You correctly talk about kinematic quantities most of the time (acceleration, velocity, time etc.) but then you suddenly invoke a unit of measurement, namely Joules (it should be capitalized). Do you mean the work done? The kinetic energy? The potential energy? Many different quantities are measured with the same unit.

Yes I'm looking for work done. Such as how many Joules does it require to get the object to move from one point to the other? When starting from a stopped position.

@ToJay By any chance, is it also given that the object is again at rest after covering a certain distance? Then you have another boundary condition, in addition to the ones you mentioned.

Use the work energy theorem. The work done is equal to the final kinetic energy.

@MariusLadegårdMeyer, in English the names of people, like James Prescott Joule, are capitalized; but the names of units named after people (like joules, watts, newtons, etc.) are not capitalized. source: [NIST: "The International System of Units (SI), p. 29]. And then again the symbols for the units (J, W, N, ...) are capitalized, even though the full names of the units are not

@ToJay, can you tell us more exactly what information you have. Do you know accelleration as a function of time ($a(t)$), for example? What do you mean by "NM"? New Mexico? Nanometers? Normal Mass? (If you mean newton-meters, the symbol is Nm, not NM, but then if you have the work done in newton-meters you are already done, because joules and newton-meters are the same thing)

@johnRennie Yes the object then comes to a complete stop. But the objective is to absorb that energy.

@ThePhoton Yes acceleration as a function of time I believe. I'm trying to see how much Force the object received. And I'm assuming it needs to be a newton meters or Newtons? I'm not too sure. But the goal is joules. I'm just trying to see how much this system is producing to be absorbed by a secondary system.

@ThePhoton you are right, I misremembered based on the capital J in the unit symbol. Too late to edit now...

@ToJay, please edit your question post to include the information needed to answer the question.

This is a modified repost of a deleted question. It’s still unclear since the acceleration as a function of time is left unspecified.

@ThePhoton sorry can you be more specific? Also the acceleration is unknown besides the fact that is variable in force to an unknown amount. All that I know is what I stated from the post.

First you said "yes" you know "acceleration as a function of time". Now you say you don't. Please update your question post with the actual and correct details of the question you want to ask.

@Ghoster yes I'm trying to pose the question better, and still need assistance. I can tell you the time period is .01875sec that the object traveled from point A to point B.

And what are you asked to calculate? "Joules" could measure work done on the object, or gravitational potential energy gained by the object, or various other things. Be specific.

@ToJay: Your comment "The object then comes to a complete stop" contradicts the "Assuming no losses" remark in your question.

@Amit, The applied force can first work "forwards" and then work "backwards" until the object stops.

@ThePhoton You're right of course. But then OP also writes "The acceleration is constantly being applied but at variable amounts", and that would suggest the direction is varying, too. I guess the upshot again is that there's a lot to be clarified here.

@Amit The post says “in a straight line”.

solve for joules The correct way to say this is “calculate the work done” or “calculate the final kinetic energy”. You are confusing units with physical quantities.