Newton's Second Law Elevator Problem

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Ryuky
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A student stands on a bathroom scale in an elevator at rest on the 64th floor of a building. The scale reads 836 N.

(a) As the elevator moves up, the scale reading increases to 936 N, then decreases back to 836 N. Find the acceleration of the elevator (ms^-2).

(b) As the elevator approaches the 74th floor, the scale reading drops to 782 N. What is the acceleration of the elevator?

(c) Using your results from parts a and b, explain which change in velocity, starting or stopping, would take the longer time.(d) What changes would you expect in the scale readings on the ride back down?

2. Well, I do not understand the concept behind the problem. Is ΣF equal to Motive force - Weight? If so, what is the motive force in the problem? Moreover, I do not really get c and d
3. Well, I have some ideas about a and b. Considering that the reading when at equilibrium was 836, we know that m=83.6 kg and then we solve for a and plug in the numbers. Thus I found (a) a=11.1 ms^-2 and (b) a=9.3 ms^-2 . Is that correct?EDIT: I found out that ΣF=Tension - Weight. Therefore (a) will be: a=(T-W)/m => a=100/83.6=1.2 ms^-2 (b) a=(T-W)/m=-0.6 ms^-2 Τhus, (c) will be stopping would take longer time (since it decelerates at a rate of -0.6ms^-2 whereas it accelerates at a rate of 1.2ms^-2 ) and (d), will be the exact reversed readings. (Since when going downwards it will be ΣF=W-T instead of T-W.)

Am I correct?
 
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All correct, I think. It is difficult to compare your answers with mine because I used g = 9.81 and I don't know what value you used. We agree to 1 or 2 digit accuracy.