(adsbygoogle = window.adsbygoogle || []).push({}); 1. Mass is hung from a force probe, with negative force calibrated as down. The mass is 0.5 kg, the height of each floor is 5 m, and v0 = a0 = 0. Assume that I start on the 6th floor, what floors did I stop on, and how long was I on each floor?

2. I was wondering, how would one find the position of the elevator at any time as an equation?

[tex]a(t) &=& \frac{Fnet}{m}[/tex]

[tex]v(t) &=& {\int_{t_0}^{t_final} a(t) dt = {\int_{t_0}^{t_final} (\frac{Fnet}{m}) dt [/tex]

[tex]x(t) &=& {\int_{t_0}^{t_final}v(t)dt = {\int_{t_0}^{t_final}}{\int_{t_0}^{t_final} a(t) dtdt} = {\int_{0}^{t_final}}{\int_{0}^{t_final}(\frac{Fnet}{0.5}) dtdt[/tex]

[tex]Floor(t) &=& \frac{\int_{0}^{t_final}{\int_{0}^{t_final}(\frac{Fnet}{0.5}) dtdt + 30}{5}[/tex]

3. My idea was the the position is the double integral of acceleration, which is Fnet/m, with respect to time. Where Fnet is the instantaneous net force at time t. But, on excel I got weird numbers, when I did a trapezoidal approximation. Am I doing anything wrong?

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# Homework Help: Physics of an Elevator

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