Solve for force, mass, or acceleration using Newton's second law, with every substitution and unit verified, or explain the law through a worked example.
You are a patient physics tutor who never trusts a calculated force, mass, or acceleration until its units check out and the number itself is physically reasonable. I want you to [MODE:select:solve for force,solve for mass,solve for acceleration,explain the law with a worked example] using Newton's second law, F = m x a. If I've described an actual situation in [WORD_PROBLEM?], read it first and pull the known values out of that instead of guessing at abstract numbers. Otherwise, work directly from [KNOWN_VALUES], the two quantities I already have. Before solving anything, sanity-check what you're given. Mass must be a positive number, since negative or zero mass isn't physical. If you're solving for mass or acceleration, the value you're dividing by can't be zero, so say so plainly if [KNOWN_VALUES] would require dividing by zero instead of forcing a calculation. If a word problem gives you a force in a unit other than newtons, such as pounds-force, or a mass in a unit other than kilograms, such as grams or pounds, convert it to standard SI units first and show that conversion as its own visible step before touching the main formula. If I chose solve for force, write F = m x a with the known mass and acceleration substituted in, then multiply them to get force in newtons, and state explicitly that one newton equals one kilogram-meter-per-second-squared so the unit itself is traceable back to the inputs. If I chose solve for mass, start from F = m x a, isolate mass algebraically as m = F / a before substituting any numbers, then substitute and divide to get mass in kilograms. If I chose solve for acceleration, isolate acceleration algebraically as a = F / m before substituting any numbers, then substitute and divide to get acceleration in meters per second squared. In every case, keep the algebraic isolation step and the numeric substitution step visibly separate instead of jumping straight from the formula to a final number. Once you have a value, verify it. Substitute all three quantities, the two you started with and the one you just solved for, back into F = m x a, recalculate both sides independently, and confirm they match. If they don't match, say so, trace back through the isolation and substitution steps to find where the error happened, and redo that step instead of adjusting the final number to make it fit. If I chose explain the law with a worked example, start with the law itself in one plain sentence: the net force on an object equals its mass times its acceleration, and for a fixed force, more mass means less acceleration. Then pick a concrete example, using [KNOWN_VALUES] if I gave you real numbers, or falling back to a simple scenario like a 2 kg cart pushed with 10 newtons of force if I left that generic, and tell me which one you picked. Walk through that example with the same discipline described above, algebraic isolation on its own line, substitution on its own line, and a final verification check, so the explanation and the worked proof of it reinforce each other. If the original input was a word problem, translate the final number back into that problem's own language, such as "the shopping cart accelerates at 1.5 meters per second squared," instead of leaving it as a bare value with no connection to what was actually being asked.
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Get Early AccessPush an empty shopping cart and a full one with the same amount of effort and the empty cart shoots forward while the loaded one barely budges. That's the whole law in one sentence: for a fixed force, more mass means less acceleration, and the two trade off in exact proportion, not roughly.
Give it your own [WORD_PROBLEM] or a set of [KNOWN_VALUES] and it solves for force, mass, or acceleration, whichever one you're missing. A force in pounds-force or a mass in grams gets converted to newtons and kilograms before the formula runs, and if a calculation would mean dividing by zero mass or zero acceleration, it says so instead of forcing a number out. The algebraic rearrangement stays visibly separate from the number-crunching, and the final answer gets checked by substituting all three values back into F = m x a.
Ask for the explain mode instead and it builds a worked example from scratch, a cart pushed across a floor, showing the isolation step and the verification step in full. Keep a running record in the Dock Editor, then combine multiple forces first with the net force calculation practice generator, or pull an acceleration out of a motion scenario with the kinematics equations solver before this formula ever sees it.
This runs fine in the Dock Editor, ChatGPT, Claude, or Gemini. Set [MODE] to solve for force, mass, or acceleration depending on which one is missing, or pick explain the law with a worked example if you want to see it demonstrated first.
Paste a real scenario into [WORD_PROBLEM] and the known values get pulled from it automatically, or drop your two known numbers directly into [KNOWN_VALUES] if you're working from an abstract problem.
If your problem uses pounds-force, grams, or another non-SI unit, the output converts everything to newtons, kilograms, and meters per second squared before solving, and shows that conversion as its own step.
The algebra that rearranges F equals m times a for your unknown happens on its own line, separate from plugging in the actual numbers, so you can see exactly how the formula was rearranged.
The output plugs all three values back into F equals m times a and recalculates both sides independently, so a wrong answer surfaces immediately instead of slipping through.
Paste your homework word problem and pick whichever variable is missing to get a fully worked solution with units tracked at every step, not just a final number.
Give a problem with pounds-force or grams and watch the conversion to SI units happen as its own visible step before the main F equals m times a calculation runs.
Run practice problems from an SAT Physics, AP Physics, or MCAT review packet through solve mode to build speed at rearranging the formula for whichever variable the question actually asks for.
Generate a model solution for any F equals m times a problem before class, with the algebra and the verification check both visible for students to follow.
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