\endgroup$
    – Sal
    Commented Nov 27, 2021 at 17:15
  • $\begingroup$ Thanks Sal for replying. I was thinking, if we assign a masks to the object, then we could at least figure out the force relative to that mass, correct? Since we know that the object of some mass m is moving through time and space at some velocity and then it experiences a directional change, I assume then it's possible to calculate what force that must have been applied to change it's direction. How would something like that be calculated if it's possible. $\endgroup$
    – Blackbird
    Commented Nov 28, 2021 at 2:30
  • $\begingroup$ Yes, if $m$ is known then you can find $F$, but note that the 'force relative to the mass' is the acceleration! $F/m=a$ $\endgroup$
    – Sal
    Commented Nov 28, 2021 at 21:26

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