We launched a mass into the air and documented the velocity and the time for it to start falling down.
I have a mass of 0.005 kg. I have a change in velocity of 4.14 m/s and my change in time is 42.3 deciseconds.
When I do the equation for force, F = ma, I get the answer F = 0.000489361702.
I have the same mass, but a different velocity and time, v = 4.5, and t = 46 deciseconds. I get
F = 0.000489130435.
Why is the second force weaker when it has a higher velocity, and it took the mass longer to fall down? Shouldn't the force be stronger?
I have a mass of 0.005 kg. I have a change in velocity of 4.14 m/s and my change in time is 42.3 deciseconds.
When I do the equation for force, F = ma, I get the answer F = 0.000489361702.
I have the same mass, but a different velocity and time, v = 4.5, and t = 46 deciseconds. I get
F = 0.000489130435.
Why is the second force weaker when it has a higher velocity, and it took the mass longer to fall down? Shouldn't the force be stronger?
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Your calculation is shown as F = ma obscuring the point at which all the variance occurs. In my experience with teachers and textbooks, this kind of thing is always intentional and meant to aggravate the student, errr... rather to explore more deeply the student's knowledge and understanding. In here, Yahoo, it's pretty much usually to twit the responders.
The force and mass are the same. The accelerations are not. "a" = change in velocity divided by time in your calculations. So drop the "a" and gloss over the "change in" part of that to get "v/t" and the reason will become apparent:
The change in velocities is slightly lower than the change in times. 4.5/4.14 = 1.086956522 while 46/42.3 = 1.087470449.
So your denominator, time, goes up more (very slightly more, but by more) than your numerator. That means the fraction "v/t" goes down in value. Therefore, finding the force is very slightly less in the second experiment is precisely to be expected.
Lesson to be learned? When you see all the variance in something is in a single variable which can be decomposed into parts that each change, decompose it! You don't have to leave it simplified just because it was presented that way! I didn't and I had the explanation in seconds...
The force and mass are the same. The accelerations are not. "a" = change in velocity divided by time in your calculations. So drop the "a" and gloss over the "change in" part of that to get "v/t" and the reason will become apparent:
The change in velocities is slightly lower than the change in times. 4.5/4.14 = 1.086956522 while 46/42.3 = 1.087470449.
So your denominator, time, goes up more (very slightly more, but by more) than your numerator. That means the fraction "v/t" goes down in value. Therefore, finding the force is very slightly less in the second experiment is precisely to be expected.
Lesson to be learned? When you see all the variance in something is in a single variable which can be decomposed into parts that each change, decompose it! You don't have to leave it simplified just because it was presented that way! I didn't and I had the explanation in seconds...
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Cool! Hopefully it will help with lots of others.
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