Gravitation force downward, lifting force upward, gravitation acceleration is around 9.8 m/s^2, and the acceleration might be calculated via: (Vf-Vi)/t = (5m/s-0m/s)/1s = 5m/s^2. This acceleration of 5 m/s^2 may be plugged into the force equation of F = ma. Newton’s 2nd law tells us a net force...
It would be helpful to carry units throughout the math/arithmetic. Also, the parantheses in the second-to-last step should encompass the whole numerator term, (840-705.6)/72 which will get you to around just below 2 or 1.9 m/s^2.
The double negative which arises where you plug-in values for Fg (-9.8) for the downward gravitational acceleration is incorrect because you already account for directionality by minusing Fg from Fn (Fn - Fg).
I realize that there is a downward force of gravity weighing the object toward earth’s surface, equaling F = mg (downward). The upward force would have to be something at least as much as the downward force in order to lift the object up ”such that it is accelerated from rest to a velocity of 5...
This is a good method of solving for the spring constant, k. It is more direct than creating multiple equations and solving them together. By plugging in the different of masses between the two boxes and the difference in length for m and x, respectively, you effectively reduce number of steps...