Question

Please help:

Answer 1

In this picture, we are asked to use the delta x equation that's to solve for the different accelerations and to choose how acceleration scales best with mass.

For some reason when I solve for acceleration in the 2nd example \[\large 10 = \frac{1}{2}a * 6.32^2\]
\[\large a = 2\]

It is not the right answer, seems I'm still solving for force (which scales directly to acceleration). What do I need to do to solve acceleration in regard to mass.
I hope that made sense. Thanks a lot!

Answer 2

ughh I figured it out.. I'm sorry !!

Answer 3

Just my stupid algebra

Answer 4

Hi Josh,

I wasn't sure I understood your question clearly, but here's my thoughts on what you are being asked to do. I hope it helps!

Here is what we know about your scenario.
We know we will eventually conclude that F = ma (mass x acceleration) since this is Newton's 2nd law. We also know that there is a constant force being applied in all cases (F = 10 N) but that mass is changing so we can make guesses about the relationship between m and a since they have some mathematically fixed relationship to each other that will be constant with a constant force. So to demonstrate that a = 1/m is the appropriate relationship (and not the others) we want to show that using this relationship to generate new accelerations in our table will yield balanced kinematic equation for delta x with the other given information, but any other mathematical relationship will yield inequalities.

What the problem doesn't state is our initial velocity Vo however, using the first example experiment (where a = 1m/s and m = 10kg) we can solve for Vo is approximately 0. Then we can use our kinematic equation for delta x to "test out" the different possibilities.

For example, if we test out the relationship that a = 1/m (and we remember to multiply by a factor 10 since our force is 10N) then using the kinematic equation for delta x that given delta t, we can calculate the given distance of delta x (10m) whereas for any other relationship between m and a, the calculated acceleration will not provide the correct distance of delta x = 10 when plugged into the kinematic equation for delta x with the given t and Vo as zero, or put another way, the equation will not balance.

For the 2nd example, the first place I would start is by calculating a new acceleration based ONLY on the given mass (20kg) and if a = 1/m then the new acceleration is 1/20 (but multiply by a factor of 10 since force is 10N) so this will yield an acceleration of 1/2. Then try solving the kinematic equation using this value (.5) for your acceleration and your known time (6.32) and your Vo = 0 and delta x = 10 and everything should be balanced, since 6.32^2 is approximately 40 and 1/2*1/2*40 is 10.

Of course if you were solving for any other relationship between a and m you would get a different acceleration that when plugged into the kinematic equation, you can't balance it.
I hope this makes sense!

Answer 5

Oh well nevermind - it was fun to figure out anyways!

Answer 6

hey @lizcody1, Thank you for your reply! Actually that's a huge help. I'm taking intro to physics online and need all the help I can get, so I will use your answer for reference. Thanks again :)